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3 and 4 .Determinants and Matrices
hard
Let $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j} \neq 0$ for all $i, j$ and $A^2=I$. Let a be the sum of all diagonal elements of $A$ and $b =| A |$, then $3 a ^2+4 b ^2$ is equal to
A
$7$
B
$14$
C
$3$
D
$4$
(JEE MAIN-2023)
Solution
$\begin{aligned}& \text { Let } A=\left[\begin{array}{ll} p & l \\r & s\end{array}\right] \\& A^2=\left[\begin{array}{ll} p ^2+ qr & pq + qs \\pr + rs & qs + s ^2\end{array}\right] \\& \Rightarrow p ^2+ qr =1(1) pq + qs =0 \Rightarrow q ( p + s )=0 \\& \Rightarrow s ^2+ qr =1(2) pr + rs =0 \Rightarrow r(p+s)=0\end{aligned}$
Equation $(1)$ – equation $(2)$
$p ^2= s ^2 \Rightarrow p + s =0$
Now $3 a^2+4 b^2$
$=3(p+s)^2+4(p s-q r)^2$
$=3.0+4\left(-p^2-q r\right)^2=4\left(p^2+q r\right)^2=4$
Standard 12
Mathematics