Solution Given: A2−5A+4I−X=0,A=[2012131−10]A^2-5A+4I-X=0,\qquad A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}A2−5A+4I−X=0,A=22101−1130 Rearrange: X=A2−5A+4IX=A^2-5A+4IX=A2−5A+4I Step 1: Compute A2A^2A2 A2=A⋅A=[2012131−10][2012131−10]=[5−129−250−1−2]A^2=A\cdot A= \begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix} \begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix} = \begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}A2=A⋅A=22101−113022101−1130=590−1−2−125−2 Step 2: Compute −5A-5A−5A −5A=[−100−5−10−5−15−550]-5A= \begin{bmatrix}-10&0&-5\\-10&-5&-15\\-5&5&0\end{bmatrix}−5A=−10−10−50−55−5−150 Step 3: Compute 4I4I4I 4I=[400040004]4I=\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}4I=400040004 Step 4: Add to get XXX X=A2−5A+4I=[5−129−250−1−2]+[−100−5−10−5−15−550]+[400040004]=[−1−1−3−1−3−10−542]\begin{aligned} X&=A^2-5A+4I\\ &= \begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix} + \begin{bmatrix}-10&0&-5\\-10&-5&-15\\-5&5&0\end{bmatrix} + \begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}\\ &= \begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix} \end{aligned}X=A2−5A+4I=590−1−2−125−2+−10−10−50−55−5−150+400040004=−1−1−5−1−34−3−102 X=[−1−1−3−1−3−10−542]\boxed{X=\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}}X=−1−1−5−1−34−3−102