NCERT Solutions for Class 12 Maths Chapter 4 Determinants

NCERT – Exercise 4.1

Evaluate the determinants in Exercises 1 and 2.

1. \( \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = 2(-1) – (-5)(4) = -2 + 20 = 18 \).

2. \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \)

Solution:

\( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = \cos \theta \cdot \cos \theta – (\sin \theta)(-\sin \theta) \)

= \( \cos^2 \theta + \sin^2 \theta = 1 \).

3. \( \begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix} = (x^2 – x + 1)(x + 1) – (x – 1)(x + 1) \)

= \( (x^3 + 1) – (x^2 – 1) = x^3 + 1 – x^2 + 1 = x^3 – x^2 + 2 \).

4. If \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), then show that \( |2A| = 4|A| \).

Solution:

Given \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), so \( 2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} \).

L.H.S. \( = |2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = 8 – 32 = -24 \).

R.H.S. \( = 4|A| = 4 \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = 4(2 – 8) = -24 \).

Hence, \( |2A| = 4|A| \).

5. If \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} \), then show that \( |3A| = 27|A| \).

Solution:

Given \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} \), so \( 3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix} \).

L.H.S. \( = |3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix} \)

= \( 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} – 0 \begin{vmatrix} 0 & 6 \\ 0 & 12 \end{vmatrix} + 3 \begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix} \)

= \( 3 \times 36 – 0 + 3 \times 0 = 108 \).

R.H.S. \( = 27|A| = 27 \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{vmatrix} \)

= \( 27 \left[ 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} – 0 \begin{vmatrix} 0 & 2 \\ 0 & 4 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} \right] \)

= \( 27[1(4) – 0 + 0] = 108 \).

Hence, \( |3A| = 27|A| \).

Evaluate the determinants:

6. (i) \( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} = 3(0 – 5) + 1(0 + 3) – 2 \times 0 \)

= \( -15 + 3 = -12 \).

6. (ii) \( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \)

= \( 3(1 + 6) + 4(1 + 4) + 5(3 – 2) = 21 + 20 + 5 = 46 \).

6. (iii) \( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} – 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} \)

= \( 0 – 1(0 – 6) + 2(-3) = 6 – 6 = 0 \).

6. (iv) \( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} = 2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} + 1 \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} – 2 \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix} \)

= \( 2(0 – 5) + 1(0 + 3) – 2(0 – 6) = -10 + 3 + 12 = 5 \).

7. If \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \), find \( |A| \).

Solution:

Given \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \),

\( |A| = \begin{vmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{vmatrix} \)

= \( 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} – 2 \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} \)

= \( 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5) = 3 + 3 – 6 = 0 \).

8. Find the values of \( x \), if:

(i) \( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)

\( \Rightarrow 2 – 20 = 2x^2 – 24 \)

\( \Rightarrow -18 = 2x^2 – 24 \)

\( \Rightarrow 2x^2 = 6 \)

\( \Rightarrow x^2 = 3 \)

\( \Rightarrow x = \pm \sqrt{3} \).

(ii) \( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)

Solution:

\( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)

\( \Rightarrow 2 \times 5 – 4 \times 3 = 5x – 6x \)

\( \Rightarrow 10 – 12 = -x \)

\( \Rightarrow x = 2 \).

9. If \( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \), then \( x \) is equal to:

1. A. 6
2. B. \( \pm 6 \)
3. C. -6
4. D. 0

Solution:

\( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \)

\( \Rightarrow x^2 – 36 = 36 – 36 \)

\( \Rightarrow x^2 = 36 \)

\( \Rightarrow x = \pm 6 \).

Hence, the correct answer is B.

Exercise – 4.2

Using the property of determinants and without expanding in Exercises 1 to 5, prove that:

1. \( \begin{vmatrix} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{vmatrix} = 0 \)

Solution:

L.H.S. \( = \begin{vmatrix} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{vmatrix} = \begin{vmatrix} x & a & x \\ y & b & y \\ z & c & z \end{vmatrix} + \begin{vmatrix} x & a & a \\ y & b & b \\ z & c & c \end{vmatrix} = 0 \).

[If any two rows or columns of a determinant are identical (all corresponding elements are same), the value of determinant is zero].

2. \( \begin{vmatrix} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{vmatrix} = 0 \)

Solution:

L.H.S. \( = \begin{vmatrix} a – b & b – c & c – a \\ b – c & c – a & a – b \\ c – a & a – b & b – c \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get

\( \begin{vmatrix} a – b + b – c + c – a & b – c & c – a \\ b – c + c – a + a – b & c – a & a – b \\ c – a + a – b + b – c & a – b & b – c \end{vmatrix} = \begin{vmatrix} 0 & b – c & c – a \\ 0 & c – a & a – b \\ 0 & a – b & b – c \end{vmatrix} = 0 \).

[as \( C_1 = 0 \)]

3. \( \begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix} = 0 \)

Solution:

L.H.S. \( = \begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix} \).

Applying \( C_1 \to C_1 + 9C_2 \), we get

\( \begin{vmatrix} 65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86 \end{vmatrix} = 0 \).

[If any two rows or columns of a determinant are identical (all corresponding elements are same), the value of determinant is zero].

4. \( \begin{vmatrix} 1 & bc & a(b + c) \\ 1 & ca & b(c + a) \\ 1 & ab & c(a + b) \end{vmatrix} = 0 \)

Solution:

L.H.S. \( = \begin{vmatrix} 1 & bc & a(b + c) \\ 1 & ca & b(c + a) \\ 1 & ab & c(a + b) \end{vmatrix} = \begin{vmatrix} 1 & bc & ab + ac \\ 1 & ca & bc + ab \\ 1 & ab & ca + bc \end{vmatrix} = 0 \).

Applying \( C_3 \to C_3 + C_2 \), we get

\( \begin{vmatrix} 1 & bc & ab + bc + ac \\ 1 & ca & bc + ca + ab \\ 1 & ab & ca + bc + ab \end{vmatrix} = (ab + bc + ca) \begin{vmatrix} 1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1 \end{vmatrix} \).

\( = (ab + bc + ca) \times 0 = 0 \).

[as \( C_1 \sim C_3 \)]

5. \( \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \)

Solution:

L.H.S. \( = \begin{vmatrix} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} \).

Applying \( R_1 \to R_1 + R_2 + R_3 \), we get

\( \begin{vmatrix} 2(a + b + c) & 2(p + q + r) & 2(x + y + z) \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} = 2 \begin{vmatrix} a + b + c & p + q + r & x + y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \), we get

\( 2 \begin{vmatrix} b & q & y \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{vmatrix} \).

Applying \( R_3 \to R_3 – R_1 \), we get

\( 2 \begin{vmatrix} b & q & y \\ c + a & r + p & z + x \\ a & p & x \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_3 \), we get

\( 2 \begin{vmatrix} b & q & y \\ c & r & z \\ a & p & x \end{vmatrix} \).

Interchanging \( R_1 \leftrightarrow R_2 \), we get

\( -2 \begin{vmatrix} a & p & x \\ c & r & z \\ b & q & y \end{vmatrix} \).

Interchanging \( R_2 \leftrightarrow R_3 \), we get

\( 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \) R.H.S.

Using properties of determinants, in Exercises 6 to 14 show that:

6. \( \begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix} = 0 \)

Solution:

Applying \( R_1 \leftrightarrow R_2, R_2 \leftrightarrow R_3 \), we get

\( \begin{vmatrix} -a & 0 & -c \\ b & c & 0 \\ 0 & a & -b \end{vmatrix} \).

Applying \( R_1 \leftrightarrow \frac{-R_1}{a} \), we get

\( \begin{vmatrix} 1 & 0 & \frac{c}{a} \\ b & c & 0 \\ 0 & a & -b \end{vmatrix} \).

Applying \( R_2 \to R_2 – bR_1 \), we get

\( \begin{vmatrix} 1 & 0 & \frac{c}{a} \\ 0 & c & -\frac{bc}{a} \\ 0 & a & -b \end{vmatrix} \).

Applying \( R_2 \to \frac{R_2}{c} \), we get

\( \begin{vmatrix} 1 & 0 & \frac{c}{a} \\ 0 & 1 & -\frac{b}{a} \\ 0 & a & -b \end{vmatrix} \).

Applying \( R_3 \to R_3 – aR_2 \), we get

\( \begin{vmatrix} 1 & 0 & \frac{c}{a} \\ 0 & 1 & -\frac{b}{a} \\ 0 & 0 & 0 \end{vmatrix} = 0 \).

(since each element of \( R_3 \) is 0)

7. \( \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} = 4a^2b^2c^2 \)

Solution:

\( \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} \).

Taking a, b and c common from \( R_1, R_2 \) and \( R_3 \) respectively, we get

\( abc \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} \).

Applying \( R_1 \to R_1 + R_3 \), we get

\( abc \begin{vmatrix} 0 & 2b & 0 \\ a & -b & c \\ a & b & -c \end{vmatrix} \).

Expanding along \( R_1 \), we get

\( (abc)(-2b) \begin{vmatrix} a & c \\ a & -c \end{vmatrix} \).

\( = (abc)(-2b)[-ac – ac] = (abc)(4abc) = 4a^2b^2c^2 \).

8. (i) \( \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a – b)(b – c)(c – a) \)

Solution:

L.H.S. \( = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \) and \( R_2 \to R_2 – R_3 \), we get

\( \begin{vmatrix} 0 & a – b & a^2 – b^2 \\ 0 & b – c & b^2 – c^2 \\ 1 & c & c^2 \end{vmatrix} = \begin{vmatrix} 0 & a – b & (a – b)(a + b) \\ 0 & b – c & (b – c)(b + c) \\ 1 & c & c^2 \end{vmatrix} \).

Taking out \( (a – b) \) and \( (b – c) \) common from \( R_1 \) and \( R_2 \) respectively, we get

\( (a – b)(b – c) \begin{vmatrix} 0 & 1 & a + b \\ 0 & 1 & b + c \\ 1 & c & c^2 \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \), we get

\( (a – b)(b – c) \begin{vmatrix} 0 & 0 & a – c \\ 0 & 1 & b + c \\ 1 & c & c^2 \end{vmatrix} \).

Taking out \( (a – c) \) common from \( R_1 \), we get

\( (a – b)(b – c)(a – c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & b + c \\ 1 & c & c^2 \end{vmatrix} \).

Expanding along \( R_1 \), we get

\( (a – b)(b – c)(a – c) \cdot 1 \cdot (0 – 1) = (a – b)(b – c)(c – a) = \) R.H.S.

8. (ii) \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a – b)(b – c)(c – a)(a + b + c) \)

Solution:

L.H.S. \( = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \).

Applying \( C_1 \to C_1 – C_2 \) and \( C_2 \to C_2 – C_3 \), we get

\( \begin{vmatrix} 0 & 0 & 1 \\ a – b & b – c & c \\ a^3 – b^3 & b^3 – c^3 & c^3 \end{vmatrix} \).

Taking out \( (a – b) \) and \( (b – c) \) common from \( C_1 \) and \( C_2 \) respectively, we get

\( (a – b)(b – c) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a^2 + ab + b^2 & b^2 + bc + c^2 & c^3 \end{vmatrix} \).

Applying \( C_1 \to C_1 – C_2 \), we get

\( (a – b)(b – c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ a^2 – c^2 + b(a – c) & b^2 + bc + c^2 & c^3 \end{vmatrix} \).

\( = (a – b)(b – c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ (a – c)(a + b + c) & b^2 + bc + c^2 & c^3 \end{vmatrix} \).

Taking out \( (a – c) \times (a + b + c) \) common from \( C_1 \), we get

\( (a – b)(b – c)(a – c)(a + b + c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ 1 & b^2 + bc + c^2 & c^3 \end{vmatrix} \).

Expanding along \( R_1 \), we get

\( (a – b)(b – c)(a – c)(a + b + c)(-1) = (a – b)(b – c)(c – a)(a + b + c) = \) R.H.S.

9. \( \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} = (x – y)(y – z)(z – x)(xy + yz + zx) \)

Solution:

Let L.H.S. \( = \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} \).

Multiplying \( R_1, R_2 \) and \( R_3 \) by x, y and z respectively, we get

\( \Delta = \frac{1}{xyz} \begin{vmatrix} x^2 & x^3 & xyz \\ y^2 & y^3 & xyz \\ z^2 & z^3 & xyz \end{vmatrix} = \frac{xyz}{xyz} \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix} \).

[Taking xyz common from \( C_3 \)]

Applying \( C_2 \leftrightarrow C_3 \), we get

\( – \begin{vmatrix} x^2 & 1 & x^3 \\ y^2 & 1 & y^3 \\ z^2 & 1 & z^3 \end{vmatrix} \).

