Matrices and Determinants JEE Mathematics Worksheets Set 02

Subjective Questions

Question. By using the principle of matrix, show that the following system of equations has infinite solution:
\( 5x + 3y + 7z = 4 \); \( 3x + 26y + 2z = 9 \); \( 7x + 2y + 10z = 5 \).
Answer: For infinite solutions,
\( D = D_1 = D_2 = D_3 = 0 \)
Now, \( D = \begin{vmatrix} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{vmatrix} \)
\( \Rightarrow D = 5(260 - 4) - 3(30 - 14) + 7(6 - 182) = 0 \)
\( \& \ D_1 = \begin{vmatrix} 4 & 3 & 7 \\ 9 & 26 & 2 \\ 5 & 2 & 10 \end{vmatrix} \)
\( \Rightarrow D_1 = 4(260 - 4) - 3(90 - 10) + 7(18 - 130) = 0 \)
\( \& \ D_2 = \begin{vmatrix} 5 & 4 & 7 \\ 3 & 9 & 2 \\ 7 & 5 & 10 \end{vmatrix} \)
\( \Rightarrow D_2 = 5(90 - 10) - 4(30 - 14) + 7(15 - 63) = 0 \)
\( \& \ D_3 = \begin{vmatrix} 5 & 3 & 4 \\ 3 & 26 & 9 \\ 7 & 2 & 5 \end{vmatrix} \)
\( \Rightarrow D_3 = 5(130 - 18) - 3(15 - 63) + 4(6 - 182) = 0 \)

Question. If the determinant \( \begin{vmatrix} \sin \theta & 1 & 0 \\ 1 & \cos \phi & -\cos \theta \\ \sin \phi & 0 & 1 \end{vmatrix} \) is a symmetric determinant then find minimum and maximum value of determinant.
Answer: Given : \( \Delta = \begin{vmatrix} \sin \theta & 1 & 0 \\ 1 & \cos \phi & -\cos \theta \\ \sin \phi & 0 & 1 \end{vmatrix} \)
\( \Rightarrow \Delta = \sin \theta (\cos \phi - 0) - 1(1 + \sin \phi \cos \theta) + 0 \)
\( \Rightarrow \Delta = \sin \theta \cos \phi - \sin \phi \cos \theta - 1 \)
\( \Rightarrow \Delta = \sin (\theta - \phi) - 1 \)
\( \therefore \quad \Delta_{\max} = 1 - 1 = 0 \)
\( \& \ \Delta_{\min} = -1 - 1 = (-2) \)

Question. If \( \begin{vmatrix} e^x & \sin x \\ \cos x & \ln(1 + x) \end{vmatrix} = A + Bx + Cx^2 + \dots \), then find the value of A and B.
Answer: Given : \( \begin{vmatrix} e^x & \sin x \\ \cos x & \ln(1 + x) \end{vmatrix} = A + Bx + Cx^2 + \dots \)
\( \Rightarrow e^x \cdot \ln(1 + x) - \cos x \cdot \sin x = A + Bx + Cx^2 + \dots \)
\( \Rightarrow (1 + x + \dots) \left( x - \frac{x^2}{2} + \dots \right) - \frac{1}{2} \sin 2x = A + Bx + Cx^2 + \dots \)
\( \Rightarrow (1 + x + \dots) \left( x - \frac{x^2}{2} + \dots \right) - \frac{1}{2} \left( 2x - \frac{(2x)^3}{3!} + \dots \right) = A + Bx + Cx^2 + \dots \)
On comparing constant term, A = 0
On comparing coefficient of x, B = 1 - 1 = 0

Question. Show that \( \Delta = \begin{vmatrix} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2 \end{vmatrix} = 4a^2b^2c^2 \)
Answer: Given : \( \begin{vmatrix} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2 \end{vmatrix} \)
\( \Rightarrow \frac{1}{abc} \begin{vmatrix} ab^2 + ac^2 & ab^2 & ac^2 \\ a^2b & bc^2 + a^2b & bc^2 \\ ca^2 & cb^2 & a^2c + b^2c \end{vmatrix} \)
Applying \( C_1 \Rightarrow C_1 - C_2 - C_3 \)
\( \Rightarrow \frac{1}{abc} \begin{vmatrix} 0 & ab^2 & ac^2 \\ -2bc^2 & bc^2 + a^2b & bc^2 \\ -2b^2c & cb^2 & a^2c + b^2c \end{vmatrix} \)
\( \Rightarrow \frac{abc}{abc} \begin{vmatrix} 0 & b^2 & c^2 \\ -2c^2 & c^2 + a^2 & c^2 \\ -2b^2 & b^2 & a^2 + b^2 \end{vmatrix} \)
\( \Rightarrow -b^2 (-2c^2a^2 - 2b^2c^2 + 2b^2c^2) + c^2 (-2c^2b^2 + 2b^2c^2 + 2b^2a^2) \)
\( \Rightarrow 2a^2b^2c^2 + 2a^2b^2c^2 = 4a^2b^2c^2 \)

Question. Show that \( \begin{vmatrix} a_1\ell_1 + b_1m_1 & a_1\ell_2 + b_1m_2 & a_1\ell_3 + b_1m_3 \\ a_2\ell_1 + b_2m_1 & a_2\ell_2 + b_2m_2 & a_2\ell_3 + b_2m_3 \\ a_3\ell_1 + b_3m_1 & a_3\ell_2 + b_3m_2 & a_3\ell_3 + b_3m_3 \end{vmatrix} = 0 \)
Answer: L.H.S. \( = \begin{vmatrix} a_1\ell_1 + b_1m_1 & a_1\ell_2 + b_1m_2 & a_1\ell_3 + b_1m_3 \\ a_2\ell_1 + b_2m_1 & a_2\ell_2 + b_2m_2 & a_2\ell_3 + b_2m_3 \\ a_3\ell_1 + b_3m_1 & a_3\ell_2 + b_3m_2 & a_3\ell_3 + b_3m_3 \end{vmatrix} \)
\( = \begin{vmatrix} a_1 & b_1 & k_1 \\ a_2 & b_2 & k_2 \\ a_3 & b_3 & k_3 \end{vmatrix} \times \begin{vmatrix} \ell_1 & \ell_2 & \ell_3 \\ m_1 & m_2 & m_3 \\ 0 & 0 & 0 \end{vmatrix} \)
where \( k_1, k_2, k_3 \in \mathbb{R} \)
\( = \begin{vmatrix} a_1 & b_1 & k_1 \\ a_2 & b_2 & k_2 \\ a_3 & b_3 & k_3 \end{vmatrix} \times 0 = 0 = \text{R.H.S.} \)

Question. Investigate for what values of \( \lambda, \mu \) the simultaneous equations \( x + y + z = 6 \); \( x + 2y + 3z = 10 \) & \( x + 2y + \lambda z = \mu \) have;
(a) A unique solution
(b) An infinite number of solutions
(c) No solution.
Answer: Given equations are
\( x + y + z = 6 \)
\( x + 2y + 3z = 10 \)
\( x + 2y + \lambda z = \mu \)
Here \( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) = \lambda - 3 \)
\( D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \mu & 2 & \lambda \end{vmatrix} = 2\lambda + \mu - 16 \)
\( D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \mu & \lambda \end{vmatrix} = 2\lambda - \mu + 4 \)
\( D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 2 & \mu \end{vmatrix} = \mu - 10 \)
By cramer's rule : \( x = \frac{D_1}{D}, \ y = \frac{D_2}{D}, \ z = \frac{D_3}{D} \)
(a) A unique solution : \( D \neq 0 \) i.e. \( \lambda \neq 3 \)
(b) Infinite number of solutions :
\( D = D_1 = D_2 = D_3 = 0 \)
i.e. \( \lambda = 3, \ \mu = 10 \)
(c) No solution : \( D = 0 \) and at least one \( D_1, D_2, D_3 \) is not zero
i.e. \( \lambda = 3, \ \mu \neq 10 \)

Advanced Subjective Questions

Question. Find the inverse of the matrix :
(i) \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w^2 & w \end{bmatrix} \) where w is the cube root of unity.
(iii) \( A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \)
Answer: (i) \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( |A| = \begin{vmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 \)
Cofacter elements matrix = \( \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \text{Adj } A = (cij)^T = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( A^{-1} = \frac{\text{Adj } A}{| A |} = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

