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If A= \( \begin{bmatrix} 1& 2 & 2\\ 2 & 1& 2\\ 2&2&1 \end{bmatrix}\)   and A-1 exists and A≠0 , then (A2 -4A)A-1

 

\(A) \begin{bmatrix} -3 & 2 & 2\\ 2 & -3 & 2\\ 2&2&-3 \end{bmatrix} \\B) \begin{bmatrix} 3 & 2 & 2\\ 2 & 3 & 2\\ 2&2&3 \end{bmatrix} \\ C) \begin{bmatrix} 5 & 2 & 0\\ 2 & 5 & 0\\ 0&2&5 \end{bmatrix} \\ D) \begin{bmatrix} 5& 2 & 5\\ 2& 5 & 5\\ 5&5&2 \end{bmatrix}\)

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The correct answer is option a) \(\begin{bmatrix} -3& 2 & 2\\ 2 & -3& 2\\ 2&2&-3 \end{bmatrix}\)

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