Applying \( C_1 \leftrightarrow C_2 \), we get

\( \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix} \).

Applying \( R_1 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_1 \), we get

\( \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & y^2 – x^2 & y^3 – x^3 \\ 0 & z^2 – x^2 & z^3 – x^3 \end{vmatrix} \).

Taking \( (y – x) \) and \( (z – x) \) common from \( R_2 \) and \( R_3 \) respectively, we get

\( (y – x)(z – x) \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & 1 & y + x + x^2 \\ 0 & 1 & z + x + x^2 \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_3 \), we get

\( (y – x)(z – x) \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & 0 & y – z \\ 0 & 1 & z + x + x^2 \end{vmatrix} \).

Taking \( (y – z) \) common from \( R_2 \), we get

\( (y – x)(z – x)(y – z) \begin{vmatrix} 1 & x^2 & x^3 \\ 0 & 0 & 1 \\ 0 & 1 & z + x + x^2 \end{vmatrix} \).

Expanding along \( C_1 \), we get

\( (y – x)(z – x)(y – z) \cdot 1 \cdot [0 – 1] = (x – y)(y – z)(z – x)(xy + yz + zx) = \) R.H.S.

10. (i) \( \begin{vmatrix} x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4 \end{vmatrix} = (5x + 4)(4 – x)^2 \)

Solution:

L.H.S. \( = \begin{vmatrix} x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4 \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get

\( \begin{vmatrix} 5x + 4 & 2x & 2x \\ 5x + 4 & x + 4 & 2x \\ 5x + 4 & 2x & x + 4 \end{vmatrix} = (5x + 4) \begin{vmatrix} 1 & 2x & 2x \\ 1 & x + 4 & 2x \\ 1 & 2x & x + 4 \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \), we get

\( (5x + 4) \begin{vmatrix} 0 & x – 4 & 0 \\ 1 & x + 4 & 2x \\ 1 & 2x & x + 4 \end{vmatrix} \).

Expanding along \( R_1 \), we get

\( (5x + 4) \left[ -(x – 4) \begin{vmatrix} 1 & 2x \\ 1 & x + 4 \end{vmatrix} \right] \).

\( = (5x + 4) \left[ -(x – 4)(x + 4 – 2x) \right] \).

\( = (5x + 4) \left[ -(x – 4)(-x + 4) \right] = (5x + 4)(4 – x)(4 – x) \).

\( = (5x + 4)(4 – x)^2 = \) R.H.S.

10. (ii) \( \begin{vmatrix} y + k & y & y \\ y & y + k & y \\ y & y & y + k \end{vmatrix} = k^2(3y + k) \)

Solution:

L.H.S. \( = \begin{vmatrix} y + k & y & y \\ y & y + k & y \\ y & y & y + k \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get

\( \begin{vmatrix} 3y + k & y & y \\ 3y + k & y + k & y \\ 3y + k & y & y + k \end{vmatrix} = (3y + k) \begin{vmatrix} 1 & y & y \\ 1 & y + k & y \\ 1 & y & y + k \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \), we get

\( (3y + k) \begin{vmatrix} 0 & -k & 0 \\ 1 & y + k & y \\ 1 & y & y + k \end{vmatrix} = (3y + k) \left[ k \begin{vmatrix} 1 & y \\ 1 & y + k \end{vmatrix} \right] \).

\( = (3y + k)k[y + k – y] = (3y + k)k^2 = \) R.H.S.

11. (i) \( \begin{vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end{vmatrix} = (a + b + c)^3 \)

Solution:

L.H.S. \( = \begin{vmatrix} a – b – c & 2a & 2a \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end{vmatrix} \).

Applying \( R_1 \to R_1 + R_2 + R_3 \), we get

\( \begin{vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end{vmatrix} \).

Taking \( (a + b + c) \) common from \( R_1 \), we get

\( (a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b – c – a & 2b \\ 2c & 2c & c – a – b \end{vmatrix} \).

Applying \( C_1 \to C_1 – C_2 \), we get

\( (a + b + c) \begin{vmatrix} 0 & 1 & 1 \\ b + c + a & b – c – a & 2b \\ 0 & 2c & c – a – b \end{vmatrix} \).

Taking \( (a + b + c) \) common from \( C_1 \), we get

\( (a + b + c)^2 \begin{vmatrix} 0 & 1 & 1 \\ 1 & b – c – a & 2b \\ 0 & 2c & c – a – b \end{vmatrix} \).

Applying \( C_2 \to C_2 – C_3 \), we get

\( (a + b + c)^2 \begin{vmatrix} 0 & 0 & 1 \\ 1 & -(a + b + c) & 2b \\ 0 & (a + b + c) & c – a – b \end{vmatrix} \).

Taking \( (a + b + c) \) common from \( C_2 \), we get

\( (a + b + c)^3 \begin{vmatrix} 0 & 0 & 1 \\ 1 & -1 & 2b \\ 0 & 1 & c – a – b \end{vmatrix} \).

Expanding along \( R_1 \), we get

\( (a + b + c)^3 = \) R.H.S.

11. (ii) \( \begin{vmatrix} x + y + 2z & x & y \\ z & y + z + 2x & y \\ z & x & z + x + 2y \end{vmatrix} = 2(x + y + z)^3 \)

Solution:

L.H.S. \( = \begin{vmatrix} x + y + 2z & x & y \\ z & y + z + 2x & y \\ z & x & z + x + 2y \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \) and \( R_2 \to R_2 – R_3 \), we get

\( \begin{vmatrix} x + y + z & -(x + y + z) & 0 \\ 0 & x + y + z & -(x + y + z) \\ z & x & z + x + 2y \end{vmatrix} \).

Taking \( (x + y + z) \) common from \( R_1 \) and \( R_2 \), we get

\( (x + y + z)^2 \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ z & x & z + x + 2y \end{vmatrix} \).

Applying \( C_2 \to C_1 + C_2 \), we get

\( (x + y + z)^2 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ z & x + z & z + x + 2y \end{vmatrix} \).

Applying \( C_3 \to C_2 + C_3 \), we get

\( (x + y + z)^2 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ z & x + z & 2(x + y + z) \end{vmatrix} \).

\( = 2(x + y + z)^2 \cdot (x + y + z) \).

[Determinant of triangular matrix is product of its diagonal elements]

\( = 2(x + y + z)^3 = \) R.H.S.

12. \( \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (1 – x^3)^2 \)

Solution:

L.H.S. \( = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get

\( \begin{vmatrix} 1 + x + x^2 & x & x^2 \\ 1 + x + x^2 & 1 & x \\ 1 + x + x^2 & x^2 & 1 \end{vmatrix} = (1 + x + x^2) \begin{vmatrix} 1 & x & x^2 \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{vmatrix} \).

Applying \( C_1 \to C_1 – C_2 \), we get

\( (1 + x + x^2) \begin{vmatrix} 0 & x & x^2 \\ 0 & 1 & x \\ 1 – x^2 & x^2 & 1 \end{vmatrix} \).

Taking \( (1 – x) \) common from \( C_1 \), we get

\( (1 + x + x^2)(1 – x) \begin{vmatrix} 0 & x & x^2 \\ 0 & 1 & x \\ 1 + x & x^2 & 1 \end{vmatrix} \).

Expanding along \( C_1 \), we get

\( (1 + x + x^2)(1 – x) \left[ 1 \begin{vmatrix} 1 & x \\ x^2 & 1 \end{vmatrix} + (1 + x) \begin{vmatrix} x & x^2 \\ 1 & x \end{vmatrix} \right] \).

\( = (1 – x^3) \left[ (1 – x^3) + (1 + x)(x^2 – x^2) \right] = (1 – x^3)^2 = \) R.H.S.

13. \( \begin{vmatrix} 1 + a^2 – b^2 & 2ab & -2b \\ 2ab & 1 – a^2 + b^2 & 2a \\ 2b & -2a & 1 – a^2 – b^2 \end{vmatrix} = (1 + a^2 + b^2)^3 \)

Solution:

L.H.S. \( = \begin{vmatrix} 1 + a^2 – b^2 & 2ab & -2b \\ 2ab & 1 – a^2 + b^2 & 2a \\ 2b & -2a & 1 – a^2 – b^2 \end{vmatrix} \).

Applying \( C_1 \to C_1 – bC_3 \) and \( C_2 \to C_2 + aC_3 \), we get

\( \begin{vmatrix} 1 + a^2 + b^2 & 0 & -2b \\ 0 & 1 + a^2 + b^2 & 2a \\ b(1 + a^2 + b^2) & -a(1 + a^2 + b^2) & 1 – a^2 – b^2 \end{vmatrix} \).

Taking \( (1 + a^2 + b^2) \) common from \( C_1 \) and \( C_2 \), we get

\( (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1 – a^2 – b^2 \end{vmatrix} \).

Applying \( R_3 \to R_3 – bR_1 \), we get

\( (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & -a & 1 – a^2 + b^2 \end{vmatrix} \).

Expanding along \( C_1 \), we get

\( (1 + a^2 + b^2)^2 \left[ 1((1 – a^2 + b^2 + 2a^2)) \right] \).

\( = (1 + a^2 + b^2)^2(1 + a^2 + b^2) \).

\( = (1 + a^2 + b^2)^3 = \) R.H.S.

14. \( \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ac & bc & c^2 + 1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \)

Solution:

L.H.S. \( = \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ac & bc & c^2 + 1 \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get

\( \begin{vmatrix} 1 + a(a + b + c) & ab & ac \\ 1 + b(a + b + c) & b^2 + 1 & bc \\ 1 + c(a + b + c) & bc & c^2 + 1 \end{vmatrix} \).

By property 5 and taking \( (a + b + c) \) common from \( C_1 \) in determinant II, we get

\( \begin{vmatrix} 1 & ab & ac \\ 1 & b^2 + 1 & bc \\ 1 & bc & c^2 + 1 \end{vmatrix} + (a + b + c) \begin{vmatrix} a & ab & ac \\ b & b^2 + 1 & bc \\ c & bc & c^2 + 1 \end{vmatrix} \).

Changing rows into columns, we have

\( \begin{vmatrix} 1 & 1 & 1 \\ ab & b^2 + 1 & bc \\ ac & bc & c^2 + 1 \end{vmatrix} + (a + b + c) \begin{vmatrix} a & b & c \\ ab & b^2 + 1 & bc \\ ac & bc & c^2 + 1 \end{vmatrix} \).

For determinant I we apply, \( C_1 \to C_1 – C_2, C_2 \to C_2 – C_3 \) and for determinant II we take out a common from \( C_1 \), we get

\( \begin{vmatrix} 0 & 0 & 1 \\ ab – b^2 – 1 & b^2 + 1 – bc & bc \\ ac – bc & bc – c^2 – 1 & c^2 + 1 \end{vmatrix} + a(a + b + c) \begin{vmatrix} 1 & b & c \\ b & b^2 + 1 & bc \\ c & bc & c^2 + 1 \end{vmatrix} \).

Applying \( C_2 \to C_2 – bC_1 \) and \( C_3 \to C_3 – cC_1 \) in determinant II, we get

\( \begin{vmatrix} 0 & 0 & 1 \\ ab – b^2 – 1 & b^2 + 1 – bc & bc \\ ac – bc & bc – c^2 – 1 & c^2 + 1 \end{vmatrix} + a(a + b + c) \begin{vmatrix} 1 & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{vmatrix} \).

Expanding along \( R_1 \), we have

\( 1[(ab – b^2 – 1)(bc – c^2 – 1) – (ac – bc)(b^2 + 1 – bc)] \)

\( + a(a + b + c) \).

\( = [(a b^2 c – a b c^2 – a b – b^3 c + b^2 c^2 + b^2 – b c + c^2 + 1) \)

\( – (a b^2 c + a c – a b c^2 – b^3 c – b c + b^2 c^2)] \)

\( + a(a + b + c) \).

\( = a b^2 c – a b c^2 – a b – b^3 c + b^2 c^2 + b^2 – b c + c^2 + 1 \)

\( – a b^2 c – a c + a b c^2 + b^3 c + b c – b^2 c^2 \)

\( + a(a + b + c) \).

\( = -a b + b^2 + c^2 + 1 – a c + a^2 + a b + a c \).

\( = 1 + a^2 + b^2 + c^2 = \) R.H.S.

Choose the correct answer in Exercise 15 and 16.

15. Let A be a square matrix of order 3 × 3, then \( |kA| \) is equal to:

1. A. \( k|A| \)
2. B. \( k^2|A| \)
3. C. \( k^3|A| \)
4. D. \( 3k|A| \)

Solution:

(C) If A is a square matrix of order n, then \( |kA| = k^n|A| \), where k is scalar.