(b) \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{bmatrix} \)
\( C_{11} = \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & \omega \end{vmatrix} = \omega^2 - \omega^4 = (\omega^2 - \omega) \)
\( C_{12} = (-1) \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} = \omega^2 - \omega \)
\( C_{13} = \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} = \omega^2 - \omega \)
\( C_{21} = (-1) \begin{vmatrix} 1 & 1 \\ \omega^2 & \omega \end{vmatrix} = \omega^2 - \omega \)
\( C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = \omega - 1 \)
\( C_{23} = (-1) \begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = 1 - \omega^2 \)
\( C_{31} = \begin{vmatrix} 1 & 1 \\ \omega & \omega^2 \end{vmatrix} = \omega^2 - \omega \)
\( C_{32} = (-1) \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = (\omega^2 - 1) = 1 - \omega^2 \)
\( C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = \omega - 1 \)
\( C = \begin{bmatrix} \omega^2 - \omega & \omega^2 - \omega & \omega^2 - \omega \\ \omega^2 - \omega & \omega - 1 & 1 - \omega^2 \\ \omega^2 - \omega & -(\omega^2 - 1) & \omega - 1 \end{bmatrix} \)
\( \text{adj } A = c^T = \begin{bmatrix} \omega^2 - \omega & \omega^2 - \omega & \omega^2 - \omega \\ \omega^2 - \omega & \omega - 1 & -(\omega^2 - 1) \\ \omega^2 - \omega & 1 - \omega^2 & \omega - 1 \end{bmatrix} \)
\( |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} \implies 1 \{\omega - \omega\} - 1 \{\omega - \omega\} + 1 \{\omega^2 - \omega\} \)
\( \implies \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega \)
\( \implies 3(\omega^2 - \omega) \implies 3 \omega (\omega - 1) \)
\( A^{-1} = \frac{\text{adj. } A}{| A |} \)
\( = \frac{1}{3\cos(\omega - 1)} \begin{bmatrix} \omega(\omega - 1) & \omega(\omega - 1) & \omega(\omega - 1) \\ \omega(\omega - 1) & \omega - 1 & -(\omega - 1)\omega + 1 \\ \omega(\omega - 1) & (1 - \omega)(1 + \omega) & \omega - 1 \end{bmatrix} \)
\( = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \frac{1}{\omega} & -(\omega + 1) \\ 1 & -(1 + \omega) & \frac{1}{\omega} \end{bmatrix} \)
\( = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{bmatrix} \)

(c) \( \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \) for any diagonal
matrix inverse of a diagonal matrix
\( A = \begin{bmatrix} a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{33} \end{bmatrix} \) then
\( A^{-1} = \begin{bmatrix} 1/ A_{11} & 0 & 0 \\ 0 & 1/ A_{22} & 0 \\ 0 & 0 & 1/ A_{33} \end{bmatrix} \)
so \( A^{-1} = \begin{bmatrix} 1/ a & 0 & 0 \\ 0 & 1/b & 0 \\ 0 & 0 & 1/ c \end{bmatrix} \)

Question. Show that,
\( \begin{bmatrix} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \)
Answer: A B
\( \begin{bmatrix} 1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \theta / 2 \\ -\tan \theta / 2 & 1 \end{bmatrix} \)
\( B^{-1} = \frac{\text{Adj}(B)}{| B |} \)
\( \text{Adj } (B) = \begin{bmatrix} 1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1 \end{bmatrix} \)
|B| = 1 + \tan2 \theta/2
= \sec2 \theta/2
\( B^{-1} = \begin{bmatrix} \frac{\cos^2 \theta / 2}{\sec^2 \theta / 2} & \frac{-\tan \theta / 2}{\sec^2 \theta / 2} \\ \frac{\tan \theta / 2}{\sec^2 \theta / 2} & \frac{1}{\sec^2 \theta / 2} \end{bmatrix} \)
\( A B^{-1} = \begin{bmatrix} 1 & -tn \theta / 2 \\ \tan \theta / 2 & 1 \end{bmatrix} \)
\( \begin{bmatrix} \frac{\cos^2 \theta / 2}{\sec^2 \theta / 2} & \frac{-\tan \theta / 2}{\sec^2 \theta / 2} \\ \frac{\tan \theta / 2}{\sec^2 \theta / 2} & \frac{1}{\sec^2 \theta} \end{bmatrix} \)
\( = \begin{bmatrix} \cos \theta & -\tan \theta \\ \sin \theta & \cos \theta \end{bmatrix} \)

Question. If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \) then show that
F(x). F(y) = F(x + y). Hence prove that \( [F(x)]^{-1} = F(-x) \).
Answer: \( F(x + y) = \begin{bmatrix} \cos(x + y) & -\sin(x + y) & 0 \\ \sin(x + y) & \cos(x + y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -\sin x \cos y - \cos x \sin y & 0 \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
= F(x) F(y)
Now \( F(x) . F(-x) = \begin{bmatrix} c & -s & 0 \\ s & c & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c & s & 0 \\ -s & c & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
= I
F(x) F(-x) = I
\( \implies [F(x)]^{-1} = F(-x) \) Hence proved

Question. If A is a skew symmetric matrix and I + A is non singular, then prove that the matrix \( B = (I - A) (I + A)^{-1} \) is an orthogonal matrix. Use this to find a matrix B given \( A = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} \).
Answer: A is skew symmetric matrix
\( I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} \)
\( (I + A) = \begin{bmatrix} 1 & 5 \\ -5 & 1 \end{bmatrix} \) | I + A | = 1 + 25 = 26
\( | I + A |^{-1} = \frac{\text{Adj}(I + A)}{| I + A |} \)
\( = 1/26 \begin{bmatrix} 1 & -5 \\ 5 & 1 \end{bmatrix} \)
\( [I + A]^{-1} [I - A] = 1/26 \begin{bmatrix} 1 & -5 \\ 5 & 1 \end{bmatrix} \begin{bmatrix} 1 & 5 \\ -5 & 1 \end{bmatrix} \)
\( = 1/26 \begin{bmatrix} 26 & 0 \\ 0 & 26 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) is

Question. Use matrix to solve the following system of equations.
(i) \( x + 4y + 9z = 6 \)
\( x + 2y + 3z = 4 \)
\( x + y + z = 3 \)
(ii) \( 2x + y - z = 1 \)
\( x - y + z = 2 \)
\( x + y + z = 6 \)
Answer: (i) \( x + y + z = 3 \)
\( x + 2y + 3y = 4 \)
\( x + 4y + 9z = 6 \)
\( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{vmatrix} = - 6 \)
D \( \neq \) 0
\( x = \frac{D_1}{D}, y = \frac{D_2}{D}, z \mp \frac{D_3}{D} \)
\( D_1 = \begin{vmatrix} 3 & 1 & 1 \\ 4 & 2 & 3 \\ 6 & 4 & 9 \end{vmatrix} \) \( D_2 = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 4 & 3 \\ 1 & 6 & 9 \end{vmatrix} \) \( D_3 = \begin{vmatrix} 1 & 1 & 3 \\ 1 & 2 & 4 \\ 1 & 4 & 6 \end{vmatrix} \)
x = 2
y = 1
z = 0

(ii) \( x + y + z = 6 \)
\( x - y + z = 2 \)
\( 2x + y - z = 1 \)
D \( \neq \) 0
\( \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & 1 \end{vmatrix} = 1(-2) - 1 + 1(3) \)
= -2 + 1 + 3
= 2
\( x = \frac{D_1}{D}, y = \frac{D_2}{D}, z = \frac{D_3}{D} \)
\( D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{vmatrix} \)
\( D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 1 & -1 & 2 \\ 2 & 1 & 1 \end{vmatrix} \)

