$ \begin{aligned} & x^2+x-1=0 \rightarrow \text { roots are } \alpha \text { and } \beta \\ & \alpha+\beta=-1 \quad \alpha \beta=-1 \\ & \text { Set } T=\{1, \alpha, \beta\} M=\left(a_{i j}\right)_{3 \times 3} \\ & R_i=a_{i 1}+a_{i 2}+a_{i 3} \quad C_j=a_{1 j}+a_{2 j}+a_{3 j}(P) \mathrm{R} \_\mathrm{i}=\mathrm{C} \_\mathrm{j}=0 \text { for all } \mathrm{i}, \mathrm{j} \\ & \alpha+\beta=-1 \quad T=\{1, \alpha, \beta\} \end{aligned} $ Number of matrices $ \begin{aligned} & =\underset{\downarrow}{=} \underset{\substack{3 \\ \text { Number of } \\ \text { ways to arrange }}}{2 \times 1=12} \quad \begin{array}{lll} \text { Number of } & \\ 1, \alpha, \beta \text { in } R_1 & \begin{array}{l} \text { ways to arrange } \\ 1, \alpha, \beta \text { in } R_2 \end{array} \end{array} \quad\left[\begin{array}{ccc} 1 & \alpha & \beta \\ & - & - \end{array}\right] \end{aligned} $ (Q) Number of symmetric matrices = ? $ C_j=0 \forall j $ Number of symmetric matrices $ =\underline{3} \times 1=6\left[\begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{array}\right] $ (R) $M \rightarrow$ skew symmetric of $3 \times 3$ $ \begin{aligned} & |M|=0 \quad a_{i j} \in T \text { for } i>j \\ & M\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} a_{12} \\ 0 \\ -a_{23} \end{array}\right) \\ & {\left[\begin{array}{ccc} 0 & -a_{21} & -a_{31} \\ a_{21} & 0 & -a_{32} \\ a_{31} & a_{32} & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a_{12} \\ 0 \\ -a_{23} \end{array}\right]} \end{aligned} $ As $x, y, z \in R$ and $a_{12} \& a_{23} \in R \quad \&|M|=0$ ∴ System has infinite solutions $ \begin{aligned} & \text { (S) } R_i=0 \forall i \\ & M=\left[\begin{array}{lll} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{array}\right] C_1 \rightarrow C_1+C_2+C_3|M|=\left|\begin{array}{lll} 1+\alpha+\beta & \alpha & \beta \\ 1+\alpha+\beta & \beta & 1 \\ 1+\alpha+\beta & 1 & \alpha \end{array}\right|=0 \end{aligned} $ $(P) \rightarrow(2)(Q) \rightarrow(4)(R) \rightarrow(3)(S) \rightarrow(5)$