16. Which of the following is correct?

1. A. Determinant is a square matrix.
2. B. Determinant is a number associated to a matrix.
3. C. Determinant is a number associated to a square matrix.
4. D. None of these.

Solution:

(C) Determinant is a number associated to a square matrix.

Exercise – 4.3

1. Find the area of the triangle with vertices at the points given in each of the following:

Solution:

1. $(1, 0), (6, 0), (4, 3)$
   Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}1&0&1\\6&0&1\\4&3&1\end{array}} \right|$
   $= \frac{1}{2}\left[ {1\left| {\begin{array}{*{20}{r}}0&1\\3&1\end{array}} \right| – 0 + 1\left| {\begin{array}{*{20}{r}}6&0\\4&3\end{array}} \right|} \right]$
   $= \frac{1}{2}[1(0 – 3) + 1(18 – 0)] = \frac{15}{2} = 7.5$ sq. units.
2. $(2, 7), (1, 1), (10, 8)$
   Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}2&7&1\\1&1&1\\10&8&1\end{array}} \right|$
   $= \frac{1}{2}\left[ {2\left| {\begin{array}{*{20}{r}}1&1\\8&1\end{array}} \right| – 7\left| {\begin{array}{*{20}{r}}1&1\\10&1\end{array}} \right| + 1\left| {\begin{array}{*{20}{r}}1&1\\10&8\end{array}} \right|} \right]$
   $= \frac{1}{2}[2(1 – 8) – 7(1 – 10) + 1(8 – 10)] = \frac{47}{2} = 23.5$ sq. units.
3. $(-2, -3), (3, 2), (-1, -8)$
   Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}-2&-3&1\\3&2&1\\-1&-8&1\end{array}} \right|$
   $= \frac{1}{2}\left[ {-2\left| {\begin{array}{*{20}{r}}2&1\\-8&1\end{array}} \right| + 3\left| {\begin{array}{*{20}{r}}3&1\\-1&1\end{array}} \right| + 1\left| {\begin{array}{*{20}{r}}3&2\\-1&-8\end{array}} \right|} \right]$
   $= \frac{1}{2}[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)] = \frac{-30}{2} = -15$
   Area $= 15$ square units. (Absolute value)

2. Show that points $A(a, b + c), B(b, c + a), C(c, a + b)$ are collinear.

Solution:

Consider $\left| {\begin{array}{*{20}{r}}a&b+c&1\\b&c+a&1\\c&a+b&1\end{array}} \right|$
Applying $C_1 \rightarrow C_1 + C_2$, we get
$= \left| {\begin{array}{*{20}{r}}a+b+c&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1\end{array}} \right|$
Taking $(a + b + c)$ common from $C_1$, we get
$= (a + b + c)\left| {\begin{array}{*{20}{r}}1&b+c&1\\1&c+a&1\\1&a+b&1\end{array}} \right|$
$= (a + b + c) \times 0 = 0$
Hence, the points are collinear.

3. Find values of k if the area of the triangle is 4 sq. units and vertices are:

Solution:

1. $(k, 0), (4, 0), (0, 2)$
   Area of triangle $= 4$ sq. units
   Area $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}k&0&1\\4&0&1\\0&2&1\end{array}} \right|$
   $= \frac{1}{2}[k(0 – 2) – 0 + 1(8 – 0)] = -k + 4$
   Now, $-k + 4 = \pm 4 \Rightarrow -k + 4 = 4$ or $-k + 4 = -4$
   $\Rightarrow k = 0$ or $k = 8$
   $\Rightarrow k = 0, 8$
2. $(-2, 0), (0, 4), (0, k)$
   Area of triangle $= 4$ sq. units
   Area $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}-2&0&1\\0&4&1\\0&k&1\end{array}} \right|$
   $= \frac{1}{2}[-2(4 – k) – 0 + 1(0)] = -4 + k$
   Now, $-4 + k = \pm 4 \Rightarrow -4 + k = 4$ or $-4 + k = -4$
   $\Rightarrow k = 0$ or $k = 8$
   $\Rightarrow k = 0, 8$

4. Find the equation of the line joining the points using determinants:

Solution:

1. $(1, 2)$ and $(3, 6)$
   Equation of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left| {\begin{array}{*{20}{r}}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}} \right| = 0$
   $\Rightarrow \left| {\begin{array}{*{20}{r}}x&y&1\\1&2&1\\3&6&1\end{array}} \right| = 0$
   $\Rightarrow x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0$
   $\Rightarrow -4x + 2y = 0 \Rightarrow 2x – y = 0$
   Hence, $y = 2x$ is the required line.
2. $(3, 1)$ and $(9, 3)$
   Equation of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is $\left| {\begin{array}{*{20}{r}}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}} \right| = 0$
   $\Rightarrow \left| {\begin{array}{*{20}{r}}x&y&1\\3&1&1\\9&3&1\end{array}} \right| = 0$
   $\Rightarrow x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0 \Rightarrow -2x + 6y = 0$
   $\Rightarrow -x + 3y = 0 \Rightarrow x – 3y = 0$
   Hence, $x – 3y = 0$ is the required line.

5. If the area of a triangle is 35 sq. units with vertices $(2, -6), (5, 4)$ and $(k, 4)$, then k is:

Solution:

(D) 12, -2
Area of triangle $= \frac{1}{2}\left| {\begin{array}{*{20}{r}}2&-6&1\\5&4&1\\k&4&1\end{array}} \right| = \pm 35$
$\Rightarrow \frac{1}{2}[2(4 – 4) + 6(5 – k) + 1(20 – 4k)] = \pm 35$
$\Rightarrow \frac{1}{2}[30 – 6k + 20 – 4k] = \pm 35$
$\Rightarrow \frac{1}{2}[50 – 10k] = \pm 35 \Rightarrow 25 – 5k = \pm 35$
$\Rightarrow 25 – 5k = 35$ or $25 – 5k = -35$
$\Rightarrow 5k = 10$ or $5k = 60$
$\Rightarrow k = -2$ or $k = 12$

Exercise – 4.4

1. Write Minors and Cofactors of the elements of the following determinants:

Solution:

1. $\left| {\begin{array}{*{20}{r}}2& -4\\0& 3\end{array}} \right|$
   Let $P = \left| {\begin{array}{*{20}{r}}2& -4\\0& 3\end{array}} \right|$
   Minor of the element $a_{ij}$ is $M_{ij}$. Here,
   $M_{11} = 3$, $M_{12} = 0$, $M_{21} = -4$, $M_{22} = 2$
   For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
   $\therefore P_{11} = 3$, $P_{12} = 0$, $P_{21} = 4$, $P_{22} = 3$
2. $\left| {\begin{array}{*{20}{r}}a& c\\b& d\end{array}} \right|$
   Let $P = \left| {\begin{array}{*{20}{r}}a& c\\b& d\end{array}} \right|$, Minor of the element $a_{ij}$ is $M_{ij}$. Here,
   $M_{11} = d$, $M_{12} = b$, $M_{21} = c$, $M_{22} = a$
   For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
   $\therefore P_{11} = d$, $P_{12} = -b$, $P_{21} = -c$, $P_{22} = a$

2. Write Minors and Cofactors of the elements of the following determinants:

Solution:

1. $\left| {\begin{array}{*{20}{r}}1& 0& 0\\0& 1& 0\\0& 0& 1\end{array}} \right|$
   Let $P = \left| {\begin{array}{*{20}{r}}1& 0& 0\\0& 1& 0\\0& 0& 1\end{array}} \right|$, we have
   $M_{11} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$, $M_{12} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{13} = \left| {\begin{array}{*{20}{r}}0& 1\\0& 1\end{array}} \right| = 0$
   $M_{21} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{22} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$, $M_{23} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 0$
   $M_{31} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{32} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 0$, $M_{33} = \left| {\begin{array}{*{20}{r}}1& 1\\0& 1\end{array}} \right| = 1$
   For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
   $P_{11} = 1$, $P_{12} = 0$, $P_{13} = 0$, $P_{21} = 0$, $P_{22} = 1$, $P_{23} = 0$, $P_{31} = 0$, $P_{32} = 0$, $P_{33} = 1$
2. $\left| {\begin{array}{*{20}{r}}1& 0& 4\\3& 5& -1\\0& 1& 2\end{array}} \right|$
   Let $P = \left| {\begin{array}{*{20}{r}}1& 0& 4\\3& 5& -1\\0& 1& 2\end{array}} \right|$, we have
   $M_{11} = \left| {\begin{array}{*{20}{r}}5& -1\\1& 2\end{array}} \right| = 10 + 1 = 11$, $M_{12} = \left| {\begin{array}{*{20}{r}}3& -1\\0& 2\end{array}} \right| = 6$, $M_{13} = \left| {\begin{array}{*{20}{r}}3& 5\\0& 1\end{array}} \right| = 3$
   $M_{21} = \left| {\begin{array}{*{20}{r}}0& 4\\1& 2\end{array}} \right| = -4$, $M_{22} = \left| {\begin{array}{*{20}{r}}1& 4\\0& 2\end{array}} \right| = 2$, $M_{23} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$
   $M_{31} = \left| {\begin{array}{*{20}{r}}0& 4\\5& -1\end{array}} \right| = -20$, $M_{32} = \left| {\begin{array}{*{20}{r}}1& 4\\3& -1\end{array}} \right| = -13$, $M_{33} = \left| {\begin{array}{*{20}{r}}1& 0\\3& 5\end{array}} \right| = 5$
   For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
   $P_{11} = 11$, $P_{12} = -6$, $P_{13} = 3$, $P_{21} = 4$, $P_{22} = 2$, $P_{23} = -1$, $P_{31} = -20$, $P_{32} = 13$, $P_{33} = 5$

3. Using Cofactors of elements of the second row, evaluate $\Delta = \left| {\begin{array}{*{20}{r}}5& 3& 8\\2& 0& 1\\1& 2& 3\end{array}} \right|$.

Solution:

$\Delta = \left| {\begin{array}{*{20}{r}}5& 3& 8\\2& 0& 1\\1& 2& 3\end{array}} \right|$
Cofactors of elements of the second row are
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3& 8\\2& 3\end{array}} \right| = -(9 – 16) = 7$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}5& 8\\1& 3\end{array}} \right| = 15 – 8 = 7$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}5& 3\\1& 2\end{array}} \right| = -(10 – 3) = -7$
Now, $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2 \times 7 + 0 \times 7 + 1 \times (-7) = 14 + 0 – 7 = 7$

4. Using Cofactors of elements of the third column, evaluate $\Delta = \left| {\begin{array}{*{20}{r}}1& x& yz\\1& y& zx\\1& z& xy\end{array}} \right|$.

Solution:

Let $\Delta = \left| {\begin{array}{*{20}{r}}1& x& yz\\1& y& zx\\1& z& xy\end{array}} \right|$
Cofactors of elements of the third column are
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1& y\\1& z\end{array}} \right| = z – y$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1& x\\1& z\end{array}} \right| = -(z – x)$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1& x\\1& y\end{array}} \right| = y – x = -(x – y)$
Now, $\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$\Delta = -yz(y – z) – zx(z – x) – xy(x – y)$
$= zy(z – y) + zx(x – z) + xy(y – x)$
$= yz^2 – y^2z + zx^2 – z^2x + xy^2 – x^2y$
$= x^2z – x^2y + xy^2 – xz^2 + yz^2 – y^2z$
$= x^2(z – y) + x(y^2 – z^2) + yz(z – y)$
$= (z – y)[x^2 – x(y + z) + yz] = (z – y)[x(x – y) – z(x – y)]$
$= (z – y)(x – y)(x – z) = (x – y)(y – z)(z – x)$

5. If $\Delta = \left| {\begin{array}{*{20}{r}}a_{11}& a_{12}& a_{13}\\a_{21}& a_{22}& a_{23}\\a_{31}& a_{32}& a_{33}\end{array}} \right|$ and $A_{ij}$ is the Cofactor of $a_{ij}$, then the value of $\Delta$ is given by:

Solution:

(D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$
We know that the value of the determinant is given by:
$a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$ (where $A_{ij}$ is the Cofactor of $a_{ij}$)

Exercise – 4.5

1. Find the adjoint of each of the matrices:

Solution:

1. $\left[ {\begin{array}{*{20}{r}}1&2\\3&4\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}1&2\\3&4\end{array}} \right]$. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}(4) = 4$
   $A_{12} = (-1)^{1+2}(3) = -3$
   $A_{21} = (-1)^{2+1}(2) = -2$
   $A_{22} = (-1)^{2+2}(1) = 1$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}4&-2\\-3&1\end{array}} \right]$