(b) \( x + y + z = 3 \)
\( x + 2y + 3z = 4 \)
\( 2x + 3y + 4z = 9 \)
D \( \neq \) 0
D = 0
No solution
D1 = D2 = D3 = 0
\( D_1 = \begin{vmatrix} 3 & 1 & 1 \\ 4 & 2 & 3 \\ 9 & 3 & 4 \end{vmatrix} = 3(8 - 9) - 1(16 - 27) + 1(12 - 1) \)
= no zero
D2 = |
D1 \( \neq \) 0 so no solution

Question. Given that \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{bmatrix}, C = \begin{bmatrix} 2 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{bmatrix}, D = \begin{bmatrix} 10 \\ 13 \\ 9 \end{bmatrix} \) and that Cb = D. Solve the matrix equation Ax = b.
Answer: \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{bmatrix}, C = \begin{bmatrix} 2 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{bmatrix}, D = \begin{bmatrix} 10 \\ 13 \\ 9 \end{bmatrix} \)
Cb = D Let \( b = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)
\( \begin{bmatrix} 2 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 13 \\ 9 \end{bmatrix} \)
x = 1
y = 3
z = 5
Ax = b
\( \begin{bmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} \)
\( a = 1, b = -1, c = 1 \)
\( x = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \)
\( x = \frac{D_1}{D} ; y = \frac{D_2}{D} ; z = \frac{D_3}{D} \)
\( \implies x = 1, y = 3, z = 5 \)
\( \begin{bmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 1 & -1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} \)
\( \implies \) multiply & compare
\( x_1 = 1 ; x_2 = -1 ; x_3 = 1 \)

Question. Find the matrix A satisfying the matrix equation, \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \cdot A \cdot \begin{bmatrix} 3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 3 & -1 \end{bmatrix} \).
Answer: \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} 3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 3 & -1 \end{bmatrix} \)
\( A = \begin{bmatrix} x & y \\ z & w \end{bmatrix} \)
\( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x & y \\ z & w \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 3 & -1 \end{bmatrix} \)
\( x = \frac{48}{19} \)
\( y = \frac{- 25}{19} \)
\( z = \frac{- 70}{19} \)
\( w = \frac{42}{19} \)

Question. If \( A = \begin{bmatrix} k & m \\ l & n \end{bmatrix} \) and \( kn \neq lm \) ; then show that \( A^2 - (k + n) A + (kn - lm) I = O \). Hence find \( A^{-1} \).
Answer: \( A = \begin{bmatrix} k & m \\ \lambda & n \end{bmatrix} \) \( kn \neq \lambda m \)
\( A^2 - (k + 4) A + (kn - \lambda m) I = \)
\( A^2 = \begin{bmatrix} k & m \\ \lambda & n \end{bmatrix} \begin{bmatrix} k & m \\ \lambda & n \end{bmatrix} = \begin{bmatrix} k^2 + m\lambda & km + mn \\ k\lambda + n\lambda & m\lambda + n^2 \end{bmatrix} \)
\( \begin{bmatrix} k^2 + m\lambda & km + mn \\ k\lambda + n\lambda & m\lambda + n^2 \end{bmatrix} - (k + n) \begin{bmatrix} k & m \\ \lambda & n \end{bmatrix} \)
\( + \begin{bmatrix} kn - \lambda n & 0 \\ 0 & kn - \lambda n \end{bmatrix} = 0 \)
\( A^2 - (k + n) A + (kn - \lambda n) I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) proceed
Multiply by \( A^{-1} \)
\( A - (k + n) I + (kn - lm) A^{-1} = A^{-1} 0 \)
\( A^{-1} = \frac{1}{kn - \ell m} \begin{bmatrix} n & -m \\ -\ell & k \end{bmatrix} \)

Question. Given \( A = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}; B = \begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} \). I is a unit matrix of order 2. Find all possible matrix X in the following cases.
(i) AX = A
(ii) XA = I
(iii) XB = O but \( BX \neq O \).
Answer: (a) \( A = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} B = \begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} \)
(i) AX = A
\( \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 2a + c & 2b + d \\ 2a + c & 2b + d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \)
2a + c .....(1)
2b + d = 1 .....(2)
c = 2 - 2a
d = 1 - 2b
\( \begin{bmatrix} a & b \\ 2 - 2a & 1 - 2b \end{bmatrix} \)

(b) A = I
\( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} = I \)
\( \begin{bmatrix} 2a + 2d & a + d \\ 2c + 2d & c + d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
2(a + b) = 1
a + b = 0 Does not exist
2(c + d) = 1
c + d = 0
(c) XB = 0 but Bx \( \neq \) 0
\( \begin{bmatrix} a & b \\ c & a \end{bmatrix} \begin{bmatrix} a & 3 \\ 3 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 9a + 3b & 3a + b \\ 9c + 3d & 3c + d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Bx \( \neq \) 0
9a + 3b = 0
3a + b = 0
ac + 3d = 0
3c + d = 0
\( \begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 9a + 3c & 9b + 3d \\ 3a + c & 3b + d \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} a & - 3a \\ c & - 3c \end{bmatrix} \) 3a + c \( \neq \) 0

Question. Find the product of two matrices A & B, where \( A = \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix} \) & \( B = \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \) and use it to solve the following system of linear equations
\( x + y + 2z = 1; 3x + 2y + z = 7 ; 2x + y + 3z = 2 \).
Answer: AB = \( \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \)
= \( \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = 4 I \)
x + y + 2z = 1
3x + 2y + z = 7
2x + 2y + 3z = 2
\( \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix}_{3 \times 3} \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix}_{3 \times 1} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} \)

Question. If \( A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \) then, find a non-zero square matrix X of order 2 such that AX = O. Is XA = O.
If \( A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \), is it possible to find a square matrix X such that AX = O. Give reasons for it.
Answer: \( A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 0 \)
\( \begin{bmatrix} a + 2c & b + 2d \\ 2a + 4c & 2b + 4d \end{bmatrix} = 0 \)
a + 2c = 0 b + 2d = 0
a = - 2c b = - 2d
\( x = \begin{bmatrix} - 2c & - 2d \\ c & d \end{bmatrix} \)

Question. Determine the value of a and b for which the system \( \begin{bmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b \\ 3 \\ -1 \end{bmatrix} \)
(i) has a unique solution;
(ii) has no solution and
(iii) has infinitely may solutions
Answer: \( \begin{bmatrix} 3 & - 2 & 1 \\ 5 & 8 & 9 \\ 2 & 1 & a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b \\ 3 \\ -1 \end{bmatrix} \)
(i) unique solution
\( \begin{vmatrix} 3 & - 2 & 1 \\ 5 & 8 & 9 \\ 2 & 1 & a \end{vmatrix} \neq 0 \)
a \( \neq \) 3
(ii) has no solution
D = 0 so a should a = 3 at least one form D1, D2, D3 non zero b \( \neq \) 1/3
\( \begin{vmatrix} b & - 2 & 1 \\ 3 & - 8 & 9 \\ - 1 & 1 & 3 \end{vmatrix} \neq 0 \)

Question. If \( A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}; B= \begin{bmatrix} 3 & 1 \\ 1 & 0 \end{bmatrix}; C= \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \) and \( X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} \) then sole the following matrix equation.
(a) AX = B = I
(b) (B - I) X = IC
(c) CX = A
Answer: (a) \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 3 & 1 \\ 1 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( x = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} \)
(b) (B - I) × = IC
\( \left( \begin{bmatrix} 3 & 1 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( \begin{bmatrix} 2 & 1 \\ 1 & - 1 \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( \begin{bmatrix} 2x_1 + x_2 & 2x_3 + x_4 \\ x_1 - x_2 & x_3 - x_4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( x = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ - 1 & - 2 \end{bmatrix} \)
(c) Cx = A
\( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
\( \begin{bmatrix} x_1 + 2x_3 & 2x_1 + 4x_3 \\ x_2 + 2x_4 & 2x_2 + 4x_4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
x1 + 2x3 = 1
2x1 + 4x2 = 2 No solution

Question. If A is an orthogonal matrix and B = AP where P is non singular matrix then show that the matrix \( PB^{-1} \) is also orthogonal.
Answer: B = AP AT A = I
BB-1 = AP B-1
I = APB-1
(AAT)T = (APB-1)T
AAT = (B-1)T (P)T AT
A = (B-1)T (P)T
AT = PB-1
(PB-1) (PB-1)T = I
So PB-1 is orthogonal

