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Verify that $A^2=I\;when\;A=\begin{bmatrix}0 & 1 & 1\\4 & 3 & 4\\3 & 3 & 4\end{bmatrix}.$

Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Given
$A=\begin{bmatrix}0 & 1 &-1\\4 & -3 & 4\\3 & -3 & 4\end{bmatrix}$

$A^2=A.A=\begin{bmatrix}0 & 1 &-1\\4 & -3 & 4\\3 & -3 & 4\end{bmatrix}\begin{bmatrix}0 & 1 &-1\\4 & -3 & 4\\3 & -3 & 4\end{bmatrix}$

$\Rightarrow \begin{bmatrix}0.0+1(4)+(-1)(3)& 0+1(-3)+-1(-3) &0+1(4)+(-1)(4)\\4.0+(-3)(4)+4(3) & 4+(-3)(-3)+4(-3) & 4(-1)+(-3)(4)+4(4)\\3.0+(-3)(4)+4(3) & 3+(-3)(-3)+4(-3)& 3(-1)+(-3)(4)+4(4)\end{bmatrix}$

$\Rightarrow \begin{bmatrix}0+4-3& 0-3+3&0+4-4\\0-12+12 & 4+9-12 & -4-12+16\\0-12+12 & 3+9-12& -3-12+16\end{bmatrix}$

$\Rightarrow\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$

$\Rightarrow I.$

Hence $A^2=I.$