2. Find the adjoint of the matrix:

Solution:

$\left[ {\begin{array}{*{20}{r}}1&-1&2\\2&3&-2\\-2&0&1\end{array}} \right]$
Let $A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\2&3&-2\\-2&0&1\end{array}} \right]$. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then:
$A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}3&-2\\0&1\end{array}} \right] = 3$
$A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}2&-2\\-2&1\end{array}} \right] = -12$
$A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}2&3\\-2&0\end{array}} \right] = 6$
$A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&1\end{array}} \right] = 1$
$A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&2\\-2&1\end{array}} \right] = 5$
$A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&-1\\-2&0\end{array}} \right] = 2$
$A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}-1&2\\3&-2\end{array}} \right] = -11$
$A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&2\\2&-2\end{array}} \right] = -1$
$A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&-1\\2&3\end{array}} \right] = 5$
$\therefore adj A = \left[ {\begin{array}{*{20}{r}}3&-11&0\\1&5&-1\\-2&-1&5\end{array}} \right]^T = \left[ {\begin{array}{*{20}{r}}3&1&-2\\-11&5&-1\\0&-1&5\end{array}} \right]$

3. Verify: $A(adj A) = (adj A)A = |A|I$:

Solution:

1. $\left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]$
   $|A| = -12 + 12 = 0$
   Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}(-6) = -6$
   $A_{12} = (-1)^{1+2}(-4) = 4$
   $A_{21} = (-1)^{2+1}(3) = -3$
   $A_{22} = (-1)^{2+2}(2) = 2$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right]$
   $A(adj A) = \left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
   $(adj A)A = \left[ {\begin{array}{*{20}{r}}-6&-3\\4&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&3\\-4&-6\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
   Since $|A| = 0$, $|A|I = O$
   Hence, $A(adj A) = (adj A)A = |A|I$
2. $\left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]$
   Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}0&-2\\0&3\end{array}} \right] = 0$
   $A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}3&-2\\1&3\end{array}} \right] = -11$
   $A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}3&0\\1&0\end{array}} \right] = 0$
   $A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&3\end{array}} \right] = 3$
   $A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&2\\1&3\end{array}} \right] = 1$
   $A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&-1\\1&0\end{array}} \right] = -1$
   $A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}-1&2\\0&-2\end{array}} \right] = 2$
   $A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&2\\3&-2\end{array}} \right] = 8$
   $A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&-1\\3&0\end{array}} \right] = 3$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right]$
   $A(adj A) = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}11&0&0\\0&11&0\\0&0&11\end{array}} \right] = 11I$
   $(adj A)A = \left[ {\begin{array}{*{20}{r}}0&3&2\\-11&1&8\\0&-1&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&-1&2\\3&0&-2\\1&0&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}11&0&0\\0&11&0\\0&0&11\end{array}} \right] = 11I$
   Also, $|A| = 11$
   Hence, $A(adj A) = (adj A)A = |A|I$

4. Find the inverse of each of the matrices (if it exists):

Solution:

1. $\left[ {\begin{array}{*{20}{r}}2&-2\\4&3\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}2&-2\\4&3\end{array}} \right]$
   $|A| = 6 + 8 = 14 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}(3) = 3$
   $A_{12} = (-1)^{1+2}(4) = -4$
   $A_{21} = (-1)^{2+1}(-2) = 2$
   $A_{22} = (-1)^{2+2}(2) = 2$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}3&2\\-4&2\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{14}\left[ {\begin{array}{*{20}{r}}3&2\\-4&2\end{array}} \right]$
2. $\left[ {\begin{array}{*{20}{r}}-1&5\\-3&2\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}-1&5\\-3&2\end{array}} \right]$
   $|A| = -2 + 15 = 13 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}(2) = 2$
   $A_{12} = (-1)^{1+2}(-3) = 3$
   $A_{21} = (-1)^{2+1}(5) = -5$
   $A_{22} = (-1)^{2+2}(-1) = -1$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}2&-5\\3&-1\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{13}\left[ {\begin{array}{*{20}{r}}2&-5\\3&-1\end{array}} \right]$
3. $\left[ {\begin{array}{*{20}{r}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}1&2&3\\0&2&4\\0&0&5\end{array}} \right]$
   $|A| = 10 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}2&4\\0&5\end{array}} \right] = 10$
   $A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}0&4\\0&5\end{array}} \right] = 0$
   $A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}0&2\\0&0\end{array}} \right] = 0$
   $A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}2&3\\0&5\end{array}} \right] = -10$
   $A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&3\\0&5\end{array}} \right] = 5$
   $A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&2\\0&0\end{array}} \right] = 0$
   $A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}2&3\\2&4\end{array}} \right] = 2$
   $A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&3\\0&4\end{array}} \right] = -4$
   $A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&2\\0&2\end{array}} \right] = 2$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}10&-10&2\\0&5&-4\\0&0&2\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}10&-10&2\\0&5&-4\\0&0&2\end{array}} \right]$
4. $\left[ {\begin{array}{*{20}{r}}1&0&0\\3&3&0\\5&2&-1\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&3&0\\5&2&-1\end{array}} \right]$
   $|A| = -3 – 0 = -3 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}3&0\\2&-1\end{array}} \right] = -3$
   $A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}3&0\\5&-1\end{array}} \right] = 3$
   $A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}3&3\\5&2\end{array}} \right] = -9$
   $A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}0&0\\2&-1\end{array}} \right] = 0$
   $A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&0\\5&-1\end{array}} \right] = -1$
   $A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&0\\5&2\end{array}} \right] = -2$
   $A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}0&0\\3&0\end{array}} \right] = 0$
   $A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&0\\3&0\end{array}} \right] = 0$
   $A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&0\\3&3\end{array}} \right] = 3$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}-3&0&0\\3&-1&0\\-9&-2&3\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{3}\left[ {\begin{array}{*{20}{r}}-3&0&0\\3&-1&0\\-9&-2&3\end{array}} \right]$
5. $\left[ {\begin{array}{*{20}{r}}2&1&3\\4&-1&0\\-7&2&1\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}2&1&3\\4&-1&0\\-7&2&1\end{array}} \right]$
   $|A| = -3 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}-1&0\\2&1\end{array}} \right] = -1$
   $A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}4&0\\-7&1\end{array}} \right] = -4$
   $A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}4&-1\\-7&2\end{array}} \right] = 1$
   $A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}1&3\\2&1\end{array}} \right] = 5$
   $A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}2&3\\-7&1\end{array}} \right] = 23$
   $A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}2&1\\-7&2\end{array}} \right] = -11$
   $A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}1&3\\-1&0\end{array}} \right] = 3$
   $A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}2&3\\4&0\end{array}} \right] = 12$
   $A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}2&1\\4&-1\end{array}} \right] = -6$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}-1&-9&-6\\5&-2&-1\\3&12&2\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{3}\left[ {\begin{array}{*{20}{r}}-1&-9&-6\\5&-2&-1\\3&12&2\end{array}} \right]$
6. $\left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$
   Let $A = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$
   $|A| = -(\cos^2\alpha + \sin^2\alpha) = -1 \neq 0$
   So, A is a non-singular matrix and therefore, it is invertible. Let $A_{ij}$ be cofactors of $a_{ij}$ in A. Then, the cofactors of elements of A are given by:
   $A_{11} = (-1)^{1+1}\left[ {\begin{array}{*{20}{r}}\cos \alpha&\sin \alpha\\\sin \alpha&-\cos \alpha\end{array}} \right] = -1$
   $A_{12} = (-1)^{1+2}\left[ {\begin{array}{*{20}{r}}0&\sin \alpha\\0&-\cos \alpha\end{array}} \right] = 0$
   $A_{13} = (-1)^{1+3}\left[ {\begin{array}{*{20}{r}}0&\cos \alpha\\0&\sin \alpha\end{array}} \right] = 0$
   $A_{21} = (-1)^{2+1}\left[ {\begin{array}{*{20}{r}}0&0\\\sin \alpha&-\cos \alpha\end{array}} \right] = 0$
   $A_{22} = (-1)^{2+2}\left[ {\begin{array}{*{20}{r}}1&0\\0&-\cos \alpha\end{array}} \right] = -\cos \alpha$
   $A_{23} = (-1)^{2+3}\left[ {\begin{array}{*{20}{r}}1&0\\0&\sin \alpha\end{array}} \right] = -\sin \alpha$
   $A_{31} = (-1)^{3+1}\left[ {\begin{array}{*{20}{r}}0&0\\\cos \alpha&\sin \alpha\end{array}} \right] = 0$
   $A_{32} = (-1)^{3+2}\left[ {\begin{array}{*{20}{r}}1&0\\0&\sin \alpha\end{array}} \right] = -\sin \alpha$
   $A_{33} = (-1)^{3+3}\left[ {\begin{array}{*{20}{r}}1&0\\0&\cos \alpha\end{array}} \right] = \cos \alpha$
   $\therefore adj A = \left[ {\begin{array}{*{20}{r}}-1&0&0\\0&-\cos \alpha&-\sin \alpha\\0&-\sin \alpha&\cos \alpha\end{array}} \right]$
   Hence, $A^{-1} = \frac{1}{|A|} adj A = -1 \cdot \left[ {\begin{array}{*{20}{r}}-1&0&0\\0&-\cos \alpha&-\sin \alpha\\0&-\sin \alpha&\cos \alpha\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&\cos \alpha&\sin \alpha\\0&\sin \alpha&-\cos \alpha\end{array}} \right]$

5. Let $A = \left[ {\begin{array}{*{20}{r}}3&7\\2&5\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{r}}6&8\\7&9\end{array}} \right]$. Verify that $(AB)^{-1} = B^{-1}A^{-1}$.

Solution:

We have, $AB = \left[ {\begin{array}{*{20}{r}}3&7\\2&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}6&8\\7&9\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}67&87\\47&61\end{array}} \right]$
Since, $|AB| = 67 \times 61 – 47 \times 87 = -2 \neq 0$
So, AB is a non-singular matrix and therefore, $(AB)^{-1}$ exists and is given by $(AB)^{-1} = \frac{1}{|AB|} adj(AB) = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}61&-87\\-47&67\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{r}}-61&87\\47&-67\end{array}} \right]$
Further, $|A| = 15 – 14 = 1 \neq 0$ and $|B| = 54 – 56 = -2 \neq 0$. So, A and B are both non-singular matrices and therefore, $A^{-1}$ and $B^{-1}$ both exist and are given by:
$A^{-1} = \left[ {\begin{array}{*{20}{r}}5&-7\\-2&3\end{array}} \right]$, $B^{-1} = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}9&-8\\-7&6\end{array}} \right]$
$\therefore B^{-1}A^{-1} = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}9&-8\\-7&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5&-7\\-2&3\end{array}} \right] = -\frac{1}{2}\left[ {\begin{array}{*{20}{r}}61&-87\\-47&67\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{r}}-61&87\\47&-67\end{array}} \right]$
Hence, $(AB)^{-1} = B^{-1}A^{-1}$.

6. If $A = \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]$, show that $A^2 – 5A + 7I = O$. Hence, find $A^{-1}$.

Solution:

$A = \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]$
L.H.S. $= A^2 – 5A + 7I$
$= \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] – 5\left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] + 7\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8&5\\-5&3\end{array}} \right] – \left[ {\begin{array}{*{20}{r}}15&5\\-5&10\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}7&0\\0&7\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8-15+7&5-5+0\\-5+5+0&3-10+7\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right] = O$
$\Rightarrow A^2 – 5A + 7I = O$
Hence, proved.
Now, multiplying by $A^{-1}$ on both sides, we get
$(A^{-1}A)A – 5AA^{-1} – 7IA^{-1} = O \Rightarrow IA – 5I + 7A^{-1} = O$
$\Rightarrow A – 5I + 7A^{-1} = O \Rightarrow 7A^{-1} = 5I – A$
$\Rightarrow 7A^{-1} = 5\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right] – \left[ {\begin{array}{*{20}{r}}3&1\\-1&2\end{array}} \right] \Rightarrow 7A^{-1} = \left[ {\begin{array}{*{20}{r}}2&-1\\1&3\end{array}} \right]$
$\Rightarrow A^{-1} = \frac{1}{7}\left[ {\begin{array}{*{20}{r}}2&-1\\1&3\end{array}} \right]$

7. For the matrix $A = \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]$, find the numbers a and b such that $A^2 + aA + bI = O$.

