Question

Let $A=\begin{bmatrix}1&-1&1\\ 2&1&-3\\ 1&1&1\end{bmatrix} and 10B = \begin{bmatrix}4&2&2\\ -5&0&\alpha\\ 1&-2&3\end{bmatrix} . I$
$B$ 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, $10A^{-1} = \begin{bmatrix}4&2&2\\ -5&0&\alpha\\ 1&-2&3\end{bmatrix}$

$\Rightarrow 10A^{-1}\cdot A =\begin{bmatrix}4&2&2\\ -5&0&\alpha \\ 1&-2&3\end{bmatrix}\begin{bmatrix}1&-1&1\\ 2&1&-3\\ 1&1&1\end{bmatrix}$

$\Rightarrow 10I = \begin{bmatrix}4+4+2&-4+2+2&4-6+2\\ -5+0+\alpha&5+0+\alpha &-5+0+\alpha \\ 1-4+3&-1-2+3&1+6+3\end{bmatrix}$

$\Rightarrow \begin{bmatrix}10&0&0\\ 0&10&0\\ 0&0&10\end{bmatrix}=\begin{bmatrix}10&0&0\\ -5+\alpha&5+\alpha &-5+\alpha \\ 0&0&10\end{bmatrix}$

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