JEE Advanced 2019 — Paper 1 — Question 36

$\mathrm{M} \operatorname{adj} \mathrm{M}=|\mathrm{M}| \mathrm{I} \Rightarrow \mathrm{a}=2, \mathrm{~b}=1$

$\Rightarrow M=\left[\begin{array}{lll}0 & 1 & 2\\ 1 & 2 & 3\\ 3 & 1 & 1\end{array}\right] \Rightarrow|M|=-2$

(A) $(\operatorname{adj} M)^{-1}+\left(\operatorname{adj} M^{-1}\right)=\left(|M| M^{-1}\right)^{-1}+\left|M^{-1}\right| M=\frac{M}{|M|}+\frac{M}{|M|}=\frac{2 M}{|M|}$

$=-M$

(B) $\mathrm{M}\left[\begin{array}{l}\alpha\\ \beta\\ \gamma\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right] \Rightarrow\left[\begin{array}{l}\alpha\\ \beta\\ \gamma\end{array}\right]=-\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & -1\\ 8 & -6 & 2\\ -5 & 3 & -1\end{array}\right]\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$

$\alpha=1, \beta=-1, \gamma=1 \Rightarrow \alpha-\beta+\gamma=3$

(C) $\left|\operatorname{adj} \mathrm{M}^{2}\right|=\left|\mathrm{M}^{2}\right|^{2}=|\mathrm{M}|^{4}=16$

(D) $\mathrm{a}=2, \mathrm{~b}=1, \mathrm{a}+\mathrm{b}=3$