Solution:

We are given that, $A = \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]$
Now, $A^2 + aA + bI = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right] + a\left[ {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right] + b\left[ {\begin{array}{*{20}{r}}1&0\\0&1\end{array}} \right] = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}9+2&6+2\\3+1&2+1\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}3a&2a\\a&a\end{array}} \right] + \left[ {\begin{array}{*{20}{r}}b&0\\0&b\end{array}} \right] = O$
$\Rightarrow \left[ {\begin{array}{*{20}{r}}11+3a+b&8+2a\\4+a&3+a+b\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&0\\0&0\end{array}} \right]$
$\Rightarrow 4 + a = 0 \Rightarrow a = -4$
Also, $3 + a + b = 0 \Rightarrow b = -3 + 4 \Rightarrow b = 1$
Hence, $a = -4, b = 1$

8. For the matrix $A = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]$, show that $A^3 – 6A^2 + 5A + 11I = O$. Hence, find $A^{-1}$.

Solution:

We have $A = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]$
$\therefore A^2 = AA = \left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right]$
and $A^3 = A^2A = \left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}8&7&1\\-23&27&-69\\32&-13&58\end{array}} \right]$
Now, L.H.S. $= A^3 – 6A^2 + 5A + 11I$
$= \left[ {\begin{array}{*{20}{r}}8&7&1\\-23&27&-69\\32&-13&58\end{array}} \right] – 6\left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right] + 5\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] + 11\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}8-24+5+11&7-12+5+0&1-6+5+0\\-23+18+5+0&27-48+10+11&-69+84-15+0\\32-42+10+0&-13+18-5+0&58-84+15+11\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O \Rightarrow A^3 – 6A^2 + 5A + 11I = O$
Hence, proved.
Now, $A^3 – 6A^2 + 5A + 11I = O$
$\Rightarrow 11I = -A^3 + 6A^2 – 5A$
Multiplying (i) by $A^{-1}$, we get
$11A^{-1}I = -A^{-1}A^3 + 6A^{-1}A^2 – 5A^{-1}A \Rightarrow 11A^{-1} = -A^2 + 6A – 5I$
$\Rightarrow A^{-1} = -\frac{1}{11}A^2 + \frac{6}{11}A – \frac{5}{11}I$
$= -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&2&1\\-3&8&-14\\7&-3&14\end{array}} \right] + \frac{6}{11}\left[ {\begin{array}{*{20}{r}}1&1&1\\1&2&-3\\2&-1&3\end{array}} \right] – \frac{5}{11}\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-4+6-5&-2+6+0&-1+6+0\\3+6+0&-8+12-5&14-18+0\\-7+12+0&3-6+0&-14+18-5\end{array}} \right]$
$= \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-3&4&5\\9&-1&-4\\5&-3&-1\end{array}} \right]$

9. If $A = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]$, verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence, find $A^{-1}$.

Solution:

We have $A = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]$
$\therefore A^2 = AA = \left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right]$
and $A^3 = A^2A = \left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}22&-21&21\\-21&22&-21\\21&-21&22\end{array}} \right]$
Now, $A^3 – 6A^2 + 9A – 4I$
$= \left[ {\begin{array}{*{20}{r}}22&-21&21\\-21&22&-21\\21&-21&22\end{array}} \right] – 6\left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right] + 9\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] – 4\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}22-36+18-4&-21+30-9+0&21-30+9+0\\-21+30-9-0&22-36+18-4&-21+30-9+0\\21-30+9+0&-21+30-9+0&22-36+18-4\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{r}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = O$
Hence, $A^3 – 6A^2 + 9A – 4I = O$
Now, $A^3 – 6A^2 + 9A – 4I = O \Rightarrow 4I = A^3 – 6A^2 + 9A$
Multiplying both sides by $A^{-1}$, we get
$4A^{-1}I = A^{-1}A^3 – 6A^{-1}A^2 + 9A^{-1}A \Rightarrow 4A^{-1} = A^2 – 6A + 9I$
$A^{-1} = \frac{1}{4}A^2 – \frac{6}{4}A + \frac{9}{4}I$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6&-5&5\\-5&6&-5\\5&-5&6\end{array}} \right] – \frac{6}{4}\left[ {\begin{array}{*{20}{r}}2&-1&1\\-1&2&-1\\1&-1&2\end{array}} \right] + \frac{9}{4}\left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6-12+9&-5+6+0&5-6+0\\-5+6+0&6-12+9&-5+6+0\\5-6+0&-5+6+0&6-12+9\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}3&1&-1\\1&3&-1\\-1&-1&3\end{array}} \right]$

10. Let A be a non-singular square matrix of order $3 \times 3$. Then, $|adj A|$ is equal to:

Solution:

(B) $|A|^2$
For any n × n matrix A, $\det(adj A) = |A|^{n-1}$ (It holds for singular and non-singular matrices.)

11. If A is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to:

Solution:

(B) $\frac{1}{\det(A)}$
When A is an invertible matrix of order 2, $AA^{-1} = I_2 = A^{-1}A$, where $I_2$ is the identity matrix of order 2.
$\Rightarrow \det(AA^{-1}) = \det I \Rightarrow \det A \cdot \det(A^{-1}) = 1$
$\Rightarrow \det(A^{-1}) = \frac{1}{\det A}, \det A \neq 0$

Exercise – 4.6

1. Examine the consistency of the system of equations: $x + 2y = 2, 2x + 3y = 3$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&2\\2&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}2\\3\end{array}} \right]$
Now, $|A| = 3 – 4 = -1 \neq 0$
Hence, the system of equations is consistent.

2. Examine the consistency of the system of equations: $2x – y = 5, x + y = 4$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-1\\1&1\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\4\end{array}} \right]$
Now, $|A| = 2 + 1 = 3 \neq 0$
Hence, the system of equations is consistent.

3. Examine the consistency of the system of equations: $x + 3y = 5, 2x + 6y = 8$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&3\\2&6\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\8\end{array}} \right]$
Now, $|A| = 6 – 6 = 0$
Hence, A is a singular matrix. So, we calculate $(adj A)B$.
$adj A = \left[ {\begin{array}{*{20}{r}}6&-3\\-2&1\end{array}} \right]$
Now, $(adj A)B = \left[ {\begin{array}{*{20}{r}}6&-3\\-2&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5\\8\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}30-24\\-10+8\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}6\\-2\end{array}} \right] \neq \left[ {\begin{array}{*{20}{r}}0\\0\end{array}} \right]$
Hence, the equations are inconsistent with no solution.

4. Examine the consistency of the system of equations: $x + y + z = 1, 2x + 3y + 2z = 2, ax + ay + 2az = 4$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&1&1\\2&3&2\\a&a&2a\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}1\\2\\4\end{array}} \right]$
Now, $|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 3a = a \neq 0$
Two conditions arise:
I: If $a \neq 0$, then $|A| \neq 0$, hence the system of equations is consistent and has a unique solution.
II: If $a = 0$, then $|A| = 0$. So, we need to calculate $adj A$. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}3&2\\a&2a\end{array}} \right| = 4a$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&2\\a&2a\end{array}} \right| = 2a$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&3\\a&a\end{array}} \right| = -a$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}1&1\\a&2a\end{array}} \right| = -a$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&1\\a&2a\end{array}} \right| = a$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&1\\a&a\end{array}} \right| = 0$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}1&1\\3&2\end{array}} \right| = -1$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&1\\2&2\end{array}} \right| = 0$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&1\\2&3\end{array}} \right| = 1$
Now, $(adj A) \cdot B = \left[ {\begin{array}{*{20}{r}}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1\\2\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2a-4\\0\\-a+4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-4\\0\\4\end{array}} \right] \neq 0$
Hence, the equations are inconsistent with no solution because if $a = 0$, then the third system of equations is not possible.

5. Examine the consistency of the system of equations: $3x – y – 2z = 2, 2y – z = -1, 3x – 5y = 3$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}3&-1&-2\\0&2&-1\\3&-5&0\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}2\\-1\\3\end{array}} \right]$
Now, $|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = -15 + 3 + 12 = 0$
Here, A is a singular matrix, so we will compute $(adj A)B$.
For $adj A$, cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}2&-1\\-5&0\end{array}} \right| = -5$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}0&-1\\3&0\end{array}} \right| = -3$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}0&2\\3&-5\end{array}} \right| = -6$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&-2\\-5&0\end{array}} \right| = 10$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}3&-2\\3&0\end{array}} \right| = 6$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}3&-1\\3&-5\end{array}} \right| = 12$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&-2\\2&-1\end{array}} \right| = 5$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}3&-2\\0&-1\end{array}} \right| = 3$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}3&-1\\0&2\end{array}} \right| = 6$
Now, $(adj A) \cdot B = \left[ {\begin{array}{*{20}{r}}-5&10&5\\-3&6&3\\-6&12&6\end{array}} \right]\left[ {\begin{array}{*{20}{r}}2\\-1\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-10-10+15\\-6-6+9\\-12-12+18\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-5\\-3\\-6\end{array}} \right] \neq 0$
Hence, the system of equations is inconsistent with no solution.

6. Examine the consistency of the system of equations: $5x – y + 4z = 5, 2x + 3y + 5z = 2, 5x – 2y + 6z = -1$.

Solution:

The system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&-1&4\\2&3&5\\5&-2&6\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\2\\-1\end{array}} \right]$
Now, $|A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15) = 140 – 13 – 76 = 51 \neq 0$
Hence, the equations are consistent with a unique solution.

7. Solve the system of linear equations using the matrix method: $5x + 2y = 4, 7x + 3y = 5$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&2\\7&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}4\\5\end{array}} \right]$
Now, $|A| = 15 – 14 = 1 \neq 0$
$\Rightarrow$ A is non-singular and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(3) = 3$
$A_{12} = (-1)^{1+2}(7) = -7$
$A_{21} = (-1)^{2+1}(2) = -2$
$A_{22} = (-1)^{2+2}(5) = 5$
$adj A = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&-2\\-7&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}4\\5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}12-10\\-28+25\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\-3\end{array}} \right]$
Hence, $x = 2, y = -3$.

8. Solve the system of linear equations using the matrix method: $2x – y = -2, 3x + 4y = 3$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-1\\3&4\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}-2\\3\end{array}} \right]$
Now, $|A| = 8 + 3 = 11 \neq 0$
$\Rightarrow$ A is non-singular and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(4) = 4$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(-1) = 1$
$A_{22} = (-1)^{2+2}(2) = 2$
$adj A = \left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}4&1\\-3&2\end{array}} \right]\left[ {\begin{array}{*{20}{r}}-2\\3\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-8+3\\6+6\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5\\12\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-\frac{5}{11}\\\frac{12}{11}\end{array}} \right]$
Hence, $x = -\frac{5}{11}, y = \frac{12}{11}$.

9. Solve the system of linear equations using the matrix method: $4x – 3y = 3, 3x – 5y = 7$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}4&-3\\3&-5\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}3\\7\end{array}} \right]$
Now, $|A| = -20 + 9 = -11 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(-5) = -5$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(-3) = 3$
$A_{22} = (-1)^{2+2}(4) = 4$
$adj A = \left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-5&-3\\3&4\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3\\7\end{array}} \right] = -\frac{1}{11}\left[ {\begin{array}{*{20}{r}}-15+21\\-9+28\end{array}} \right] = \frac{1}{11}\left[ {\begin{array}{*{20}{r}}-6\\-19\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-\frac{6}{11}\\-\frac{19}{11}\end{array}} \right]$
Hence, $x = -\frac{6}{11}, y = -\frac{19}{11}$.

10. Solve the system of linear equations using the matrix method: $5x + 2y = 3, 3x + 2y = 5$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}5&2\\3&2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}3\\5\end{array}} \right]$
Now, $|A| = 10 – 6 = 4 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution. Cofactors of elements of A are given by:
$A_{11} = (-1)^{1+1}(2) = 2$
$A_{12} = (-1)^{1+2}(3) = -3$
$A_{21} = (-1)^{2+1}(2) = -2$
$A_{22} = (-1)^{2+2}(5) = 5$
$adj A = \left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}2&-2\\-3&5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}3\\5\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}6-10\\-9+25\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}-4\\16\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}-1\\4\end{array}} \right]$
Hence, $x = -1, y = 4$.

11. Solve the system of linear equations using the matrix method: $2x + y + z = 1, x – 2y – z = \frac{3}{2}, 3y – 5z = 9$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&1&1\\1&-2&-1\\0&3&-5\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}1\\\frac{3}{2}\\9\end{array}} \right]$
Now, $|A| = 26 + 8 = 34 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}-2&-1\\3&-5\end{array}} \right| = 13$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}1&-1\\0&-5\end{array}} \right| = 5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1&-2\\0&3\end{array}} \right| = 3$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}1&1\\3&-5\end{array}} \right| = 8$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&1\\0&-5\end{array}} \right| = -10$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&1\\0&3\end{array}} \right| = -6$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}1&1\\-2&-1\end{array}} \right| = 1$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&1\\1&-1\end{array}} \right| = 3$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&1\\1&-2\end{array}} \right| = -5$
$adj A = \left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13&8&1\\5&-10&3\\3&-6&-5\end{array}} \right]\left[ {\begin{array}{*{20}{r}}1\\\frac{3}{2}\\9\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}13+12+9\\5-15+27\\3-9-45\end{array}} \right] = \frac{1}{34}\left[ {\begin{array}{*{20}{r}}34\\17\\-51\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\\frac{1}{2}\\-\frac{3}{2}\end{array}} \right]$
Hence, $x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$.

