QJEEM24Mabaa2: JEE Main 2024 Mathematics - Class 12 JEEM24Mabaa2 Solution: Let be a symmetric matrix such that and the determinant of be 1 …

Let $A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]$

$|A|=1 \Rightarrow a c-b^2=0 \quad \text{... (i)}$

$\text { Given }\left[\begin{array}{ll}a & b \\b & c\end{array}\right]\left[\begin{array}{l}1 \\1\end{array}\right]=\left[\begin{array}{l}3 \\7\end{array}\right]$

$\begin{aligned}& \Rightarrow \quad a+b=3 \quad \text{... (ii)}\\& \text { and } b+c=7 \quad \text{... (iii)}\end{aligned}$

from (i), (ii) and (iii) $a=1, b=2, c=5$

$\begin{aligned}\Rightarrow & A=\left[\begin{array}{ll}1 & 2 \\2 & 5\end{array}\right] \Rightarrow A^{-1}=\left[\begin{array}{cc}5 & -2 \\-2 & 1\end{array}\right] \\& \text { Given } A^{-1}=\alpha A+\beta I \\\Rightarrow & {\left[\begin{array}{cc}5 & -2 \\-2 & 1\end{array}\right]=\alpha\left[\begin{array}{ll}1 & 2 \\2 & 5\end{array}\right]+\beta\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] } \\\Rightarrow & \alpha=-1 \text { and } \beta=6 \\& \alpha+\beta=5\end{aligned}$