Question

Let $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]and10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right].I$
$B$ is the inverse of matrix $A$, then $\alpha$ is

• A. $-2$
• B. $1$
• C. $2$
• D. $5$

Answer 1

Answer: (D)

Since, $B$ is the inverse of matrix $A$.

so, $10A^{-1}=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$

$\Rightarrow 10A^{-1}\cdot A=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$

$\Rightarrow 10I=\left[\begin{array}{ccc}4+4+2& -4+2+2& 4-6+2\\ -5+0+\alpha & 5+0+\alpha & -5+0+\alpha \\ 1-4+3& -1-2+3& 1+6+3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right]=\left[\begin{array}{ccc}10& 0& 0\\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0& 0& 10\end{array}\right]$

$\Rightarrow -5+\alpha =0\Rightarrow \alpha =5$