Question

Let $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$ and $10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$
If B is the inverse of matrix A, then $\alpha$ is

• A. 2
• B. $-1$
• C. 3
• D. 5

Answer 1

The correct option is D 5

$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$
and $10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$
Here, $|A|=4+5+1=10$
Also given, $A^{-1}=B$
$\Rightarrow \frac{adjA}{10}=B$
$\Rightarrow adjA=10B$
Now,$adjA=C^T={\left[\begin{array}{ccc}4& -5& -1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}4& -5& 2\\ 2& 0& 5\\ 2& 5& 3\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}4& -5& 2\\ 2& 0& 5\\ 2& 5& 3\end{array}\right]=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$
On comparing ,we get $\alpha =5$