Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $|\mathrm{A}-x \mathrm{I}|=0$ be $-1$ and 3, then the sum of the diagonal elements of the matrix $\mathrm{A}^2$ is

<p>$|A-x I|=0$</p> <p>Roots are $-$1 and 3</p> <p>Sum of roots $=\operatorname{tr}(A)=2$</p> <p>Product of roots $=|\mathrm{A}|=-3$</p> <p>Let $$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$</p> <p>We have $\mathrm{a}+\mathrm{d}=2$</p> <p>$\mathrm{ad}-\mathrm{bc}=-3$</p> <p>$$A^2=\left[\begin{array}{ll}a & b \\ c & d \end{array}\right] \times\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]$$</p> <p>We need $a^2+b c+b c+d^2$</p> <p>$$\begin{aligned} & =a^2+2 b c+d^2 \\ & =(a+d)^2-2 a d+2 b c \\ & =4-2(a d-b c) \\ & =4-2(-3) \\ & =4+6 \\ & =10 \end{aligned}$$</p>