Question

If $A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 2& 0& 3\end{array}\right]$ then $\left(adjA\right)A=$

• A. $\left[\begin{array}{ccc}13& 0& 0\\ 0& 13& 0\\ 0& 0& 13\end{array}\right]$
• B. $\left[\begin{array}{ccc}7& 0& 0\\ 0& 7& 0\\ 0& 0& 7\end{array}\right]$
• C. $\left[\begin{array}{ccc}-7& 0& 0\\ 0& 0& -7\\ 0& 0& -7\end{array}\right]$
• D. $\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right]$

Answer 1

Answer: (A)

$\left[\begin{array}{ccc}13& 0& 0\\ 0& 13& 0\\ 0& 0& 13\end{array}\right]$

We have,

$\begin{array}{c}A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 2& 0& 3\end{array}\right]\\ \left(adjA\right)=?\\ \left(adjA\right)=\left|A\right|I\\ \left|A\right|=\left|\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 2& 0& 3\end{array}\right|=1\left(0-0\right)+1\left(9+4\right)+2\left(0-0\right)\\ \left(adjA\right)A=\left[\begin{array}{ccc}13& 0& 0\\ 0& 13& 0\\ 0& 0& 13\end{array}\right]\end{array}$

Hence,

Option $A$ is correct answer.