Solve this

Solution:

Given: $A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$

$A^{2}=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$

$=\left[\begin{array}{lll}4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0\end{array}\right]$

$=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]$

Now,

$A^{2}-5 A+4 I=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]+4\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4\end{array}\right]$

$=\left[\begin{array}{ccc}-1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2\end{array}\right]$

Now, $A^{2}-5 A+4 l+X=0$

$A^{2}-5 A+4 I=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]+4\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4\end{array}\right]$

$=\left[\begin{array}{ccc}-1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2\end{array}\right]$

Now, $A^{2}-5 A+4 l+X=0$

$\Rightarrow X=-\left(A^{2}-5 A+4 l\right)$

$\therefore X=-\left[\begin{array}{ccc}-1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2\end{array}\right]$