12. Solve the system of linear equations using the matrix method: $x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&-1&1\\2&1&-3\\1&1&1\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}4\\0\\2\end{array}} \right]$
Now, $|A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10 \neq 0$
$\Rightarrow$ A is non-singular and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}1&-3\\1&1\end{array}} \right| = 4$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&-3\\1&1\end{array}} \right| = -5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&1\\1&1\end{array}} \right| = 1$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&1\\1&1\end{array}} \right| = 2$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&1\\1&1\end{array}} \right| = 0$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&-1\\1&1\end{array}} \right| = -2$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&1\\1&-3\end{array}} \right| = 2$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&1\\2&-3\end{array}} \right| = 5$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&-1\\2&1\end{array}} \right| = 3$
$adj A = \left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}4&2&2\\-5&0&5\\1&-2&3\end{array}} \right]\left[ {\begin{array}{*{20}{r}}4\\0\\2\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}16+0+4\\-20+0+10\\4-0+6\end{array}} \right] = \frac{1}{10}\left[ {\begin{array}{*{20}{r}}20\\-10\\10\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\-1\\1\end{array}} \right]$
Hence, $x = 2, y = -1, z = 1$.

13. Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&3&3\\1&-2&1\\3&-1&-2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}5\\-4\\3\end{array}} \right]$
Now, $|A| = 40 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}-2&1\\-1&-2\end{array}} \right| = 5$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}1&1\\3&-2\end{array}} \right| = 5$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1&-2\\3&-1\end{array}} \right| = 5$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3&3\\-1&-2\end{array}} \right| = 3$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&3\\3&-2\end{array}} \right| = -13$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&3\\3&-1\end{array}} \right| = 11$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}3&3\\-2&1\end{array}} \right| = 9$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&3\\1&1\end{array}} \right| = 1$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&3\\1&-2\end{array}} \right| = -7$
$adj A = \left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}5&3&9\\5&-13&1\\5&11&-7\end{array}} \right]\left[ {\begin{array}{*{20}{r}}5\\-4\\3\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}25-12+27\\25+52+3\\25-44-21\end{array}} \right] = \frac{1}{40}\left[ {\begin{array}{*{20}{r}}40\\80\\-40\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\2\\-1\end{array}} \right]$
Hence, $x = 1, y = 2, z = -1$.

14. Solve the system of linear equations using the matrix method: $x – y + 2z = 7, 3x + 4y – 5z = -5, 2x – y + 3z = 12$.

Solution:

The given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}1&-1&2\\3&4&-5\\2&-1&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}7\\-5\\12\end{array}} \right]$
Now, $|A| = 4 \neq 0$
$\Rightarrow$ A is a non-singular matrix and so the given equations have a unique solution. Here, cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}4&-5\\-1&3\end{array}} \right| = 7$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}3&-5\\2&3\end{array}} \right| = -19$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}3&4\\2&-1\end{array}} \right| = -11$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-1&2\\-1&3\end{array}} \right| = 1$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}1&2\\2&3\end{array}} \right| = -1$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1&-1\\2&-1\end{array}} \right| = -1$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-1&2\\4&-5\end{array}} \right| = -3$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}1&2\\3&-5\end{array}} \right| = 11$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1&-1\\3&4\end{array}} \right| = 7$
$adj A = \left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]$
and $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]$
Solution of the given system is given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}} \right]\left[ {\begin{array}{*{20}{r}}7\\-5\\12\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}49-5-36\\-133+5+132\\-77+5+84\end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{r}}8\\4\\12\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}2\\1\\3\end{array}} \right]$
Hence, $x = 2, y = 1, z = 3$.

15. If $A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$, find $A^{-1}$. Using $A^{-1}$, solve the system of equations $2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3$.

Solution:

We have, $A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$
$\therefore |A| = -1 \neq 0$
So, A is invertible. Cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}2&-4\\1&-2\end{array}} \right| = 0$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}3&-4\\1&-2\end{array}} \right| = 2$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}3&2\\1&1\end{array}} \right| = 1$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}-3&5\\1&-2\end{array}} \right| = -1$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}2&5\\1&-2\end{array}} \right| = -9$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}2&-3\\1&1\end{array}} \right| = -5$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}-3&5\\2&-4\end{array}} \right| = 2$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}2&5\\3&-4\end{array}} \right| = 23$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}2&-3\\3&2\end{array}} \right| = 13$
$adj A = \left[ {\begin{array}{*{20}{r}}0&-1&2\\2&-9&23\\1&-5&13\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = -\left[ {\begin{array}{*{20}{r}}0&-1&2\\2&-9&23\\1&-5&13\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&-2\\-2&9&-23\\-1&5&-13\end{array}} \right]$
Now, the given system of equations can be written in the form $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}2&-3&5\\3&2&-4\\1&1&-2\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}11\\-5\\-3\end{array}} \right]$
As, $|A| = -1 \neq 0$, so the given system of equations has a unique solution given by $X = A^{-1}B$.
$\Rightarrow \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&-2\\-2&9&-23\\-1&5&-13\end{array}} \right]\left[ {\begin{array}{*{20}{r}}11\\-5\\-3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0-5+6\\-22-45+69\\-11-25+39\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1\\2\\3\end{array}} \right]$
Hence, $x = 1, y = 2, z = 3$.

16. The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat, and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat, and 3 kg rice is Rs. 70. Find the cost of each item per kg by the matrix method.

Solution:

Let the cost of 1 kg onion = Rs. $x$, cost of 1 kg wheat = Rs. $y$, and cost of 1 kg rice = Rs. $z$.
According to the question, we have:
$4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$
This system can be written as $AX = B$, where
$A = \left[ {\begin{array}{*{20}{r}}4&3&2\\2&4&6\\6&2&3\end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{r}}x\\y\\z\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{r}}60\\90\\70\end{array}} \right]$
Now, $|A| = 50 \neq 0$
$\therefore$ A is a non-singular matrix and the system has a unique solution. Cofactors of elements of A are:
$A_{11} = (-1)^{1+1}\left| {\begin{array}{*{20}{r}}4&6\\2&3\end{array}} \right| = 0$
$A_{12} = (-1)^{1+2}\left| {\begin{array}{*{20}{r}}2&6\\6&3\end{array}} \right| = 30$
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}2&4\\6&2\end{array}} \right| = -20$
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3&2\\2&3\end{array}} \right| = -5$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}4&2\\6&3\end{array}} \right| = 0$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}4&3\\6&2\end{array}} \right| = 10$
$A_{31} = (-1)^{3+1}\left| {\begin{array}{*{20}{r}}3&2\\4&6\end{array}} \right| = 10$
$A_{32} = (-1)^{3+2}\left| {\begin{array}{*{20}{r}}4&2\\2&6\end{array}} \right| = -20$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}4&3\\2&4\end{array}} \right| = 10$
$adj A = \left[ {\begin{array}{*{20}{r}}0&-5&10\\30&0&-20\\-20&10&10\end{array}} \right]$
Hence, $A^{-1} = \frac{1}{|A|} adj A = \frac{1}{50}\left[ {\begin{array}{*{20}{r}}0&-5&10\\30&0&-20\\-20&10&10\end{array}} \right]$

As, \( AX = B \Rightarrow X = A^{-1}B \).

\(\Rightarrow \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \frac{1}{50} \left[ \begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array} \right] \left[ \begin{array}{r} 60 \\ 90 \\ 70 \end{array} \right] \).

\(= \frac{1}{5} \left[ \begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array} \right] \left[ \begin{array}{r} 6 \\ 9 \\ 7 \end{array} \right] \).

\(= \frac{1}{5} \left[ \begin{array}{r} 0 – 45 + 70 \\ 180 + 0 – 140 \\ -120 + 90 + 70 \end{array} \right] \).

\(= \frac{1}{5} \left[ \begin{array}{r} 25 \\ 40 \\ 40 \end{array} \right] = \left[ \begin{array}{r} 5 \\ 8 \\ 8 \end{array} \right] \).

Hence, the cost of 1 kg of onion = Rs. 5,

cost of 1 kg of wheat = Rs. 8,

and cost of 1 kg of rice = Rs. 8.

NCERT – Miscellaneous Exercise

1. Prove that the determinant \( \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} \) is independent of \( \theta \).

Solution:

Let \( \Delta = \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} \).

Expanding by \( R_1 \), we get:

\( = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} – \sin \theta \begin{vmatrix} -\sin \theta & 1 \\ \cos \theta & x \end{vmatrix} + \cos \theta \begin{vmatrix} -\sin \theta & -x \\ \cos \theta & 1 \end{vmatrix} \).

\( = x(-x^2 – 1) – \sin \theta(-x \sin \theta – \cos \theta) + \cos \theta(-\sin \theta + x \cos \theta) \).

\( = -x^3 – x + x \sin^2 \theta + \sin \theta \cos \theta – \sin \theta \cos \theta + x \cos^2 \theta \).

\( = -x^3 – x + x(\sin^2 \theta + \cos^2 \theta) = -x^3 – x + x(1) = -x^3 \),

which is independent of \( \theta \).

2. Without expanding the determinant, prove that:

\( \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} = \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix} \).

Solution:

Let \( \Delta = \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} \).

Multiplying \( R_1, R_2 \) and \( R_3 \) by \( a, b \) and \( c \) respectively and dividing the determinant by \( abc \), we get:

\( \Delta = \frac{1}{abc} \begin{vmatrix} a^2 & a^3 & abc \\ b^2 & b^3 & abc \\ c^2 & c^3 & abc \end{vmatrix} \).

Taking \( abc \) common from \( C_3 \), we get:

\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix} = – \begin{vmatrix} a^2 & 1 & a^3 \\ b^2 & 1 & b^3 \\ c^2 & 1 & c^3 \end{vmatrix} \).

Interchanging \( C_1 \leftrightarrow C_2 \), we get:

\( \Delta = \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix} \).

Hence, \( \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} = \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix} \).

3. Evaluate \( \begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} \).

Solution:

\( \begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} \).

Expanding along \( R_1 \), we get:

\( = \cos \alpha \cos \beta \begin{vmatrix} \cos \beta & 0 \\ \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} – \cos \alpha \sin \beta \begin{vmatrix} -\sin \beta & 0 \\ \sin \alpha \cos \beta & \cos \alpha \end{vmatrix} – \sin \alpha \begin{vmatrix} -\sin \beta & \cos \beta \\ \sin \alpha \sin \beta & \sin \alpha \sin \beta \end{vmatrix} \).

\( = \cos \alpha \cos \beta (\cos \beta \cos \alpha) – \cos \alpha \sin \beta (-\sin \beta \cos \alpha) – \sin \alpha (-\sin \alpha \sin^2 \beta – \sin \alpha \cos^2 \beta) \).

\( = \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + \sin^2 \alpha \sin^2 \beta + \sin^2 \alpha \cos^2 \beta \).

\( = \cos^2 \alpha (\cos^2 \beta + \sin^2 \beta) + \sin^2 \alpha (\sin^2 \beta + \cos^2 \beta) \).

\( = \cos^2 \alpha (1) + \sin^2 \alpha (1) = \cos^2 \alpha + \sin^2 \alpha = 1 \).

4. If \( a, b \) and \( c \) are real numbers, and \( \Delta = \begin{vmatrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} = 0 \), show that either \( a + b + c = 0 \) or \( a = b = c \).

Solution:

\( \Delta = \begin{vmatrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:

\( \Delta = \begin{vmatrix} 2(a + b + c) & c + a & a + b \\ 2(a + b + c) & a + b & b + c \\ 2(a + b + c) & b + c & c + a \end{vmatrix} \).

Taking \( 2(a + b + c) \) common from \( C_1 \), we get:

\( \Delta = 2(a + b + c) \begin{vmatrix} 1 & c + a & a + b \\ 1 & a + b & b + c \\ 1 & b + c & c + a \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_3 \) and \( R_3 \to R_3 – R_1 \), we get:

\( \Delta = 2(a + b + c) \begin{vmatrix} 1 & c + a & a + b \\ 0 & a – c & b – a \\ 0 & b – a & c – b \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( \Delta = 2(a + b + c) \left[ (a – c)(c – b) – (b – a)(b – a) \right] \).

