Roots are $-1$ and $3$

Sum of roots $=\operatorname{tr}(A)=2$

Product of roots $=|\mathrm{A}|=-3$

Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

We have $\mathrm{a}+\mathrm{d}=2$ $\mathrm{ad}-\mathrm{bc}=-3$

$A^2=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \times\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+c d & b c+d^2\end{array}\right]$

We need $a^2+b c+b c+d^2$

$ =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$