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 \(M=\left(a_{i j}\right)_{3 \times 3}\), define \(R_i=a_{i 1}+a_{i 2}+a_{i 3}\) and \(C_j=a_{1 j}+a_{2 j}+a_{3 j}\) for \(i=1,2,3\) and \(j=1,2,3\).
Match each entry in List-I to the correct entry in List-II.

The correct option is

\(\begin{aligned} & \alpha, \beta \text { are roots of } \mathrm{x}^2+\mathrm{x}-1=0 \\ & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll}\mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33}\end{array}\right]\end{aligned}\)
(P)
\(M=\left[\begin{array}{lll}1 & \alpha & \beta \\\alpha & \beta & 1 \\\beta & 1 & \alpha\end{array}\right] \Rightarrow 3!\times 2=12\)
For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3 ! ways
\(\therefore 3!\times 2=12 \text { ways }\)
(Q)
\(M=\left[\begin{array}{lll}x & a & b \\ a & y & c \\ b & c & z\end{array}\right] \Rightarrow\) Consider one such arrangement with \(a=\alpha, b=\beta, c=1\)
\(M=\left[\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}\right]\)
a, b, c can be arranged in \(3 !\) ways and corresponding entries can be arranged in 1 way.
(R)
\(\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}a \\ 0 \\ -c\end{array}\right]\)
\(\begin{aligned}& a y+b z=a \\& -a x+c z=0 \\& -b x-c y=-c\end{aligned}\)
It is observed that \(\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0\)
\(\therefore\) infinite solution
(S)
\(\left[\begin{array}{lll}1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta\end{array}\right]\)
\(\Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad(\) since \(\alpha \beta=\alpha+\beta=-1\) )