\( = 2(a + b + c) \left[ ac – ab – c^2 + cb – (b^2 + a^2 – 2ab) \right] \).

\( = 2(a + b + c) \left[ ac – ab – c^2 + cb – b^2 – a^2 + 2ab \right] \).

\( = -2(a + b + c) \left[ a^2 + b^2 + c^2 – ab – bc – ca \right] \).

\( = -(a + b + c) \left[ (a – b)^2 + (b – c)^2 + (c – a)^2 \right] \).

When \( \Delta = 0 \),

\( -(a + b + c) \left[ (a – b)^2 + (b – c)^2 + (c – a)^2 \right] = 0 \).

\( \Rightarrow a + b + c = 0 \) or \( (a – b)^2, (b – c)^2, (c – a)^2 = 0 \).

\( \Rightarrow a + b + c = 0 \) or \( a = b, b = c, c = a \).

\( \Rightarrow a + b + c = 0 \) or \( a = b = c \).

5. Solve the equation \( \begin{vmatrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{vmatrix} = 0, a \neq 0 \).

Solution:

\( \begin{vmatrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{vmatrix} = 0 \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:

\( \begin{vmatrix} 3x + a & x & x \\ 3x + a & x + a & x \\ 3x + a & x & x + a \end{vmatrix} = 0 \).

\( \Rightarrow (3x + a) \begin{vmatrix} 1 & x & x \\ 1 & x + a & x \\ 1 & x & x + a \end{vmatrix} = 0 \).

Applying \( R_2 \to R_2 – R_1, R_3 \to R_3 – R_1 \), we get:

\( (3x + a) \begin{vmatrix} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = 0 \).

Applying \( C_2 \to C_2 – C_3 \), we get:

\( (3x + a) \begin{vmatrix} 1 & 0 & x \\ 0 & a & 0 \\ 0 & -a & a \end{vmatrix} = 0 \).

Expanding along \( R_1 \), we get:

\( (3x + a)(a^2) = 0 \).

Since \( a \neq 0 \Rightarrow 3x + a = 0 \Rightarrow x = -\frac{a}{3} \).

6. Prove that \( \begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + bc & b^2 & ac \\ ab & b^2 + bc & c^2 \end{vmatrix} = 4a^2b^2c^2 \).

Solution:

L.H.S.: \( \begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + bc & b^2 & ac \\ ab & b^2 + bc & c^2 \end{vmatrix} \).

Taking \( a, b, c \) common from \( C_1, C_2 \) and \( C_3 \), we get:

\( abc \begin{vmatrix} a & c & a + c \\ a + b & b & a \\ b & b + c & c \end{vmatrix} \).

Applying \( C_3 \to C_3 – (C_1 + C_2) \), we get:

\( abc \begin{vmatrix} a & c & 0 \\ a + b & b & -2b \\ b & b + c & -2b \end{vmatrix} \).

Taking \( -2b \) common from \( C_3 \), we get:

\( -2b(abc) \begin{vmatrix} a & c & 0 \\ a + b & b & 1 \\ b & b + c & 1 \end{vmatrix} \).

Expanding the determinant along \( R_1 \), we get:

\( -2b(abc) \left[ a(b – b – c) – c(a + b – c) \right] \).

\( = -2b(abc) \left[ -ac – ac \right] = 4a^2b^2c^2 = \) R.H.S.

7. If \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \), find \( (AB)^{-1} \).

Solution:

Given, \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \).

Now, \( |B| = 1(3) – 2(-1) – 2(2) = 3 + 2 – 4 = 1 \neq 0 \).

\( \therefore B^{-1} \) exists. Since \( (AB)^{-1} = B^{-1}A^{-1} \), we need to calculate \( B^{-1} \).

Cofactors of elements of B are:

\( B_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} = 3 \),

\( B_{12} = (-1)^{1+2} \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \),

\( B_{13} = (-1)^{1+3} \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = 2 \),

\( B_{21} = (-1)^{2+1} \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} = 2 \),

\( B_{22} = (-1)^{2+2} \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1 \),

\( B_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix} = 2 \),

\( B_{31} = (-1)^{3+1} \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} = 6 \),

\( B_{32} = (-1)^{3+2} \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = 2 \),

\( B_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 5 \).

\( \therefore adj B = \begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{bmatrix}^T = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).

Hence, \( B^{-1} = \frac{1}{|B|}(adj B) = \frac{1}{1} \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).

Now, \( B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \).

\( = \begin{bmatrix} 9 – 30 + 30 & -3 + 12 – 12 & 3 – 10 + 12 \\ 3 – 15 + 10 & -1 + 6 – 4 & 1 – 5 + 4 \\ 6 – 30 + 25 & -2 + 12 – 10 & 2 – 10 + 10 \end{bmatrix} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} \).

8. Let \( A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} \). Verify that:

(i) \( (adj A)^{-1} = adj(A^{-1}) \).

(ii) \( (A^{-1})^{-1} = A \).

Solution:

\( A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} \).

Now, \( |A| = 1(15 – 1) + 2(-10 – 1) + 1(-2 – 3) \).

\( = 14 – 22 – 5 = -13 \neq 0 \).

\( \therefore A^{-1} \) exists.

Cofactors of elements of A are:

\( A_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = 15 – 1 = 14 \),

\( A_{12} = (-1)^{1+2} \begin{vmatrix} -2 & 1 \\ 1 & 5 \end{vmatrix} = -(-10 – 1) = 11 \),

\( A_{13} = (-1)^{1+3} \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} = -3 – 2 = -5 \),

\( A_{21} = (-1)^{2+1} \begin{vmatrix} -2 & 1 \\ 1 & 5 \end{vmatrix} = -(-10 – 1) = 11 \),

\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 1 & 5 \end{vmatrix} = 5 – 1 = 4 \),

\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = -(1 + 2) = -3 \),

\( A_{31} = (-1)^{3+1} \begin{vmatrix} -2 & 1 \\ 3 & 1 \end{vmatrix} = -2 – 3 = -5 \),

\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -(1 + 2) = -3 \),

\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} = 3 – 4 = -1 \).

\( \therefore adj A = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}^T = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} \).

So, \( A^{-1} = \frac{1}{|A|}(adj A) = \frac{1}{-13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{bmatrix} \).

(i) Now, \( |adj A| = \begin{vmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{vmatrix} \).

\( = 14(-4 – 9) – 11(-11 – 15) – 5(-33 + 20) \).

\( = -182 + 286 + 65 = 269 \neq 0 \).

\( \therefore adj A \) is invertible and \( (adj A)^{-1} = \frac{1}{|adj A|} \{ adj(adj A) \} \).

Now, to obtain \( adj(adj A) \), the cofactors of adj A are:

\( adj A_{11} = (-1)^{1+1} \begin{vmatrix} 4 & -3 \\ -3 & -1 \end{vmatrix} = -4 – 9 = -13 \),

\( adj A_{12} = (-1)^{1+2} \begin{vmatrix} 11 & -3 \\ -5 & -1 \end{vmatrix} = -(-11 – 15) = 26 \),

\( adj A_{13} = (-1)^{1+3} \begin{vmatrix} 11 & 4 \\ -5 & -3 \end{vmatrix} = -33 + 20 = -13 \),

\( adj A_{21} = (-1)^{2+1} \begin{vmatrix} 11 & -5 \\ -3 & -1 \end{vmatrix} = -(-11 – 15) = 26 \),

\( adj A_{22} = (-1)^{2+2} \begin{vmatrix} 14 & -5 \\ -5 & -1 \end{vmatrix} = -14 – 25 = -39 \),

\( adj A_{23} = (-1)^{2+3} \begin{vmatrix} 14 & 11 \\ -5 & -3 \end{vmatrix} = -(-42 + 55) = -13 \),

\( adj A_{31} = (-1)^{3+1} \begin{vmatrix} 11 & -5 \\ 4 & -3 \end{vmatrix} = -33 + 20 = -13 \),

\( adj A_{32} = (-1)^{3+2} \begin{vmatrix} 14 & -5 \\ 11 & -3 \end{vmatrix} = -(-42 + 55) = -13 \),

\( adj A_{33} = (-1)^{3+3} \begin{vmatrix} 14 & 11 \\ 11 & 4 \end{vmatrix} = 56 – 121 = -65 \).

\( \therefore adj(adj A) = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix}^T = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} \).

So, \( (adj A)^{-1} = \frac{1}{|adj A|} \{ adj(adj A) \} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} \).

\( = \frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix} \).

Also, \( adj(A^{-1}) = adj \left( \frac{1}{13} \begin{bmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{bmatrix} \right) \).

Similarly, \( adj \begin{bmatrix} -\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{bmatrix} = \begin{bmatrix} -\frac{13}{169} & \frac{26}{169} & -\frac{13}{169} \\ \frac{26}{169} & -\frac{39}{169} & -\frac{13}{169} \\ -\frac{13}{169} & -\frac{13}{169} & -\frac{65}{169} \end{bmatrix} \).

\( \Rightarrow adj(A^{-1}) = \frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix} \).

From (1) and (2), we find that \( adj(A^{-1}) = (adj A)^{-1} \).

(ii) \( |A^{-1}| = \left| \frac{1}{13} \begin{bmatrix} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{bmatrix} \right| \).

\( = \frac{1}{13^2} \{ -14(-4 – 9) + 11(-11 – 15) + 5(-33 + 20) \} \).

\( = \frac{1}{13 \times 13 \times 13} (-169) = -\frac{1}{13} \neq 0 \).

\( \therefore (A^{-1})^{-1} \) exists and \( (A^{-1})^{-1} = \frac{1}{|A^{-1}|} \{ adj(A^{-1}) \} \).

\( = \frac{1}{-\frac{1}{13}} \cdot \frac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{bmatrix} = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} = A \).

9. Evaluate \( \begin{vmatrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{vmatrix} \).

Solution:

\( \begin{vmatrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:

\( \begin{vmatrix} 2(x + y) & y & x + y \\ 2(x + y) & x + y & x \\ 2(x + y) & x & y \end{vmatrix} \).

Taking \( 2(x + y) \) common from \( C_1 \), we get:

\( 2(x + y) \begin{vmatrix} 1 & y & x + y \\ 1 & x + y & x \\ 1 & x & y \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_1 \), we get:

\( 2(x + y) \begin{vmatrix} 1 & y & x + y \\ 0 & x & -y \\ 0 & x – y & -x \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( 2(x + y) \left[ x(-x) – (-y)(x – y) \right] = 2(x + y) \left[ -x^2 + xy – y^2 \right] \).

\( = -2(x + y)(x^2 – xy + y^2) = -2(x^3 + y^3) \).

10. Evaluate \( \begin{vmatrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{vmatrix} \).

Solution:

Let \( \Delta = \begin{vmatrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_1 \), we get:

\( \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( 1 \cdot y \cdot x = xy \).

Using properties of determinants in questions 11 to 15, prove that:

11. \( \begin{vmatrix} \alpha & \alpha^2 & \beta + \gamma \\ \beta & \beta^2 & \gamma + \alpha \\ \gamma & \gamma^2 & \alpha + \beta \end{vmatrix} = (\beta – \gamma)(\gamma – \alpha)(\alpha – \beta)(\alpha + \beta + \gamma) \).

Solution:

L.H.S. \( = \begin{vmatrix} \alpha & \alpha^2 & \beta + \gamma \\ \beta & \beta^2 & \gamma + \alpha \\ \gamma & \gamma^2 & \alpha + \beta \end{vmatrix} \).

Applying \( C_3 \to C_3 + C_1 \), we get:

\( \begin{vmatrix} \alpha & \alpha^2 & \alpha + \beta + \gamma \\ \beta & \beta^2 & \alpha + \beta + \gamma \\ \gamma & \gamma^2 & \alpha + \beta + \gamma \end{vmatrix} \).

Taking \( (\alpha + \beta + \gamma) \) common from \( C_3 \), we get:

\( (\alpha + \beta + \gamma) \begin{vmatrix} \alpha & \alpha^2 & 1 \\ \beta & \beta^2 & 1 \\ \gamma & \gamma^2 & 1 \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_2 \), we get:

\( (\alpha + \beta + \gamma) \begin{vmatrix} \alpha & \alpha^2 & 1 \\ \beta – \alpha & \beta^2 – \alpha^2 & 0 \\ \gamma – \beta & \gamma^2 – \beta^2 & 0 \end{vmatrix} \).

