Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}$, where $\mathrm{a}_{\mathrm{ij}} \neq 0$ for all $\mathrm{i}, \mathrm{j}$ and $\mathrm{A}^{2}=\mathrm{I}$. Let a be the sum of all diagonal elements of $\mathrm{A}$ and $\mathrm{b}=|\mathrm{A}|$. Then $3 a^{2}+4 b^{2}$ is equal to :

Given, $A^2=I$ <br/><br/>and $b=|A|$ <br/><br/>Let $$ A=\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] $$ <br/><br/>$$ \begin{aligned} \therefore \quad A^2 & =\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right]\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] \\\\ & =\left[\begin{array}{cc} a_1^2+b_1 a_2 & a_1 b_1+b_1 b_2 \\ a_1 a_2+a_2 b_2 & b_1 a_2+b_2^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \left(\because A^2=I\right) \end{aligned} $$ <br/><br/>By equality of matrices, we get <br/><br/>$$ \begin{aligned} & a_1 a_2+a_2 b_2=0 \text { and } a_1 b_1+b_1 b_2=0 \\\\ &\Rightarrow a_2\left(a_1+b_2\right)=0 \text { and } b_1\left(a_1+b_2\right)=0 \\\\ &\Rightarrow a_1+b_2=0 \text { (since, } b_1, a_2 \neq 0 \text { ) }\\\\ &\Rightarrow \text { Sum of diagonal elements }=0 \\\\ &\Rightarrow a=0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Now, } A^2=I \\\\ & \Rightarrow \left|A^2\right|=1 \\\\ & \Rightarrow |A|^2=1 ~~~~~~~\left(\because\left|A^n\right|=|A|^n\right)\\\\ & \Rightarrow b^2=1 ~~~~~~~~~~~(\because|A|=b)\\\\ & \therefore 3 a^2+4 b^2=3(0)+4(1)=4 \end{aligned} $$