JEE Problems

Question. (a) If \( f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix} \) then \( f(100) \) is equal to
(a) 0
(b) 1
(c) 100
(d) – 100
Answer: (a) 0
Solution:
(a) Given \( f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix} \)
Applying \( C_3 \rightarrow C_3 - (C_1 + C_2) \)
\( = \begin{vmatrix} 1 & x & 0 \\ 2x & x(x-1) & 0 \\ 3x(x-1) & x(x-1)(x-2) & 0 \end{vmatrix} = 0 \)
\( \therefore f(x) = 0 \Rightarrow f(100) = 0 \)

Question. (b) Let \( a, b, c, d \) be real numbers in G.P. If \( u, v, w \) satisfy the system of equations \( u + 2v + 3w = 6 \) ; \( 4u + 5v + 6w = 12 \) then show that the roots of the equation \( 6u + 9v = 4 \left( \frac{1}{u} + \frac{1}{v} + \frac{1}{w} \right) x^2 + [(b - c)^2 + (c - a)^2 + (d - b)^2] x + u + v + w = 0 \) and \( 20x^2 + 10 (a - d)^2 x - 9 = 0 \) are reciprocals of each other.
Answer:
Solution:
(b) Given system of equation,
\( u + 2v + 3w = 6 \)
\( 4u + 5v + 6w = 12 \)
\( 6u + 9v = 4 \)
Augmented matrix, \( \begin{pmatrix} 1 & 2 & 3 & 6 \\ 4 & 5 & 6 & 12 \\ 6 & 9 & 0 & 4 \end{pmatrix} \)
Applying \( R_2 \rightarrow R_2 - 4R_1 \) & \( R_3 \rightarrow R_3 - 6R_1 \)
\( \begin{pmatrix} 1 & 2 & 3 & 6 \\ 0 & -3 & -6 & -12 \\ 0 & -3 & -18 & -32 \end{pmatrix} \xrightarrow{R_3 \rightarrow R_3 - R_2} \begin{pmatrix} 1 & 2 & 3 & 6 \\ 0 & -3 & -6 & -12 \\ 0 & 0 & -12 & -20 \end{pmatrix} \)
\( \Rightarrow u + 2v + 3w = 6 \)
\( -3v - 6w = -12 \)
\( -12w = -20 \)
\( \therefore u = \left(-\frac{1}{3}\right) \); \( v = \frac{2}{3} \); \( w = \frac{5}{3} \)
\( \Rightarrow (u + v + w) = 2 \) ; \( \left( \frac{1}{u} + \frac{1}{v} + \frac{1}{w} \right) = \frac{9}{10} \)
Now, \( a, b, c, d \) are in GP, then
\( b^2 = ac \); \( c^2 = bd \); \( ad = bc \)
\( \therefore [(b - c)^2 + (c - a)^2 + (d - b)^2] \)
\( = b^2 + c^2 + c^2 + a^2 + d^2 + b^2 - 2bc - 2ca - 2bd \)
\( = (a - d)^2 \)
Observing given equation,
\( \left(\frac{1}{u} + \frac{1}{v} + \frac{1}{w}\right) x^2 + [(b - c)^2 + (c - a)^2 + (d - b)^2] x + (u + v + w) = 0 \)
& \( 20 x^2 + 10 (a - d)^2 x - 9 = 0 \)
we can say, equations have reciprocal roots.

Question. If the system of equations \( x - ky - z = 0 \), \( kx - y - z = 0 \), \( x + y - z = 0 \) has a non-zero solution then the possible values of \( k \) are
(a) –1, 2
(b) 1, 2
(c) 0, 1
(d) –1, 1
Answer: (d) –1, 1
Solution:
Since, the given system has non-zero solution
\( \therefore \begin{vmatrix} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1+k & -k-1 & -1 \\ 1+k & -2 & -1 \\ 0 & 0 & -1 \end{vmatrix} = 0 \)
(\( \because \) Applying \( C_1 \rightarrow C_1 - C_2 \), \( C_2 \rightarrow C_2 + C_3 \))
\( \Rightarrow 2(k + 1) - (k + 1)^2 = 0 \Rightarrow (k + 1) (2 - k - 1) = 0 \)
\( \Rightarrow k = \pm 1 \)

Question. Prove that for all values of \( \theta \)
\( \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin(\theta + 2\pi/3) & \cos(\theta + 2\pi/3) & \sin(2\theta + 4\pi/3) \\ \sin(\theta - 2\pi/3) & \cos(\theta - 2\pi/3) & \sin(2\theta - 4\pi/3) \end{vmatrix} = 0 \)
Answer:
Solution:
Let \( \Delta = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 + R_3 \)
\( = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta + \frac{4\pi}{3}\right) + \sin\left(2\theta - \frac{4\pi}{3}\right) \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} \)
\( = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ 2\sin\theta\cos\frac{2\pi}{3} & 2\cos\theta\cos\frac{2\pi}{3} & 2\sin 2\theta\cos\frac{4\pi}{3} \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} \)
\( = \begin{vmatrix} \sin \theta & \cos \theta & \sin 2\theta \\ -\sin\theta & -\cos\theta & -\sin 2\theta \\ \sin\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(2\theta - \frac{4\pi}{3}\right) \end{vmatrix} = 0 \)
(since \( R_1 \) and \( R_2 \) are proportional)

Question. Find the real values of \( r \) for which the following system of linear equations has a non-trivial solutions. Also find the non-trivial solutions
\( 2rx - 2y + 3z = 0 \)
\( x + ry + 2z = 0 \)
\( 2x + rz = 0 \)
Answer:
Solution:
For non – trivial solution,
\( D = \begin{vmatrix} 2r & -2 & 3 \\ 1 & r & 2 \\ 2 & 0 & r \end{vmatrix} = 0 \)
\( \Rightarrow 2r (r^2 - 0) + 2(r - 4) + 3(0 - 2r) = 0 \)
\( \Rightarrow r = 2 \)
\( \therefore \) system of equations become,
\( 4x - 2y + 3z = 0 \) ....(1)
\( x + 2y + 2z = 0 \) .....(2)
\( 2x + 2z = 0 \) ....(3)
Let \( x = k \)
from equation (3),
\( z = (-k) \) .....(4)
from equation (1) & (4),
\( y = \frac{k}{2} \)

Question. Solve for \( x \) the equation \( \begin{vmatrix} a^2 & a & 1 \\ \sin(n+1)x & \sin nx & \sin(n-1)x \\ \cos(n+1)x & \cos nx & \cos(n-1)x \end{vmatrix} = 0 \)
Answer:
Solution:
Given : \( \begin{vmatrix} a^2 & a & 1 \\ \sin(n+1)x & \sin nx & \sin(n-1)x \\ \cos(n+1)x & \cos nx & \cos(n-1)x \end{vmatrix} = 0 \)
\( \Rightarrow a^2 [\sin nx \cdot \cos(n - 1)x - \cos nx \cdot \sin(n - 1)x] - a [\sin(n + 1)x \cdot \cos(n - 1)x - \cos(n + 1)x \cdot \sin(n - 1)x] + 1 [\cos nx \cdot \sin(n + 1)x - \sin nx \cdot \cos(n + 1)x] = 0 \)
\( \Rightarrow a^2 \sin[nx - (n - 1)x] - a \sin[(n + 1)x - (n - 1)x] + \sin[(n + 1)x - nx] = 0 \)
\( \Rightarrow \sin x (a^2 - a \cos x + 1) = 0 \Rightarrow \sin x = 0 \)
\( \Rightarrow x = n\pi; n \in I \)

