Question

Let $P=\left[\begin{array}{ccc}3& -1& -2\\ 2& 0& \alpha \\ 3& -5& 0\end{array}\right]$, where $\alpha \in R$. Suppose $Q=\left[q_{ij}\right]$ is a 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 $\det (Q)=\frac{k^2}{2}$, then

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

Answer 1

Answer: (B), (C)

$PQ=kI$
$Q=k{P}^{-1}=\frac{k\text{adj}P}{|\overline{P}|}$
$=\frac{k}{12\alpha +20}=\left[\begin{array}{ccc}5\alpha & 10& -\alpha \\ 3\alpha & 6& -(3\alpha +4)\\ -10& 12& \alpha \end{array}\right]$
$q_{23}=-\frac{k}{8}=\frac{k}{12\alpha +20}(-3\alpha -4)$
$12\alpha +20=24\alpha +32$
$-12=12\alpha$
$\alpha =-1$
$Q=\frac{k}{8}\left[\begin{array}{ccc}-5& 10& 1\\ -3& 6& -1\\ -10& 12& 2\end{array}\right]$
$|Q|=\frac{k^2}{2}=\frac{k^3}{8^3}(-5(12+12)-10(-6-10)+1(-36+60))$
$\frac{1}{2}=\frac{k}{\mathrm{8.8.8}}(-120+160+24)$
$\frac{1}{2}=\frac{k}{8}$
$k=4$
$|P\text{adj}Q|$
$=\left|P\text{adj}\left(4P^{-1}\right)\right|$
$=\left|16P\text{adj}{P}^{-1}\right|$
$=\left|16P\cdot \frac{P}{\mid P\Vert }\right|=\frac{16^3}{|P{|}^3}\left|P^2\right|$
$=\frac{16^3}{|P|}=\frac{16^3}{8}=2^9$
$\mid Q\text{adj}P\mid$
$=\mid Q\text{adj}(k{Q}^{-1}\mid$
$=\left|k^{2}Q\cdot \frac{Q}{|Q|}\right|$
$=2^3|Q{|}^2=2^3\cdot 2^6=2^9$