JEE Main 2024 — 4 April, Shift 2 — Question 25

Let A=\left[ \begin{array}{*{35}{l}}a & b \\b & d \\\end{array} \right]

\left[ \begin{array}{*{35}{l}}\text{a} & \text{b} \\{b} & \text{d} \\\end{array} \right]\left[ \begin{array}{*{35}{l}}1 \\1 \\\end{array} \right]=\left[ \begin{array}{*{35}{l}}3 \\7 \\\end{array} \right],\text{ad}-{{\text{b}}^{2}}=1

$a+b=3,b+d=7,\left( 3-b \right)\left( 7-b \right)-{{b}^{2}}=1$

$21-10b=1\to b=2,a=1,d=5$

A=\left[ \begin{array}{*{35}{l}}1 & 2 \\2 & 5 \\\end{array} \right],{{A}^{-1}}=\left[ \begin{matrix}5 & -2 \\-2 & 1 \\\end{matrix} \right]

${{\text{A}}^{-1}}=\alpha \text{A}+\beta \text{I}$

$\left[ \begin{matrix}5 & -2 \\-2 & 1 \\\end{matrix} \right]=\left[ \begin{matrix}\alpha +\beta & 2\alpha \\2\alpha & 5\alpha +\beta \\\end{matrix} \right]$

$\alpha =-1,\beta =6\to \alpha +\beta =5$