Question. Test the consistency and solve them when consistent, the following system of equations for all values of \( \lambda \)
\( x + y + z = 1 \)
\( x + 3y - 2z = \lambda \)
\( 3x + (\lambda + 2)y - 3z = 2\lambda + 1 \)
Answer:
Solution:
Augmented matrix,
\( \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 3 & -2 & \lambda \\ 3 & \lambda+2 & -3 & 2\lambda+1 \end{pmatrix} \xrightarrow{R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - 3R_1} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 2 & -3 & \lambda-1 \\ 0 & \lambda-1 & -6 & 2\lambda-2 \end{pmatrix} \)
\( \xrightarrow{R_3 \rightarrow R_3 - 2R_2} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 2 & -3 & \lambda-1 \\ 0 & \lambda-5 & 0 & 0 \end{pmatrix} \)
Hence, consistent for all values of \( \lambda \),
Case – I : When \( \lambda = 5 \),
\( x + y + z = 1 \)
\( 2y - 3z = 4 \)
Let \( z = k \)
\( \Rightarrow 2y - 3k = 4 \Rightarrow y = \left(\frac{3k+4}{2}\right) \)
& \( x = 1 - \frac{3k+4}{2} - k \Rightarrow x = \left(\frac{-5k-2}{2}\right) \)
Case – 2 : when \( \lambda \neq 5 \)
\( x + y + z = 1 \)
\( 2y - 3z = \lambda - 1 \)
\( (\lambda - 5)y = 0 \Rightarrow y = 0 \)
\( \Rightarrow -3z = \lambda - 1 \Rightarrow z = \left(\frac{1-\lambda}{3}\right) \)
& \( x = 1 - \frac{1-\lambda}{3} \Rightarrow x = \left(\frac{2+\lambda}{3}\right) \)

Question. Let \( a, b, c \) be real number with \( a^2 + b^2 + c^2 = 1 \). Show that the equation \( \begin{vmatrix} ax-by-c & bx+ay & cx+a \\ bx+ay & -ax+by-c & cy+b \\ cx+a & cy+b & -ax-by+c \end{vmatrix} = 0 \) represents a straight line.
Answer:
Solution:
Given, \( \begin{vmatrix} ax-by-c & bx+ay & cx+a \\ bx+ay & -ax+by-c & cy+b \\ cx+a & cy+b & -ax-by+c \end{vmatrix} = 0 \)
\( \Rightarrow \frac{1}{a} \begin{vmatrix} a^2 x - aby - ac & bx + ay & cx + a \\ abx + a^2 y & -ax + by - c & cy + b \\ acx + a^2 & cy + b & -ax - by + c \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 + bC_2 + cC_3 \)
\( \Rightarrow \frac{1}{a} \begin{vmatrix} (a^2+b^2+c^2)x & bx + ay & cx + a \\ (a^2+b^2+c^2)y & -ax + by - c & cy + b \\ (a^2+b^2+c^2) & cy + b & -ax - by + c \end{vmatrix} = 0 \)
\( \Rightarrow \frac{1}{a} \begin{vmatrix} x & ay + bx & cx + a \\ y & by - c - ax & b + cy \\ 1 & b + cy & c - ax - by \end{vmatrix} = 0 \)
(\( \because a^2 + b^2 + c^2 = 1 \) given)
Applying \( C_2 \rightarrow C_2 + bC_1 \) and \( C_3 \rightarrow C_3 - cC_1 \)
\( \Rightarrow \frac{1}{a} \begin{vmatrix} x & ay & a \\ y & -c - ax & b \\ 1 & cy & -ax - by \end{vmatrix} = 0 \)
\( \Rightarrow \frac{1}{ax} \begin{vmatrix} x^2 & axy & ax \\ y & -c - ax & b \\ 1 & cy & -ax - by \end{vmatrix} = 0 \)
Applying \( R_1 \rightarrow R_1 + yR_2 + R_3 \)
\( \Rightarrow \frac{1}{ax} \begin{vmatrix} x^2 + y^2 + 1 & 0 & 0 \\ y & -c - ax & b \\ 1 & cy & -ax - by \end{vmatrix} = 0 \)
\( \Rightarrow \frac{1}{ax} [(x^2 + y^2 + 1) \{(-c - ax)(-ax - by) - b(y)\}] = 0 \)
\( \Rightarrow \frac{1}{ax} [(x^2 + y^2 + 1)(acx + bcy + a^2 x^2 + abxy - bcy)] = 0 \)
\( \Rightarrow \frac{1}{ax} [(x^2 + y^2 + 1)(acx + a^2 x^2 + abxy)] = 0 \)
\( \Rightarrow \frac{1}{ax} [ax(x^2 + y^2 + 1)(c + ax + by)] = 0 \)
\( \Rightarrow (x^2 + y^2 + 1) (ax + by + c) = 0 \)
\( \Rightarrow ax + by + c = 0 \)
which represent a straight line.

Question. The number of values of \( k \) for which the system of equations \( (k + 1) x + 8y = 4k \), \( kx + (k + 3)y = 3k - 1 \) has infinitely many solutions is
(a) 0
(b) 1
(c) 2
(d) infinite
Answer: (b) 1
Solution:
For infinitely many solutions, we must have,
\( \frac{k+1}{k} = \frac{8}{k+3} = \frac{4k}{3k-1} \Rightarrow k = 1 \)

Question. If matrix \( A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \) where \( a, b, c \) are real positive numbers, \( abc = 1 \) and \( A^T A = I \), then find the value of \( a^3 + b^3 + c^3 \).
Answer:
Solution:
Given \( A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \), \( abc = 1 \) and \( A^T A = I \) ....(1)
Now \( A^T A = I \)
\( \Rightarrow \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} a^2+b^2+c^2 & ab+bc+ca & ab+bc+ca \\ ab+bc+ca & a^2+b^2+c^2 & ab+bc+ca \\ ab+bc+ca & ab+bc+ca & a^2+b^2+c^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \Rightarrow a^2 + b^2 + c^2 = 1 \) and \( ab + bc + ca = 0 \) ....(2)
we know \( a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca) \)
\( \Rightarrow a^3 + b^3 + c^3 = (a + b + c) (1 - 0) + 3 \) (from equation (1) and (2))
\( \Rightarrow a^3 + b^3 + c^3 = (a + b + c) + 3 \) .....(3)
Now \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1 \) ....(4)
\( \therefore \) From equation (3), \( a^3 + b^3 + c^3 = 1 + 3 = 4 \)

Question. The value of \( \lambda \) for which the system of equations \( 2x - y - z = 12 \), \( x - 2y + z = - 4 \), \( x + y + \lambda z = 4 \) has no solution is
(a) 3
(b) -3
(c) 2
(d) -2
Answer: (d) -2
Solution:
Given \( 2x - y + 2z = 2 \)
\( x - 2y + z = -4 \)
\( x + y + \lambda z = 4 \)
since, given system has no solution
\( \therefore \Delta = 0 \) and any one amongst \( \Delta_x, \Delta_y, \Delta_z \) is non-zero
Let \( \Delta = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0 \) and \( \Delta_z = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & -4 \\ 1 & 1 & 4 \end{vmatrix} = 6 \neq 0 \Rightarrow \lambda = 1 \)

Question. If \( A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} \) and \( |A^3| = 125 \) then the value of \( \alpha \) is
(a) \( \pm 3 \)
(b) \( \pm 2 \)
(c) \( \pm 5 \)
(d) 0
Answer: (a) \pm 3
Solution:
Given \( A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} \)
\( A^3 = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} = \begin{bmatrix} \alpha^2+4 & 4\alpha \\ 4\alpha & \alpha^2+4 \end{bmatrix} \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix} = \begin{bmatrix} \alpha^3+12\alpha & 6\alpha^2+8 \\ 6\alpha^2+8 & \alpha^3+12\alpha \end{bmatrix} \)
then \( |A^3| = 125 \)
\( \Rightarrow \begin{vmatrix} \alpha^3+12\alpha & 6\alpha^2+8 \\ 6\alpha^2+8 & \alpha^3+12\alpha \end{vmatrix} = 125 \)
\( \Rightarrow (\alpha^3 + 12\alpha)^2 - (6\alpha^2 + 8)^2 = 125 \)
\( \Rightarrow (\alpha^3 + 6\alpha^2 + 12\alpha + 8) (\alpha^3 - 6\alpha^2 + 12\alpha - 8) = 125 \)
\( \Rightarrow (\alpha + 2)^3 (\alpha - 2)^3 = 125 \)
\( \Rightarrow [(\alpha + 2)(\alpha - 2)]^3 = (5)^3 \Rightarrow \alpha^2 - 4 = 5 \)
\( \Rightarrow \alpha^2 = 9 \Rightarrow \alpha = \pm 3 \)

