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Question
If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]
Solution
\[\text{ We have, } A = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\]
\[ A^2 = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 6 + 0 & - 9 + 9 - 4 & 12 - 12 + 4\\6 - 6 + 0 & - 6 + 9 - 4 & 8 - 12 + 4\\0 - 2 + 0 & 0 + 3 - 1 & 0 - 4 + 1 \end{bmatrix} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\]
\[\text{ Now,} A^3 = A^2 \times A^{} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 8 & - 9 + 12 - 4 & 12 - 16 + 4\\ 0 - 2 + 0 & 0 + 3 + 0 & - 4\\ - 6 + 4 + 0 & 6 - 6 + 3 & - 8 + 8 - 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\]
\[\text{ Again, }A^3 \times A = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 - 2 + 0 & - 3 + 3 + 0 & 4 - 4 + 0\\ - 6 + 6 & 6 - 9 + 4 & - 8 + 12 - 4\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = I_3 [\text{ Identity matrix of order 3 }]\]
\[ \Rightarrow A^3 \times A = I_3 \]
\[ \Rightarrow A^3 = A^{- 1}\]
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