Let A=[a_ij]_2 2, where a_ij 0 for all i,j and A^2=I.

Question

Let $A={[{a}_{ij}]}_{2	imes 2}$, where ${a}_{ij}
e 0$ 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 :

Accepted Answer

Given, ${A}^{2}=I$

and $b=|A|$

Let
$A=[\begin{gathered} {a}_{1} {b}_{1} \ {a}_{2} {b}_{2} ]$

$\begin{gathered} 	herefore \, {A}^{2} \quad =[{a}_{1} \quad {b}_{1} \ {a}_{2} \quad {b}_{2}][{a}_{1} \quad {b}_{1} \ {a}_{2} \quad {b}_{2}] \ =[{a}_{2}^{1}+{b}_{1}{a}_{2} \quad {a}_{1}{b}_{1}+{b}_{1}{b}_{2} \ {a}_{1}{a}_{2}+{a}_{2}{b}_{2} \quad {b}_{1}{a}_{2}+{b}_{2}^{2}]=[1 \quad 0 \ 0 \quad 1 \end{gathered}](\because {A}^{2}=I) \end{gathered}$

By equality of matrices, we get

$\begin{gathered} {a}_{1}{a}_{2}+{a}_{2}{b}_{2}=0{a}_{1}{b}_{1}+{b}_{1}{b}_{2}=0 \ \Rightarrow {a}_{2}({a}_{1}+{b}_{2})=0{b}_{1}({a}_{1}+{b}_{2})=0 \ \Rightarrow {a}_{1}+{b}_{2}=0{b}_{1},{a}_{2}
e 0 \ \Rightarrow =0 \ \Rightarrow a=0 \end{gathered}$

$\begin{gathered} {A}^{2}=I \ \Rightarrow |{A}^{2}|=1 \ \Rightarrow |A{|}^{2}=1(\because |{A}^{n}|=|A{|}^{n}) \ \Rightarrow {b}^{2}=1(\because |A|=b) \ 	herefore 3{a}^{2}+4{b}^{2}=3(0)+4(1)=4 \end{gathered}$

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