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. $2$
• D. $5$

Answer 1

The correct option is D $5$

Since, $B$ is the inverse of $A$.
ie, $B=10A^{-1}$
$\therefore \left(10\right)A^{-1}=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$
$\therefore \left(10\right)A^{-1}\cdot A=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]A$
$\Rightarrow 10I=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]=\left[\begin{array}{ccc}10& 0& 0\\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0& 0& 10\end{array}\right]$
$\Rightarrow 5+\alpha =10$
$\Rightarrow \alpha =5$