Question

Let $P=\left[\begin{array}{c}3\\ 2\\ 3\end{array}\begin{array}{c}-1\\ 0\\ -5\end{array}\begin{array}{c}-2\\ \alpha \\ 0\end{array}\right]$, where $\alpha \in R$. Suppose $Q=\left[q_{ij}\right]$ is matrix such that $PQ=kI$, where $k\in R,k\ne 0$ and $I$ is the identity matrix of order $3$. If $q_{23}=-\frac{k}{8}$ and $det\left(Q\right)=\frac{k^2}{2}$, then

• A. $\alpha =0,k=8$
• B. $4\alpha -k+8=0$
• C. $det\left(Padj\right(Q\left)\right)=2^9$
• D. $det\left(Qadj\right(P\left)\right)=2^{13}$

Answer 1

Answer: (B), (C)

$det\left(Padj\right(Q\left)\right)=2^9$
$4\alpha -k+8=0$

As $PQ=kI\Rightarrow Q=k{P}^{-1}I$

$|P|=\left[\begin{array}{c}3\\ 2\\ 3\end{array}\begin{array}{c}-1\\ 0\\ -5\end{array}\begin{array}{c}-2\\ \alpha \\ 0\end{array}\right]$,

$=(20+12\alpha )$ .......$\left(1\right)$

Now $Q=\frac{k}{|P|}\left(adjP\right)I$ $\Rightarrow Q=\frac{k}{(20+12\alpha )}\left[\begin{array}{ccc}5\alpha & 10& -\alpha \\ 3\alpha & 0& (-3\alpha -4)\\ -10& -12& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\because q_{23}=\frac{-k}{8}\Rightarrow \frac{-k(3\alpha +4)}{(20+12\alpha )}=\frac{-k}{8}\Rightarrow 2(3\alpha +4)=5+3\alpha$

$3\alpha =-3\Rightarrow \alpha =-1$

Also $|Q|=\frac{k^3|I|}{|P|}\Rightarrow \frac{k^2}{2}=\frac{k^3}{(20+12\alpha )}$ $\because from\left(1\right)$

$(20+12\alpha )=2k\Rightarrow 8=2k\Rightarrow k=4$

(A) incorrect

(B) correct
putting $\alpha =-1$ and $k=4$ satisfy the equation $4\alpha -k+8$

(C) $|P\left(adjQ\right)|=|P||adjQ|=P||Q{|}^2=2^3(2^3{)}^2=2^9$ correct

(D) $|Q\left(adjP\right)|=|Q||adjP|=|Q||P{|}^2=2^3(2^3{)}^2=2^9$ incorrect