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If \(A=\left[\begin{array}{lll}1&-1&1\\ 2&-1&0\\ 1&0&0\end{array}\right]\) then \(A^{-1}\) is |
\(A\) \(A^2\) \(A^3\) \(\frac{1}{2}\left(A-2I\right)\) |
\(A^2\) |
\(A=\left[\begin{array}{lll}1&-1&1\\ 2&-1&0\\ 1&0&0\end{array}\right]\) so $|A-λI|=\begin{vmatrix}1-λ&-1&1\\2&-1-λ&0\\1&0&-λ\end{vmatrix}$ $=λ((-1+λ)(1+λ)+2)$ $=λ^3-1=0$ $⇒A^3=I$ so $A^{-1}=A^2$ |
