Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^{2}+x-1=0$. Consider the set $T = \{1, \alpha, \beta\}$ For a $3 \times 3$ matrix $M = (a_{ij})_{3\times3}$, define $R_i = a_{i1} + a_{i2} + a_{i3}$ and $C_j = a_{1j} + a_{2j} + a_{3j}$ for $i = 1, 2, 3$ and $j = 1, 2, 3$. Match each entry in List-I to the correct entry in List-II. | List-I | List-II | | --- | --- | | (P) The number of matrices $M = (a_{ij})_{3\times3}$ with all entries in $T$ such that $R_i = C_j = 0$ for all $i, j$ is | (1) $1$ | | (Q) The number of symmetric matrices $M = (a_{ij})_{3\times3}$ with all entries in $T$ such that $C_j = 0$ for all $j$, is | (2) $12$ | | (R) Let $M = (a_{ij})_{3\times3}$ be a skew symmetric matrix such that $a_{ij} \in T$ for $i > j$. Then the number of elements in the set $\left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R}, M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix} \right\}$ is | (3) infinite | | (S) Let $M = (a_{ij})_{3\times3}$ be a matrix with all entries in $T$ such that $R_i = 0$ for all $i$. Then the absolute value of the determinant of $M$ is | (4) $6$ | | | (5) $0$ | The correct option is

$x^2 + x - 1 = 0: \quad \alpha + \beta = -1, \ \alpha\beta = -1 \Rightarrow 1 + \alpha + \beta = 0$ $T = \{1, \alpha, \beta\}, \qquad M = (a_{ij})_{3\times3}$ $R_i = a_{i1}+a_{i2}+a_{i3}, \qquad C_j = a_{1j}+a_{2j}+a_{3j}$ $\text{To find: match } (P),(Q),(R),(S) \text{ with } (1)\text{-}(5)$ $1.\ \text{row sum } 0 \Rightarrow \text{row is a permutation of } (1,\alpha,\beta)$ $2.\ \text{count for } (P), (Q); \text{ determinant for } (S); \text{ system for } (R)$ $\text{row sum } 0 \text{ with entries in } T \Rightarrow \text{row is a permutation of } (1,\alpha,\beta)$ $(\text{only } 1 + \alpha + \beta = 0 \text{ works}; \text{ all other triples} \neq 0)$ $(P): \text{every row } \& \text{ column a permutation} \Rightarrow \text{Latin square of order } 3$ $\text{number of } 3\times3 \text{ Latin squares} = 12 \Rightarrow P \to (2)$ $(Q): \text{symmetric} \Rightarrow R_i = C_i; \ C_j = 0 \Rightarrow \text{rows are permutations of } T$ $\text{odd-order symmetric Latin square: diagonal uses } 1,\alpha,\beta \text{ once} \Rightarrow 3! = 6 \Rightarrow Q \to (4)$ $(S): R_i = 0 \Rightarrow M\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \det M = 0$ $|\det M| = 0 \Rightarrow S \to (5)$ $M = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}, \quad a,b,c \in T; \quad M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a \\ 0 \\ -c \end{pmatrix}$ $a y + b z = a, \quad -a x + c z = 0, \quad b x + c y = c$ $\det M = 0 \text{ (odd skew-symmetric)}, \quad \operatorname{rank} = 2 \Rightarrow \text{nullity } 1$ $\text{null vector } v = \left(\tfrac{c}{a}, -\tfrac{b}{a}, 1\right), \quad r\cdot v = a\cdot\tfrac{c}{a} - c = 0 \Rightarrow \text{consistent}$ $\Rightarrow \text{infinitely many solutions} \Rightarrow R \to (3)$ $P \to 2, \quad Q \to 4, \quad R \to 3, \quad S \to 5 \quad \Rightarrow \quad \text{option } (3)$