Let A be a 2 × 2 real matrix and I be the identity matrix of order...

Given: $| A - xI | = 0$ have roots $- 1$ and $3$.

Let, A = \left[\begin{cases}a & b \\ c & d\end{cases}\right]

⇒ A - xI = \left[\begin{cases}a & b \\ c & d\end{cases}\right] - \left[\begin{cases}x & 0 \\ 0 & x\end{cases}\right]

⇒ A - xI = \left[\begin{cases}a - x & b \\ c & d - x\end{cases}\right]

$⇒ \left|A - xI\right| = \left(a - x\right)\left(d - x\right) - bc$

$⇒ \left|A - xI\right| = ad - ax - dx + {x}^{2} - bc$

$⇒ \left|A - xI\right| = {x}^{2} - \left(a + d\right)x + \left(ad - bc\right)$

Sum of roots, $a + d = 2$

Product of roots, $ad - bc = - 3$

Now finding matrix ${A}^{2}$,

⇒ {A}^{2} = \left[\begin{cases}a & b \\ c & d\end{cases}\right] \times \left[\begin{cases}a & b \\ c & d\end{cases}\right]

⇒ {A}^{2} = \left[\begin{cases}{a}^{2} + bc & ab + bd \\ ac + cd & bc + {d}^{2}\end{cases}\right]

Sum of diagonal elements, $S = {a}^{2} + bc + bc + {d}^{2}$

$⇒ S = {a}^{2} + 2bc + {d}^{2}$

$⇒ S = ( a + d{)}^{2} - 2\left(ad - bc\right)$

$⇒ S = ( 2{)}^{2} - 2\left(- 3\right)$

$⇒ S = 4 + 6$

$⇒ S = 10$