Question 14

Given $x^2 + x - 1 = 0$, the roots $\alpha, \beta$ satisfy $\alpha + \beta = -1$, hence $1 + \alpha + \beta = 0$. Also, $\alpha, \beta \neq 0$.

(P) Since $R_i = C_j = 0$ and entries are from $\{1, \alpha, \beta\}$, each row and each column must be a permutation of $\{1, \alpha, \beta\}$. This forms a $3 \times 3$ Latin Square. The number of such matrices is $3! \times 2 = 6 \times 2 = 12$. So, (P) $\to$ (2).

(Q) For a symmetric matrix $M = \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$ with $C_j = 0$ (which implies $R_i = 0$ due to symmetry), the first row must be a permutation of $(1, \alpha, \beta)$ (6 ways). For each choice of $(a, b, c)$, the constraints $b+d+e=0$ and $c+e+f=0$ along with $d+b+e=0$ etc. result in exactly one unique symmetric matrix. Thus there are 6 such matrices. So, (Q) $\to$ (4).

(R) Let $M$ be a $3 \times 3$ skew-symmetric matrix. Thus $\det(M) = 0$. The system $M\mathbf{X} = \mathbf{B}$ where $\mathbf{B} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix}$ is consistent because $\mathbf{B}$ is the second column of $M$ (since $a_{12} = -a_{21}$ and $a_{32} = -a_{23}$). A consistent system with $\det(M) = 0$ and $x, y, z \in \mathbb{R}$ has infinitely many solutions. So, (R) $\to$ (3).

(S) $R_i = 0$ implies $a_{i1} + a_{i2} + a_{i3} = 0$ for each $i$. Summing all columns and replacing the first column with this sum makes the first column entries $R_1, R_2, R_3$, which are all zero. Thus $\det(M) = 0$. So, (S) $\to$ (5).

Correct Option: (C)