Question

Let
$M=\begin{bmatrix}0&1&a\\ 1&2&3\\ 3&b&1\end{bmatrix}$ and adj $M=\begin{bmatrix}-1&1&-1\\ 8&-6&2\\ -5&3&-1\end{bmatrix}$
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 $M=\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix},$ then $\alpha-\beta+\gamma=3$

Answer 1

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

$\left|adj\,M\right|=\left|M\right|^{2}=\begin{vmatrix}-1&1&-1\\ 8&-6&2\\ -5&3&-1\end{vmatrix}=-1\left(6-6\right)-1\left(-8+10\right)-1\left(24-30\right)$
$\left|M\right|^{2}=-2+6=4 \Rightarrow \left|M\right|=\pm2$
We know
A.(adj A) = |A| I
So $M = |M| (adj M)^{-1}$ ...(1)
So $\left(adj\,M^{-1}\right)=\begin{bmatrix}0&\frac{-1}{2}&\frac{-3}{2}\\ \frac{-1}{2}&-1&\frac{-1}{2}\\ -1&\frac{-3}{2}&\frac{-1}{2}\end{bmatrix}^T=\begin{bmatrix}0&\frac{-1}{2}&-1\\ \frac{-1}{2}&-1&\frac{-3}{2}\\ \frac{-3}{2}&\frac{-1}{2}&\frac{-1}{2}\end{bmatrix}$
Now from equation (1)
$\begin{bmatrix}0&1&a\\ 1&2&3\\ 3&b&1\end{bmatrix}=\left|M\right|\begin{bmatrix}0&\frac{-1}{2}&-1\\ \frac{-1}{2}&-1&\frac{-3}{2}\\ \frac{-3}{2}&\frac{-1}{2}&-\frac{1}{2}\end{bmatrix}$
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) $(adj M)^{-1} + adj M^{-1}$ $\because adj\,A^{-1}=\left(adj\,A\right)^{-1}$
$=2\,adj\left(M^{-1}\right)\, \because A.adj\,A=\left|A\right|I_{n}$
$=2\left(\frac{-M}{2}\right)\,adj A = A\,{}^{-1}\, |A|$
$=-M\,\,$ So $adj \left(M^{-1}\right) = \left(M^{-1}\right)^{-1} |M^{-1}|$
$adj \left(M^{-1}\right) = M|M|^{-1}$
$adj\left(M^{-1}\right)=\frac{M}{\left|M\right|}=\frac{-M}{2}$
$\left(C\right)\because M=\begin{bmatrix}0&1&2\\ 1&2&3\\ 3&1&1\end{bmatrix}$
So, $\begin{bmatrix}0&1&2\\ 1&2&3\\ 3&1&1\end{bmatrix}\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
So, $\beta+2\gamma=1..\left(2\right)$
$\alpha+2\beta+3\gamma=2...\left(3\right)$
$3\alpha+\beta+\gamma=3...\left(4\right)$
From (2), (3) and (4), we get
$\alpha=1, \beta=-1, \gamma=1$
So value of $\alpha-\beta+\gamma=1-\left(-1\right)+1=3$
$\left(D\right) |adj \left(M^{2}\right)| = |M^{2}|^{2} = |M|^{4} = |-2|^{4} = 16$