If $A=\begin{bmatrix}0&1&-1\\2&1&3\\3&2&1\end{bmatrix}$, then $(A (adj\, A) A^{-1}) A=$
Answer & explanation
Correct answer: option 1
We have,
$|A|=\begin{vmatrix}0&1&-1\\2&1&3\\3&2&1\end{vmatrix}=0+7-1=6$
$∴(A (adj\, A) A^{-1}) A=(A (adj\, A))(A^{-1}A)$
$⇒(A (adj\, A) A^{-1}) A=(|A|I)I$
$⇒(A (adj\, A) A^{-1}) A=|A|I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=2\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$