Mathematics - Matrices Question with Solution | TestHub

Given, $M=\left[\begin{array}{ccc}0& 1& a\\ 1& 2& 3\\ 3& b& 1\end{array}\right]$ ...(i)
$adj\left(M\right)=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$ ...(ii)
Option $\left(\mathrm{A}\right)$ :
Now from matrix $M$
${\left(adj M\right)}_{11}=2-3b$
As given $2-3b=-1\Rightarrow b=1$

Similarly, $a=2$

$a+b=3$
Using the values of $a$ and $b$ in matrix $M\Rightarrow \left|M\right|=-2$
$\left|adj\left(M^2\right)\right|=|M^2{|}^2=|M{|}^4=16$
Now, $M^{-1}=\frac{adjM}{\left|M\right|}$
$adj M=\left|M\right|M^{-1}$
$adj \left(M\right)= -2M^{-1}$
${\left(adj\left(M\right)\right)}^{-1}=-\frac{1}{2}{\left(M^{-1}\right)}^{-1}$ $\left({\left(KA\right)}^{-1}=\frac{1}{K}{A}^{-1}\right)$
${\left(adj\right(M\left)\right)}^{-1}=\frac{-M}{2}$
and $adj\left(M^{-1}\right)={\left(M^{-1}\right)}^{-1}\left|M^{-1}\right|$ $\left(adj A=\left|A\right|A^1\right)$
$adj\left(M^{-1}\right)=\frac{M}{\left|M\right|}$ $\left(\left|A^{-1}\right|=\frac{1}{\left|A\right|}\right)$
$adj\left(M^{-1}\right)=-\frac{M}{2}$ ...(iii)
From (iii) and (iv)
$(adj {\left(M\right)}^{-1}+\left(adj \left(M^{-1}\right)\right)=-M$
Option $\left(\mathrm{D}\right)$ :
$\mathrm{M}\mathrm{X}=B \left(\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}x=\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}B=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)$
$X=M^{-1} \left(B\right)$
$X=\frac{adj\left(M\right)}{\left|M\right|}B$
$\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=-\frac{1}{2}\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
$=-\frac{1}{2}\left[\begin{array}{c}-2\\ 2\\ -2\end{array}\right]$
$\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$
$\because \alpha =1\beta =-1\gamma =1$
$\alpha -\beta +\gamma =3$