Question. If \( M \) is a \( 3 \times 3 \) matrix, where \( \det (M) = 1 \) and \( M^T M = I \) (where 'I' is an identity matrix) then prove that \( \det (M - I) = 0 \).
Answer:
Solution:
Given: \( |M| = 1 \) & \( M^T M = I \)
Now, \( |M^T| = |M| = 1 \)
\( \therefore (M - I)^T = M^T - I^T = M^T - I \)
But, \( M^T M = I \)
\( \therefore (M - I)^T = M^T - M^T M = M^T (I - M) \)
Now, \( \det (M - I)^T = \det (M^T (I - M)) \)
\( \Rightarrow \det (M - I)^T = \det (M^T) \det (I - M) \)
\( \Rightarrow \det (M - I) = \det (M) \det (I - M) \)
\( \Rightarrow \det (M - I) = (-1) \det (M - I) \)
\( \Rightarrow 2 \det (M - I) = 0 \)
\( \Rightarrow \det (M - I) = 0 \)

Question. \( A= \begin{bmatrix} a & 1 & 0 \\ 1 & b & d \\ 1 & b & c \end{bmatrix} \), \( B= \begin{bmatrix} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{bmatrix} \), \( U= \begin{bmatrix} f \\ g \\ h \end{bmatrix} \), \( V= \begin{bmatrix} a^2 \\ 0 \\ 0 \end{bmatrix} \), \( X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)
If \( AX = U \) has infinitely many solution then prove that \( BX = V \), can not have a unique solution. If further \( afd \neq 0 \) then prove that \( BX = V \) has no solution.
Answer:
Solution:
Since \( AX = U \) has infinitely many solution
\( \Rightarrow |A| = 0 \)
\( \Rightarrow \begin{vmatrix} a & 0 & 1 \\ 1 & c & b \\ 1 & d & b \end{vmatrix} = 0 \)
\( \Rightarrow a(bc - bd) + 1(d - c) = 0 \)
\( \Rightarrow (d - c)(ab - 1) = 0 \)
\( \Rightarrow ab = 1 \) or \( d = c \)
Again \( |A_3| = \begin{vmatrix} a & 0 & f \\ 1 & c & g \\ 1 & d & h \end{vmatrix} = 0 \Rightarrow g = h \)
\( |A_2| = \begin{vmatrix} a & f & 1 \\ 1 & g & b \\ 1 & h & b \end{vmatrix} = 0 \Rightarrow g = h \)
and \( |A_1| = \begin{vmatrix} a & f & 1 \\ 1 & g & b \\ 1 & h & b \end{vmatrix} = 0 \Rightarrow g = h \)
\( \therefore g = h, c = d \) and \( ab = 1 \) .....(1)
Now \( BX = V \quad |B| = \begin{vmatrix} a & 1 & 1 \\ 0 & d & c \\ f & g & h \end{vmatrix} = 0 \)
(since, \( c_2 \) and \( c_3 \) are equal) (from equation (1)
\( \therefore BX = V \) has no solution
\( |B_1| = \begin{vmatrix} a^2 & 1 & 1 \\ 0 & d & c \\ 0 & g & h \end{vmatrix} = 0 \)
(since, \( c = d \) and \( g = h \)) (from equation (1)
\( |B_2| = \begin{vmatrix} a & a^2 & 1 \\ 0 & 0 & c \\ f & 0 & h \end{vmatrix} = a^2 cf = a^2 df \)
(since \( c = d \))
since \( adf \neq 0 \Rightarrow |B_2| \neq 0 \)
\( \therefore |B| = 0 \) and \( |B_2| \neq 0 \)
\( \therefore BX = V \) has no solution

Question. \( A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix} \), \( I= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) & \( A^{-1} = \frac{1}{6}(A^2 + cA + dI) \), then the value of \( c \) and \( d \) are
(a) (-6, -11)
(b) (6, 11)
(c) (-6, 11)
(d) (6, -11)
Answer: (c) (-6, 11)
Solution:
Every square matrix satisfied its characteristic equation
i.e. \( |A - \lambda I| = 0 \)
\( \Rightarrow \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & -2 & 4-\lambda \end{vmatrix} = 0 \)
\( \Rightarrow (1 - \lambda) \{(1 - \lambda) (4 - \lambda) + 2\} = 0 \)
\( \Rightarrow \lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0 \)
\( \Rightarrow A^3 - 6A^2 + 11A - 6I = 0 \) ....(1)
Given, \( 6A^{-1} = A^2 + cA + dI \), multiplying both sides by \( A \), we get
\( 6I = A^3 + cA^2 + dA \)
\( \Rightarrow A^3 + cA^2 + dA - 6I = 0 \) ....(2)
On comparing equation (1) and (2) we get
\( c = - 6 \) and \( d = 11 \)

Question. If \( P = \begin{bmatrix} \sqrt{3}/2 & 1/2 \\ -1/2 & \sqrt{3}/2 \end{bmatrix} \) and \( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) and \( Q = PAP^T \) and \( x = P^T Q^{2005} P \) then \( x \) is equal to
(a) \( \begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} 4 + 2005\sqrt{3} & 6015 \\ 2005 & 4 - 2005\sqrt{3} \end{bmatrix} \)
(c) \( \frac{1}{4} \begin{bmatrix} 2 + \sqrt{3} & 1 \\ -1 & 2 - \sqrt{3} \end{bmatrix} \)
(d) \( \frac{1}{4} \begin{bmatrix} 2005 & 2 - \sqrt{3} \\ 2 + \sqrt{3} & 2005 \end{bmatrix} \)
Answer: (a) \( \begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix} \)

Comprehension : Read the passage given below and answer the equations that follows.
Let \( A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \), \( U_1, U_2 \) and \( U_3 \) are columns matrices satisfying \( AU_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( AU_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \), \( AU_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} \) and \( U \) is \( 3 \times 3 \) matrix whose columns are \( U_1, U_2, U_3 \) then answer the following questions.

Question. (a) The value of \( |U| \) is
(a) 3
(b) -3
(c) 3/2
(d) 2
Answer: (a) 3
Solution:
(a) Let \( U_1 = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \), so that \( \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \)
\( \Rightarrow x = 1, y = -2, z = 1 \)
\( \therefore U_1 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \)
similarly \( U_2 = \begin{bmatrix} 2 \\ -1 \\ -4 \end{bmatrix} \), \( U_3 = \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \)
Hence \( U = \begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix} \)
\( \therefore |U| = 3 \)

Question. (b) The sum of the elements of the matrix \( U^{-1} \) is
(a) -1
(b) 0
(c) 1
(d) 3
Answer: (b) 0
Solution:
(b) More over
\( \text{adj } U = \begin{bmatrix} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{bmatrix} \)
\( \therefore U^{-1} = \frac{\text{adj } U}{3} \) and sum of the \( U^{-1} = 0 \)

Question. (c) The value of \( [3 \quad 2 \quad 0] U \begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix} \) is
(a) 5
(b) 5/2
(c) 4
(d) 3/2
Answer: (a) 5
Solution:
(c) The value of \( [3 \quad 2 \quad 0] U \begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix} \)
\( = [3 \quad 2 \quad 0] \begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix} \)
\( = [-1 \quad 4 \quad 4] \begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix} \)
\( = [5] \)

Question. (a) Consider three points \( P = (-\sin(\beta - \alpha), -\cos\beta) \), \( Q = (\cos(\beta - \alpha), \sin \beta) \) and \( R = (\cos(\beta - \alpha + \theta), \sin (\beta - \theta)) \), where \( 0 < \alpha, \beta, \theta < \pi/4 \)
(a) P lies on the line segment RQ
(b) Q lies on the line segment PR
(c) R lies on the line segment QP
(d) P, Q, R are non collinear
Answer: (d) P, Q, R are non collinear
Solution:
(a) Join P & Q,
Let T divides PQ in ratio \( \cos\theta : \sin\theta \), then
\( \left( \frac{\cos(\beta - \alpha)\cos\theta - \sin(\beta - \alpha)\sin\theta}{\cos\theta + \sin\theta}, \frac{\sin\beta\cos\theta - \cos\beta\sin\theta}{\cos\theta + \sin\theta} \right) \)
\( \because \) P, T, Q are collinear
\( \therefore \) P, Q, R are non – collinear.

