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Question
`A=[[2,0,1],[2,1,3],[1,-1,0]]` , find A2 − 5A + 4I and hence find a matrix X such that A2 − 5A + 4I + X = 0.
Solution
Given:
\[A = \begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0\end{bmatrix}\]\[A^2 = \begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0\end{bmatrix}\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0\end{bmatrix}\]
\[ = \begin{bmatrix}4 + 0 + 1 & 0 + 0 - 1 & 2 + 0 + 0 \\ 4 + 2 + 3 & 0 + 1 - 3 & 2 + 3 + 0 \\ 2 - 2 + 0 & 0 - 1 - 0 & 1 - 3 + 0\end{bmatrix}\]
\[ = \begin{bmatrix}5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2\end{bmatrix}\]
Now,
\[A^2 - 5A + 4I = \begin{bmatrix}5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2\end{bmatrix} - 5\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0\end{bmatrix} + 4\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}5 - 10 + 4 & - 1 - 0 + 0 & 2 - 5 + 0 \\ 9 - 10 + 0 & - 2 - 5 + 4 & 5 - 15 + 0 \\ 0 - 5 + 0 & - 1 + 5 + 0 & - 2 - 0 + 4\end{bmatrix}\]
\[ = \begin{bmatrix}- 1 & - 1 & - 3 \\ - 1 & - 3 & - 10 \\ - 5 & 4 & 2\end{bmatrix}\]\
Now, A2 − 5A + 4I + X = 0
⇒ X = −(A2 − 5A + 4I)
∴ `X =[[ - 1 - 1 - 3] ,[- 1 - 3 - 10],[- 5 4 2]]`
`=[[1 1 3] ,[ 1 3 10], [ 5 -4 - 2]]`
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