Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha,…

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
  1. $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(1)$
  2. $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)$
  3. $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$
  4. $(\mathrm{P}) \rightarrow(1) \quad(\mathrm{Q}) \rightarrow(5) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(4)$

Solution

$\begin{aligned} & \alpha, \beta \text { are roots of } \mathrm{x}^2+\mathrm{x}-1=0 \\ & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll}\mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33}\end{array}\right]\end{aligned}$ (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 $\alpha \beta=\alpha+\beta=-1$ )

Asked in: JEE Advanced 2024 (Paper 1)

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