Question. (b) Consider the system of equations \( x - 2y + 3z = -1 \)
\( -x + y - 2z = k \)
\( x - 3y + 4z = 1 \).
Statement-I : The system of equation has no solution for \( k \neq 3 \).
because
Statement-II : The determinant \( \begin{vmatrix} 1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1 \end{vmatrix} \neq 0, \text{for } k \neq 3 \)
(a) Statement-I is true, statement-II is true; statement-II is correct explanation for statement-I
(b) Statement-I is true, statement-II is true; statement-II is NOT correct explanation for statement-I
(c) Statement-I is true, Statement-II is False
(d) Statement-I is False, Statement-II is True
Answer: (a) Statement-I is true, statement-II is true; statement-II is correct explanation for statement-I
Solution:
(b) The given system of equation can be expressed as
\( \begin{bmatrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ k \\ 1 \end{bmatrix} \)
Applying \( R_2 \rightarrow R_2 + R_1 \), \( R_3 \rightarrow R_3 - R_1 \)
\( = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ k-1 \\ 2 \end{bmatrix} \xrightarrow{R_3 \rightarrow R_3 - R_2} \)
when \( k \neq 3 \) the given system of equations has no solution
\( \Rightarrow \) statement I is true clearly statement II is also true as it is rearrangement of rows and columns of \( \begin{bmatrix} 1 & -2 & 3 \\ 1 & -3 & 4 \\ -1 & 1 & -2 \end{bmatrix} \)
Hence option (A) is correct

Question. Match the following
Column-I
(A) The minimum value of \( \frac{x^2 + 2x + 4}{x + 2} \) is
(B) Let \( A \) and \( B \) be \( 3 \times 3 \) matrices of real numbers, where \( A \) is symmetric, \( B \) is skew-symmetric, and \( (A + B) (A - B) = (A - B) (A + B) \). If \( (AB)^t = (-1)^k AB \), where \( (AB)^t \) is the transpose of the matrix \( AB \), then the possible values of \( k \) are
(C) Let \( a = \log_3 \log_3 2 \). An integer \( k \) satisfying \( 1 < 2^{-k + 3^{-a}} < 2 \), must be
(D) If \( \sin \theta = \cos \phi \), then the possible values of \( \frac{1}{\pi} \left( \theta \pm \phi - \frac{\pi}{2} \right) \) are
Column-II
(P) 0
(Q) 1
(R) 2
(S) 3

Answer:
(A) \(\rightarrow\) (R)
(B) \(\rightarrow\) (Q), (S)
(C) \(\rightarrow\) (Q)
(D) \(\rightarrow\) (P), (R)
Solution:
(A) Let \( y = \frac{x^2 + 2x + 4}{x + 2} \)
\( \Rightarrow x^2 + (2 - y) x + 4 - 2y = 0 \)
for real x, \( D \geq 0 \)
\( \Rightarrow (2 - y)^2 - 4 (4 - 2y) \geq 0 \)
\( \Rightarrow y^2 + 4y - 12 \geq 0 \)
\( \Rightarrow (y + 6) (y - 2) \geq 0 \)
\( \Rightarrow y \leq -6 \) or \( y \geq 2 \)
\( \therefore \) Minimum Value = 2

(B) Given : \( A^t = A \); \( B^t = (-B) \)
and \( (A + B) (A - B) = (A - B) (A + B) \)
\( \Rightarrow A^2 + B^2 - AB + BA = A^2 + B^2 + AB - BA \)
\( \Rightarrow AB = BA \)
Now, \( (AB)^t = B^t A^t = -BA = -AB \)
\( \therefore k = 1, 3 \)

(C) Given, \( a = \log_3 \log_3 2 \)
\( \Rightarrow \log_3 2 = 3^a \Rightarrow 3^{-a} = \log_2 3 \)
Now, \( 1 < 2^{-k + 3^{-a}} < 2 \)
\( \Rightarrow 1 < 2^{-k} \cdot 2 \log_2 3 < 2 \)
\( \Rightarrow 1 < 3 \cdot 2^{-k} < 2 \)
\( \Rightarrow \frac{1}{3} < 2^{-k} < \frac{2}{3} \)
\( \Rightarrow \frac{3}{2} < 2^k < 3 \)
\( \Rightarrow \log_2 \left(\frac{3}{2}\right) < k < \log_2 3 \)
Solve & get value of k

(D) Given \( \sin \theta = \cos \phi \)
\( \Rightarrow \cos \left(\frac{\pi}{2} - \theta\right) = \cos \phi \Rightarrow \left(\frac{\pi}{2} - \theta\right) = 2n\pi \pm \phi \)
\( \Rightarrow \left(\theta \pm \phi - \frac{\pi}{2}\right) = -2n\pi \)
\( \Rightarrow \frac{1}{\pi} \left(\theta \pm \phi - \frac{\pi}{2}\right) = -2n \)
for \( n = 0 \); we get (0)
for \( n = 1 \); we get (–2)
for \( n = (-1) \); we get (2)

Comprehension : Read the passage given below and answer the equations that follows.
Let A be the set of all \( 3 \times 3 \) symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

Question. (a) The number of matrices in A is
(a) 12
(b) 6
(c) 9
(d) 3
Answer: (a) 12
Solution:
(a) Since A is a symmetric matrix and its five entries are 1 and 4 entries are zero. So, following cases are possible : -
(i) When 2 entries of principal diagonal are zero:-
Total matrices = \( ^3C_2 \times ^3C_1 = 3 \times 3 = 9 \)
(ii) If all entries of principal diagonal are 1
Total matrices = \( ^3C_2 = 3 \)
Hence, total matrices = 9 + 3 = 12

Question. (b) The number of matrices A in A for which the system of linear equations \( A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \) has a unique solution, is
(a) less than 4
(b) at least 4 but less than 7
(c) at least 7 but less than 10
(d) at least 10
Answer: (b) at least 4 but less than 7
Solution:
(b) For unique solution, \( |A| \neq 0 \); Let \( A= \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix} \)
Possible matrices such that \( |A| \neq 0 \) are:
Case–1: \( \begin{bmatrix} 0 & b & c \\ b & 0 & e \\ c & e & 1 \end{bmatrix} \Rightarrow \) for \( c = 0 \) or \( e = 0 \Rightarrow |A| \neq 0 \)
Hence, 2 matrices are possible.
Case–2: \( \begin{bmatrix} 1 & b & c \\ b & 0 & e \\ c & e & 0 \end{bmatrix} \Rightarrow \) for \( b = 0 \) or \( c = 0 \Rightarrow |A| \neq 0 \)
Hence, 2 matrices are possible.
Case–3: \( \begin{bmatrix} 0 & b & c \\ b & 1 & e \\ c & e & 0 \end{bmatrix} \Rightarrow \) for \( b = 0 \) or \( e = 0 \Rightarrow |A| \neq 0 \)
Hence, 2 matrices are possible.
Case–4: \( \begin{bmatrix} 1 & b & c \\ b & 1 & e \\ c & e & 1 \end{bmatrix} \)
\( \Rightarrow \) for \( b = c = 0, |A| = 0 \)
for \( c = e = 0, |A| = 0 \)
for \( b = e = 0, |A| = 0 \)
Hence, no matrix is possible
\( \therefore \) Total matrice = 2 + 2 + 2 = 6

Question. (c) The number of matrices A in A for which the system of linear equations \( A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \) is inconsistent, is
(a) 0
(b) more than 2
(c) 2
(d) 1
Answer: (b) more than 2
Solution:
(c) Six matrices (Augmented) for which \( |A| = 0 \)
(i) \( \begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 \end{pmatrix} \Rightarrow \) Inconsistent
(ii) \( \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \Rightarrow \) Inconsistent
(iii) \( \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \Rightarrow \) Infinite
(iv) \( \begin{pmatrix} 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \Rightarrow \) Inconsistent
(v) \( \begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \end{pmatrix} \Rightarrow \) Inconsistent
(vi) \( \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{pmatrix} \Rightarrow \) Infinite

Question. The number of \( 3 \times 3 \) matrices A whose entries are either 0 or 1 and for which the system \( A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \) has exactly two distinct solutions is
(a) 0
(b) \( 2^9 - 1 \)
(c) 168
(d) 2
Answer: (a) 0
Solution:
Given system will give equation of 3 planes but 3 planes cannot intersect at two distinct points. Hence, no matrix possible.

