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 $I$ 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. $det\left(P\text{adj}\right(Q\left)\right)=2^9$
• D. $det\left(Q\text{adj}\right(P\left)\right)=2^{13}$

Answer 1

Answer: (B), (C)

The correct options are
B $4\alpha -k+8=0$
C $det\left(P\text{adj}\right(Q\left)\right)=2^9$

$Q=k{P}^{-1}{P}^{-1}=\frac{\left[\begin{array}{ccc}5\alpha & 10& -\alpha \\ & & \\ 3\alpha & 6& -(3\alpha +4)\\ -10& 12& 2\end{array}\right]}{20+12\alpha }{q}_{23}=\frac{-k}{8}\frac{k(-3\alpha -4)}{20+12\alpha }=\frac{-k}{8}\alpha =-1|Q|=\frac{k^2}{2}|k{P}^{-1}|=\frac{k^2}{2}\frac{k^3}{|P|}=\frac{k^2}{2}\Rightarrow k=4|P|=8\Rightarrow |Q|=8\therefore 4\alpha -k+8=0|P.adjQ|=|P||Q{|}^2=2^3\times 2^6=2^9|Q.adjP|=|Q||P{|}^2=2^9$