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:
- A. 0
- B. 1
- C. x
- 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:
- A. \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- B. \( x y z \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- C. \( \frac{1}{x y z} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \)
- 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:
- A. \( \text{Det}(A) = 0 \)
- B. \( \text{Det}(A) \in (2, \infty) \)
- C. \( \text{Det}(A) \in (2, 4) \)
- 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 \( \sin^2 \theta \in [0, 1] \),
\( \Rightarrow 1 + \sin^2 \theta \in [1, 2] \),
\( \Rightarrow 2(1 + \sin^2 \theta) \in [2, 4] \).