<div><strong>MATRIX MATCH TYPE QUESTION</strong><br><br><table><thead><tr><th>Column-I</th><th>Column-II</th></tr></thead><tbody><tr><td>(A) Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$, where each of $a, b$ and $c$ is either $\omega$ or $\omega^2$. Then the number of distinct matrices in the set $S$ is-</td><td>(P) 0</td></tr><tr><td>(B) Let $M$ be $3 \times 3$ 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</td><td>(Q) 4</td></tr><tr><td>(C) The number of $3 \times 3$ matrices $A$ whose entries are either 0 or 1 and for which the system $A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ has exactly two distinct solutions, is</td><td>(R) 9</td></tr><tr><td>(D) Let $k$ be a positive real number and let $A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$. If $\det(\text{adj } A) + \det(\text{adj } B) = 10^6$, then $[k]$ is equal to<br>[Note: $\text{adj } M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$].</td><td>(S) 2</td></tr></tbody></table></div>
Step-by-Step Solution
Key Concept: General
<div>(A) The determinant of the matrix is $\det = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c$. For $a, c \in \{\omega, \omega^2\}$, only the pair $(a,c) = (\omega, \omega)$ results in a non-zero determinant: $\det = 1 - \omega^2 - \omega^2 + \omega^4 = 1 + \omega - 2\omega^2 = -3\omega^2 \neq 0$. For this pair, $b$ can be either $\omega$ or $\omega^2$, giving 2 distinct matrices. Thus, (A) matches (S).<br><br>(B) Let $M = [C_1 C_2 C_3]$. From the given equations, $C_2 = [-1, 2, 3]^T$, $C_1 - C_2 = [1, 1, -1]^T \implies C_1 = [0, 3, 2]^T$, and $C_1 + C_2 + C_3 = [0, 0, 12]^T \implies C_3 = [1, -5, 7]^T$. The sum of the diagonal entries is $0 + 2 + 7 = 9$. Thus, (B) matches (R).<br><br>(C) A system of linear equations $AX=B$ can have zero, one, or infinitely many solutions. It can never have exactly two distinct solutions. Therefore, the number of such matrices is 0. Thus, (C) matches (P).<br><br>(D) Calculating the determinant of $A$, we find $\det(A) = (2k+1)^3$. Since $B$ is a $3 \times 3$ skew-symmetric matrix, $\det(B) = 0$. Then $\det(\text{adj } A) = (\det A)^2 = (2k+1)^6$ and $\det(\text{adj } B) = 0$. Given $(2k+1)^6 = 10^6$, we have $2k+1 = 10$, which gives $k = 4.5$. Thus, $[k] = 4$. Thus, (D) matches (Q).</div>
Correct Answer: (A) -> (S), (B) -> (R), (C) -> (P), (D) -> (Q)