Question

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

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

Answer 1

Answer: (B), (C), (D)

The correct options are
B $a+b=3$
C If $M\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$, then $\alpha -\beta +\gamma =3$
D $(adjM{)}^{-1}+adj{M}^{-1}=-M$

$M=\left[\begin{array}{ccc}0& 1& a\\ 1& 2& 3\\ 3& b& 1\end{array}\right]$ and $adjM=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$
$(adjM{)}_{11}=2-3b,(adjM{)}_{22}=-3a$
$\Rightarrow 2-3b=-1$
$\Rightarrow b=1$ and $-3a=-6\Rightarrow a=2$
$|adjM|=-1(6-6)-1(-8+10)-1(24-30)=4$
$det\left\{adj\right(M^2\left)\right\}=|det\left(adjM\right){|}^2=16$
Now $\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\Rightarrow \beta +2\gamma =1,\alpha +2\beta +3\gamma =2,3\alpha +\beta +\gamma =3$
On solving $\alpha =1,\beta =-1,\gamma =1$ so $\alpha -\beta +\gamma =3$
Now $(adjM{)}^{-1}+(adjM{)}^{-1}=2(adjM{)}^{-1}=\frac{2adj\left(adjM\right)}{|adjM|}=\frac{1}{2}|M{|}^{3-2}M=-M$.