Comprehension : Read the passage given below and answer the equations that follows.
Let p be an odd prime number and \( T_p \) be the following set of \( 2 \times 2 \) matrices :
\( T_p = \left\{ A = \begin{bmatrix} a & b \\ c & a \end{bmatrix} ; a, b, c \in \{0, 1, 2, \dots, p-1\} \right\} \)

Question. (a) The number of A in \( T_p \) such that A is either symmetric or skew-symmetric or both, and \(\det(A)\) divisible by p is
(a) \( (p - 1)^2 \)
(b) \( 2(p - 1) \)
(c) \( (p - 1)^2 - 1 \)
(d) \( 2p - 1 \)
Answer: (d) 2p - 1

Question. (b) The number of A in \( T_p \) such that the trace of A is not divisible by p but det (A) is divisible by p is
[Note : The trace of a matrix is the sum of its diagonal entries]
(a) \( (p - 1)(p^2 - p + 1) \)
(b) \( p^3 - (p - 1)^2 \)
(c) \( (p - 1)^2 \)
(d) \( (p - 1) (p^2 - 2) \)
Answer: (c) \( (p - 1)^2 \)

Question. (c) The number of A in \( T_p \) such that \(\det(A)\) is not divisible by p is
(a) \( 2p^2 \)
(b) \( p^3 - 5p \)
(c) \( p^3 - 3p \)
(d) \( p^3 - p^2 \)
Answer: (d) \( p^3 - p^2 \)

Question. Let k be a positive real number and let
\( A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & 1 \end{bmatrix} \) & \( B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix} \)
If \( \det(\text{adj } A) + \det(\text{adj } B) = 10^6 \), then {k} is equal to
{Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k}
Answer: [k] = 4
Solution:
After applying suitable transformation,
\( |A| = (2k + 1)^3 \) and \( |B| = 0 \) (Skew symmetric of odd order)
Now, \( |\text{adj } A| = |A|^{n-1} = |A|^{3-1} = |A|^2 \)
Thus, \( \det (\text{adj } A) + \det (\text{adj } B) = 10^6 \)
\( \Rightarrow |A|^2 + |B|^2 = 10^6 \Rightarrow ((2k + 1)^3)^2 = 10^6 \)
\( \Rightarrow 2k + 1 = 10 \Rightarrow k = \frac{9}{2} = 4.5 \Rightarrow [k] = 4 \)

Question. Let \( M \) and \( N \) be two \( 3 \times 3 \) non-singular skew-symmetric matrices such that \( MN = NM \). If \( P^T \) denotes the transpose of P, then \( M^2 N^2 (M^T N)^{-1} (M N^{-1})^T \) is equal to
(a) \( M^2 \)
(b) \( -N^2 \)
(c) \( -M^2 \)
(d) MN
Answer: (c) \( -M^2 \)
Solution:
Given: \( M^T = (-M) \); \( N^T = (-N) \); \( MN = NM \)
Now, \( M^2 N^2 (M^T N)^{-1} (M N^{-1})^T \)
\( \Rightarrow M M N N (N^{-1}) (M^T)^{-1} (N^{-1})^T M^T \)
\( \Rightarrow M N M I (M^T)^{-1} (N^T)^{-1} M^T \)
\( \Rightarrow -M N M^T (M^T)^{-1} (N^T)^{-1} M^T \Rightarrow -M(N I) (N^T)^{-1} M^T \)
\( \Rightarrow M N^T (N^T)^{-1} M^T \Rightarrow (M I) M^T \)
\( \Rightarrow -M M \Rightarrow -M^2 \)

Question. Let M be a \( 3 \times 3 \) matrix satisfying \( M \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \), \( M \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \), and \( M \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} \).
Then the sum of the diagonal entries of M is
Answer: 9
Solution:
Let \( M = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \)
Now, \( M \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \)
\( \Rightarrow a_2 = (-1); b_2 = 2; c_2 = 3 \)
Now, \( M \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} a_1 & -1 & a_3 \\ b_1 & 2 & b_3 \\ c_1 & 3 & c_3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \)
\( \Rightarrow a_1 = 0 ; b_1 = 3; c_1 = 2 \)
Now, \( M \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 0 & -1 & a_3 \\ 3 & 2 & b_3 \\ 2 & 3 & c_3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} \)
\( \Rightarrow c_3 = 7 \quad \therefore \) sum of diagonal elements = \( a_1 + b_2 + c_3 = 0 + 2 + 7 = 9 \)

Question. Let \( P = [a_{ij}] \) be a \( 3 \times 3 \) matrix and let \( Q = [b_{ij}] \), where \( b_{ij} = 2^{i + j} a_{ij} \) for \( 1 \leq i, j \leq 3 \). If the determinant of P is 2, then the determinant of the matrix Q is
(a) \( 2^{10} \)
(b) \( 2^{11} \)
(c) \( 2^{12} \)
(d) \( 2^{13} \)
Answer: (d) \( 2^{13} \)
Solution:
Let \( P = \begin{bmatrix} a & b & c \\ d & e & f \\ p & q & r \end{bmatrix} \)
Now, \( P = [a_{ij}] \) & \( Q = [b_{ij}] \) and \( b_{ij} = 2^{i+j} a_{ij} \)
\( \therefore Q = \begin{bmatrix} 2^2 a & 2^3 b & 2^4 c \\ 2^3 d & 2^4 e & 2^5 f \\ 2^4 p & 2^5 q & 2^6 r \end{bmatrix} \)
\( \Rightarrow |Q| = 2^{12} |P| = 2^{12} \cdot 2 = 2^{13} \)

Question. If P is a \( 3 \times 3 \) matrix such that \( P^T = 2P + I \), where \( P^T \) is the transpose of P and I is the \( 3 \times 3 \) identity matrix, then there exists a column matrix \( X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \) such that
(a) \( PX = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \)
(b) \( PX = X \)
(c) \( PX = 2X \)
(d) \( PX = -X \)
Answer: (d) \( PX = -X \)
Solution:
Let \( P = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \)
\( P^T = 2P + I \Rightarrow \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} = \begin{bmatrix} 2a_1 & 2a_2 & 2a_3 \\ 2b_1 & 2b_2 & 2b_3 \\ 2c_1 & 2c_2 & 2c_3 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} = \begin{bmatrix} 2a_1 + 1 & 2a_2 & 2a_3 \\ 2b_1 & 2b_2 + 1 & 2b_3 \\ 2c_1 & 2c_2 & 2c_3 + 1 \end{bmatrix} \)
On comparing corresponding elements,
\( a_1 = (-1); b_2 = (-1); c_3 = (-1) \)
Also, \( b_1 = c_1 = a_2 = c_2 = a_3 = b_3 = 0 \)
Hence, \( P = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = (-I) \)
\( \Rightarrow PX = (-I)X = (-X) \)

Question. If the adjoint of a \( 3 \times 3 \) matrix P is \( \begin{bmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{bmatrix} \), then the possible value(s) of the determinant of P is (are)
(a) -2
(b) -1
(c) 1
(d) 2
Answer: (a) -2, (d) 2
Solution:
As we know that,
\( |\text{adj } A| = |A|^{n-1} \)
Here, \( |\text{adj } P| = |P|^{3-1} = |P|^2 \)
Now, \( |\text{adj } P| = \begin{vmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{vmatrix} \)
\( \Rightarrow |\text{adj } P| = 1(3 - 7) - 4(6 - 7) + 4 (2 - 1) \)
\( \Rightarrow |\text{adj } P| = 4 \Rightarrow |P|^2 = 4 \)
\( \Rightarrow |P| = \pm 2 \Rightarrow |P| = 2 \) or \( |P| = (-2) \)