JEE Main 2023 (Online) 6th April Morning ShiftQuestion 1
Let $A=[a_{ij}]_{2\times2}$, where $a_{ij}\neq0$ for all $i,j$ and $A^{2}=I$. Let $a$ be the sum of all diagonal elements of $A$ and $b=|A|$. Then $3a^{2}+4b^{2}$ is equal to
Matrices
Solution
$A=\begin{bmatrix}p&q\\r&s\end{bmatrix}$, $A^2=I$ with all entries nonzero. $q(p+s)=0\Rightarrow p+s=0$ (as $q\neq0$), so $a=0$. Then $A^2=(p^2+qr)I=I$, so $\det A=-p^2-qr=-1$, $b=-1$. $3a^2+4b^2=4$. Option (B).