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Question
`A=[[1,0,-3],[2,1,3],[0,1,1]]`then verify that A2 + A = A(A + I), where I is the identity matrix.
Solution
\[A = \begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\]\[A^2 = \begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}1 + 0 + 0 & 0 + 0 - 3 & - 3 + 0 - 3 \\ 2 + 2 + 0 & 0 + 1 + 3 & - 6 + 3 + 3 \\ 0 + 2 + 0 & 0 + 1 + 1 & 0 + 3 + 1\end{bmatrix}\]
\[ = \begin{bmatrix}1 & - 3 & - 6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{bmatrix}\]
L. H . S
\[A^2 + A = \begin{bmatrix}1 & - 3 & - 6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{bmatrix} + \begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}1 + 1 & - 3 + 0 & - 6 - 3 \\ 4 + 2 & 4 + 1 & 0 + 3 \\ 2 + 0 & 2 + 1 & 4 + 1\end{bmatrix}\]
\[ = \begin{bmatrix}2 & - 3 & - 9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{bmatrix}\]
R. H . S
\[A + I = \begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix} + \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}1 + 1 & 0 + 0 & - 3 + 0 \\ 2 + 0 & 1 + 1 & 3 + 0 \\ 0 + 0 & 1 + 0 & 1 + 1\end{bmatrix}\]
\[ = \begin{bmatrix}2 & 0 & - 3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{bmatrix}\]
\[\]
\[A\left( A + I \right) = \begin{bmatrix}1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}\begin{bmatrix}2 & 0 & - 3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{bmatrix}\]
\[ = \begin{bmatrix}2 + 0 + 0 & 0 + 0 - 3 & - 3 + 0 - 6 \\ 4 + 2 + 0 & 0 + 2 + 3 & - 6 + 3 + 6 \\ 0 + 2 + 0 & 0 + 2 + 1 & 0 + 3 + 2\end{bmatrix}\]
\[ = \begin{bmatrix}2 & - 3 & - 9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{bmatrix}\]
Therfore,LHS=RHS
Hence,`A^2+A=A(A+l)` is verified
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