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:
- A. \( k|A| \)
- B. \( k^2|A| \)
- C. \( k^3|A| \)
- 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?
- A. Determinant is a square matrix.
- B. Determinant is a number associated to a matrix.
- C. Determinant is a number associated to a square matrix.
- D. None of these.
Solution:
(C) Determinant is a number associated to a square matrix.