Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

If $A=\begin{bmatrix}0&1&-1\\2&1&3\\3&2&1\end{bmatrix}$, then $(A (adj\, A) A^{-1}) A=$

Options:

$2\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$

$\begin{bmatrix}-6&0&0\\0&-6&0\\0&0&-6\end{bmatrix}$

$\begin{bmatrix}0&1/6&-1/6\\2/6&1/6&3/6\\3/6&2/6&1/6\end{bmatrix}$

none of these

Correct Answer:

$2\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$

Explanation:

We have,

$|A|=\begin{vmatrix}0&1&-1\\2&1&3\\3&2&1\end{vmatrix}=0+7-1=6$

$∴(A (adj\, A) A^{-1}) A=(A (adj\, A))(A^{-1}A)$

$⇒(A (adj\, A) A^{-1}) A=(|A|I)I$

$⇒(A (adj\, A) A^{-1}) A=|A|I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=2\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$