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 $ofA$, then $\alpha$ is

• A. -2
• B. -1
• C. 2
• D. 5

Answer 1

The correct option is D 5

$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$

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

$\because B^{-1}=A$

$AB=I$

$|A|\cdot |B|=1$

Expanding the determinant of A, we get

$|A|=4+5+1=10$

$|B|=\frac{1}{{10}^3}(8\alpha +30+2\alpha +20)$

$=\frac{1}{10}(50+10\alpha )=\frac{1}{100}(5+\alpha )$

Now, $\left(10\right)\left(\frac{1}{100}\right(5+\alpha \left)\right)=1$

$\therefore 5+\alpha =10$

$\therefore \alpha =5$