Question

If the inverse of the matrix

$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& -1& 2\\ 2& 2& 1\end{array}\right]$ is $\frac{1}{5}\left[\begin{array}{ccc}-3& 2& 2\\ 2& -3& \alpha \\ 2& 2& -3\end{array}\right]$,

then the value of $\alpha$ is

• A. $3$
• B. $4$
• C. $2$
• D. $-2$

Answer 1

The correct option is C $2$

We know that $A{A}^{-1}=\text{I}$

$\Rightarrow \frac{1}{5}\left[\begin{array}{ccc}1& 2& 2\\ 2& -1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}-3& 2& 2\\ 2& -3& \alpha \\ 2& 2& -3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Comparing $a_{23}$, we get
$\frac{1}{5}(4-\alpha -6)=0$
$\Rightarrow \alpha =2$