Question

Let
$M=\left[\begin{array}{ccc}0& 1& a\\ 1& 2& 3\\ 3& b& 1\end{array}\right]$ and adj $M=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$
where a and b are real numbers. Which of the following options is/are correct?

• A. $a+b=3$
• B. (adj M)${}^{-1}$ + adj M${}^{-1}$ = —M
• C. det(adj M${}^2$) = 81
• D. If $M=\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right],$ then $\alpha -\beta +\gamma =3$

Answer 1

Answer: (A), (B), (D)

$\left|adjM\right|={\left|M\right|}^2=\left|\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right|=-1\left(6-6\right)-1\left(-8+10\right)-1\left(24-30\right)$
${\left|M\right|}^2=-2+6=4\Rightarrow \left|M\right|=\pm 2$
We know
A.(adj A) = |A| I
So $M=|M|(adjM{)}^{-1}$ ...(1)
So $\left(adj{M}^{-1}\right)={\left[\begin{array}{ccc}0& \frac{-1}{2}& \frac{-3}{2}\\ \frac{-1}{2}& -1& \frac{-1}{2}\\ -1& \frac{-3}{2}& \frac{-1}{2}\end{array}\right]}^T=\left[\begin{array}{ccc}0& \frac{-1}{2}& -1\\ \frac{-1}{2}& -1& \frac{-3}{2}\\ \frac{-3}{2}& \frac{-1}{2}& \frac{-1}{2}\end{array}\right]$
Now from equation (1)
$\left[\begin{array}{ccc}0& 1& a\\ 1& 2& 3\\ 3& b& 1\end{array}\right]=\left|M\right|\left[\begin{array}{ccc}0& \frac{-1}{2}& -1\\ \frac{-1}{2}& -1& \frac{-3}{2}\\ \frac{-3}{2}& \frac{-1}{2}& -\frac{1}{2}\end{array}\right]$
By comparison, $|M|=-2$
So $a=|M|(-1)=-2(-1)=2$
and $b=\left|M\right|\times \frac{-1}{2}=1$
(A) $a+b=2+1=3$
(B) $(adjM{)}^{-1}+adj{M}^{-1}$ $\because adj{A}^{-1}={\left(adjA\right)}^{-1}$
$=2adj\left(M^{-1}\right)\because A.adjA=\left|A\right|I_n$
$=2\left(\frac{-M}{2}\right)adjA=A{}^{-1}|A|$
$=-M$ So $adj\left(M^{-1}\right)={\left(M^{-1}\right)}^{-1}|M^{-1}|$
$adj\left(M^{-1}\right)=M|M{|}^{-1}$
$adj\left(M^{-1}\right)=\frac{M}{|M|}=\frac{-M}{2}$
$\left(C\right)\because M=\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$
So, $\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
So, $\beta +2\gamma =\mathrm{1..}\left(2\right)$
$\alpha +2\beta +3\gamma =\mathrm{2...}\left(3\right)$
$3\alpha +\beta +\gamma =\mathrm{3...}\left(4\right)$
From (2), (3) and (4), we get
$\alpha =1,\beta =-1,\gamma =1$
So value of $\alpha -\beta +\gamma =1-\left(-1\right)+1=3$
$\left(D\right)|adj\left(M^2\right)|=|M^2{|}^2=|M{|}^4=|-2{|}^4=16$