Q64 / 125 · BITSAT 2015 · Mathematics
Let M bea $3 \times 3 \mathrm{n} \times \mathrm{n}$ -singular matrix with $\operatorname{det}(\mathrm{M})$ $=\alpha$. If $\left[\mathrm{M}^{-1}\right.$ adj $(\mathrm{adj} (\mathrm{M})]=\mathrm{KI}$, then the value of $\mathrm{K}$ is
- A1
- Correct answerB$\alpha$
- C$\alpha^{2}$
- D$\alpha^{3}$
Solution
We know that, $M(\operatorname{adj} M)=|M|$ I
Replacing $\mathrm{M}$ by adj $\mathrm{M},$ we get $\operatorname{adj} M[\operatorname{adj}(\operatorname{adj} M)=\operatorname{det}(\operatorname{adj} M) I$
$=\operatorname{det}(M) M^{-1}\left[\operatorname{adj}(\operatorname{adj} M)=\alpha^{2} 1\right.$
$$
\left[\because \mathrm{M}^{-1}=\frac{1}{|\mathrm{M}|} \operatorname{adj}(\mathrm{M})\right]
$$
$\Rightarrow \alpha \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} \mathrm{I}$
$\Rightarrow \mathrm{M}^{-1}[\mathrm{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}$
But $M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=K I$
Hence, $\mathrm{K}=\alpha$
Hence, the correct answer is option: 'B'.