Expanding along \( C_3 \), we get:

\( (\alpha + \beta + \gamma) \begin{vmatrix} \beta – \alpha & \beta^2 – \alpha^2 \\ \gamma – \beta & \gamma^2 – \beta^2 \end{vmatrix} \).

Taking \( (\beta – \alpha) \) and \( (\gamma – \beta) \) common from \( R_1 \) and \( R_2 \) respectively, we get:

\( (\alpha + \beta + \gamma)(\beta – \alpha)(\gamma – \beta) \begin{vmatrix} 1 & \beta + \alpha \\ 1 & \gamma + \beta \end{vmatrix} \).

\( = (\alpha + \beta + \gamma)(\alpha – \beta)(\beta – \gamma)(\gamma – \alpha) = \) R.H.S.

12. \( \begin{vmatrix} x & x^2 & 1 + p x^3 \\ y & y^2 & 1 + p y^3 \\ z & z^2 & 1 + p z^3 \end{vmatrix} = (1 + p x y z)(x – y)(y – z)(z – x) \).

Solution:

Let \( \Delta = \begin{vmatrix} x & x^2 & 1 + p x^3 \\ y & y^2 & 1 + p y^3 \\ z & z^2 & 1 + p z^3 \end{vmatrix} \).

Using property 5, we get:

\( \Delta = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & p x^3 \\ y & y^2 & p y^3 \\ z & z^2 & p z^3 \end{vmatrix} = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + p x y z \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).

Taking \( x, y, z \) and \( p \) common from \( R_1, R_2, R_3 \) and \( C_3 \) in determinant II.

Interchanging \( C_2 \leftrightarrow C_3 \) in determinant I.

\( \Delta = – \begin{vmatrix} x & 1 & x^2 \\ y & 1 & y^2 \\ z & 1 & z^2 \end{vmatrix} + p x y z \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).

Interchanging \( C_1 \leftrightarrow C_2 \) in determinant I.

\( \Delta = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + p x y z \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (1 + p x y z) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_1 \), we get:

\( \Delta = (1 + p x y z) \begin{vmatrix} 1 & x & x^2 \\ 0 & y – x & y^2 – x^2 \\ 0 & z – x & z^2 – x^2 \end{vmatrix} \).

Taking \( (y – x) \) and \( (z – x) \) common from \( R_2 \) and \( R_3 \), we get:

\( \Delta = (1 + p x y z)(y – x)(z – x) \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & x + y \\ 0 & 1 & z + x \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_3 \), we get:

\( \Delta = (1 + p x y z)(y – x)(z – x) \begin{vmatrix} 1 & x & x^2 \\ 0 & 0 & y – z \\ 0 & 1 & z + x \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( \Delta = (1 + p x y z)(y – x)(z – x) \{ 0 – (y – z) \} \).

\( = (1 + p x y z)(x – y)(y – z)(z – x) \).

13. \( \begin{vmatrix} 3a & -a + b & -a + c \\ -b + a & 3b & -b + c \\ -c + a & -c + b & 3c \end{vmatrix} = 3(a + b + c)(ab + bc + ca) \).

Solution:

Let \( \Delta = \begin{vmatrix} 3a & -a + b & -a + c \\ -b + a & 3b & -b + c \\ -c + a & -c + b & 3c \end{vmatrix} \).

Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:

\( \begin{vmatrix} a + b + c & -a + b & -a + c \\ a + b + c & 3b & -b + c \\ a + b + c & -c + b & 3c \end{vmatrix} \).

Taking \( (a + b + c) \) common from \( C_1 \), we get:

\( (a + b + c) \begin{vmatrix} 1 & -a + b & -a + c \\ 1 & 3b & -b + c \\ 1 & -c + b & 3c \end{vmatrix} \).

Applying \( R_2 \to R_2 – R_1 \) and \( R_3 \to R_3 – R_1 \), we get:

\( (a + b + c) \begin{vmatrix} 1 & -a + b & -a + c \\ 0 & 2b + a & a – b \\ 0 & a – c & 2c + a \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( (a + b + c) \left[ (2b + a)(2c + a) – (a – b)(a – c) \right] \).

\( = (a + b + c)(4bc + 2ab + 2ac + a^2 – a^2 + ab + ac – bc) \).

\( = (a + b + c)(3ab + 3bc + 3ca) = 3(a + b + c)(ab + bc + ca) \).

14. \( \begin{vmatrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix} = 1 \).

Solution:

L.H.S. \( = \begin{vmatrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix} \).

Applying \( R_2 \to R_2 – 2R_1 \) and \( R_3 \to R_3 – 3R_1 \), we get:

\( \begin{vmatrix} 1 & 1 + p & 1 + p + q \\ 0 & 1 & 2 + p \\ 0 & 3 & 7 + 3p \end{vmatrix} \).

Expanding along \( C_1 \), we get:

\( = 1(7 + 3p – 6 – 3p) = 1(1) = 1 = \) R.H.S.

15. \( \begin{vmatrix} \sin \alpha & \cos \alpha & \cos(\alpha + \delta) \\ \sin \beta & \cos \beta & \cos(\beta + \delta) \\ \sin \gamma & \cos \gamma & \cos(\gamma + \delta) \end{vmatrix} = 0 \).

Solution:

Let \( \Delta = \begin{vmatrix} \sin \alpha & \cos \alpha & \cos(\alpha + \delta) \\ \sin \beta & \cos \beta & \cos(\beta + \delta) \\ \sin \gamma & \cos \gamma & \cos(\gamma + \delta) \end{vmatrix} \).

Then, using \( \cos(A + B) = \cos A \cos B – \sin A \sin B \), we get:

\( \Delta = \begin{vmatrix} \sin \alpha & \cos \alpha & \cos \alpha \cos \delta – \sin \alpha \sin \delta \\ \sin \beta & \cos \beta & \cos \beta \cos \delta – \sin \beta \sin \delta \\ \sin \gamma & \cos \gamma & \cos \gamma \cos \delta – \sin \gamma \sin \delta \end{vmatrix} \).

Applying \( C_3 \to C_3 + (\sin \delta)C_1 – (\cos \delta)C_2 \), we get:

\( \begin{vmatrix} \sin \alpha & \cos \alpha & 0 \\ \sin \beta & \cos \beta & 0 \\ \sin \gamma & \cos \gamma & 0 \end{vmatrix} = 0 \).

(As \( C_3 = 0 \)).

16. Solve the system of the following equations:

\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \),

\( \frac{4}{x} – \frac{6}{y} + \frac{5}{z} = 1 \),

\( \frac{6}{x} + \frac{9}{y} – \frac{20}{z} = 2 \).

Solution:

The equations can be written in the form \( AX = B \),

where, \( A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \), \( X = \begin{bmatrix} 1/x \\ 1/y \\ 1/z \end{bmatrix} \) and \( B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \).

Now, \( |A| = \begin{vmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{vmatrix} \).

\( = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36) \).

\( = 2(75) – 3(-110) + 10(72) = 150 + 330 + 720 = 1200 \neq 0 \).

\( \therefore A^{-1} \) exists.

Now, let \( A_{ij} \) be the cofactor of the element in \( i^{th} \) row and \( j^{th} \) column. The cofactors are:

\( A_{11} = (-1)^{1+1} \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} = 120 – 45 = 75 \),

\( A_{12} = (-1)^{1+2} \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} = -(-80 – 30) = 110 \),

\( A_{13} = (-1)^{1+3} \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} = 36 + 36 = 72 \),

\( A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 10 \\ 9 & -20 \end{vmatrix} = -(-60 – 90) = 150 \),

\( A_{22} = (-1)^{2+2} \begin{vmatrix} 2 & 10 \\ 6 & -20 \end{vmatrix} = -40 – 60 = -100 \),

\( A_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 3 \\ 6 & 9 \end{vmatrix} = -(18 – 18) = 0 \),

\( A_{31} = (-1)^{3+1} \begin{vmatrix} 3 & 10 \\ -6 & 5 \end{vmatrix} = 15 + 60 = 75 \),

\( A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 10 \\ 4 & 5 \end{vmatrix} = -(10 – 40) = 30 \),

\( A_{33} = (-1)^{3+3} \begin{vmatrix} 2 & 3 \\ 4 & -6 \end{vmatrix} = -12 – 12 = -24 \).

\( \therefore adj A = \begin{bmatrix} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{bmatrix}^T = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \).

Hence, \( A^{-1} = \frac{1}{|A|}(adj A) = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \).

As, \( AX = B \Rightarrow X = A^{-1}B \).

\( \Rightarrow \begin{bmatrix} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \).

\( = \frac{1}{1200} \begin{bmatrix} 300 + 150 + 150 \\ 440 – 100 + 60 \\ 288 + 0 – 48 \end{bmatrix} \).

\( = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix} \).

Thus, \( \frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{3}, \frac{1}{z} = \frac{1}{5} \).

Hence, \( x = 2, y = 3, z = 5 \).

Choose the correct answer in questions 17 to 19.

17. If \( a, b, c \) are in A.P, then the determinant \( \begin{vmatrix} x + 2 & x + 3 & x + 2a \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix} \) is:

1. A. 0
2. B. 1
3. C. x
4. D. 2x

Solution:

(A) \( \begin{vmatrix} x + 2 & x + 3 & x + 2a \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix} \).

Applying \( R_1 \to R_1 – R_2 \), we get:

\( \begin{vmatrix} -1 & -1 & 2a – 2b \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix} \).

Applying \( C_1 \to C_1 – C_2 \), we get:

\( \begin{vmatrix} 0 & -1 & 2a – 2b \\ -1 & x + 4 & x + 2b \\ -1 & x + 5 & x + 2c \end{vmatrix} \).

Expanding along \( R_1 \), we get:

\( 1 \begin{vmatrix} -1 & x + 2b \\ -1 & x + 2c \end{vmatrix} + (2a – 2b) \begin{vmatrix} -1 & x + 4 \\ -1 & x + 5 \end{vmatrix} \).

\( = -x – 2c + x + 2b + (2a – 2b)(-x – 5 + x + 4) \).

\( = 2b – 2c + (2a – 2b)(-1) \).

\( = 2b – 2c – 2a + 2b = 2[2b – (c + a)] \).

\( = 2 \left[ 2 \left( \frac{a + c}{2} \right) – (c + a) \right] = 0 \).

[Since \( a, b, c \) are in A.P.]

18. If \( x, y, z \) are non-zero real numbers, then the inverse of matrix \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \) is:

1. A. \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
2. B. \( x y z \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
3. C. \( \frac{1}{x y z} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \)
4. D. \( \frac{1}{x y z} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Solution:

(A) Let \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \).

\( \therefore |A| = x y z \neq 0, A^{-1} \) exists.

Now, cofactors of elements of A are:

\( A_{11} = (-1)^{1+1} \begin{vmatrix} y & 0 \\ 0 & z \end{vmatrix} = y z \),

\( A_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = 0 \),

\( A_{13} = (-1)^{1+3} \begin{vmatrix} 0 & y \\ 0 & 0 \end{vmatrix} = 0 \),

\( A_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = 0 \),

\( A_{22} = (-1)^{2+2} \begin{vmatrix} x & 0 \\ 0 & z \end{vmatrix} = x z \),

\( A_{23} = (-1)^{2+3} \begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = 0 \),

\( A_{31} = (-1)^{3+1} \begin{vmatrix} 0 & 0 \\ y & 0 \end{vmatrix} = 0 \),

\( A_{32} = (-1)^{3+2} \begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = 0 \),

\( A_{33} = (-1)^{3+3} \begin{vmatrix} x & 0 \\ 0 & y \end{vmatrix} = x y \).

\( \therefore adj A = \begin{bmatrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{bmatrix}^T = \begin{bmatrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{bmatrix} \).

Hence, \( A^{-1} = \frac{1}{x y z} \begin{bmatrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{bmatrix} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \).

19. Let \( A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix} \), where \( 0 \leq \theta \leq 2\pi \). Then:

1. A. \( \text{Det}(A) = 0 \)
2. B. \( \text{Det}(A) \in (2, \infty) \)
3. C. \( \text{Det}(A) \in (2, 4) \)
4. D. \( \text{Det}(A) \in [2, 4] \)

Solution:

(D) \( \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix} \).

Expanding along \( R_1 \), we get:

\( 1(1 + \sin^2 \theta) – \sin \theta(-\sin \theta + \sin \theta) + 1(\sin^2 \theta + 1) \).

\( = 1 + \sin^2 \theta + 1 + \sin^2 \theta = 2(1 + \sin^2 \theta) \).

As \( \sin^2 \theta \in [0, 1] \),

\( \Rightarrow 1 + \sin^2 \theta \in [1, 2] \),

\( \Rightarrow 2(1 + \sin^2 \theta) \in [2, 4] \).