If \(A=\left[\begin{array}{lll}1&-1&1\\ 2&-1&0\\ 1&0&0\end{array}\right]\) then \(A^{-1}\) is
Answer & explanation
Correct answer: option 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 $A^{-1}=A^2$