Question

Let $P=\begin{bmatrix}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{bmatrix}$, where $\alpha \in R$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in R, k \neq 0$ and $I$ is the identity matrix of order $3$ . If $q _{23}=-\frac{ k }{8}$ and $\operatorname{det}( Q )=\frac{ k ^{2}}{2}$, 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)

$PQ = kI$
$Q = kP ^{-1}=\frac{ k \text{adj} P }{|\overline{ P }|}$
$=\frac{ k }{12 \alpha+20}=\begin{bmatrix}5 \alpha & 10 & -\alpha \\ 3 \alpha & 6 & -(3 \alpha+4) \\ -10 & 12 & \alpha\end{bmatrix}$
$q _{23}=-\frac{ k }{8}=\frac{ k }{12 \alpha+20}(-3 \alpha-4)$
$12 \alpha+20=24 \alpha+32$
$-12=12 \alpha$
$\alpha=-1$
$Q =\frac{ k }{8}\begin{bmatrix}-5 & 10 & 1 \\ -3 & 6 & -1 \\ -10 & 12 & 2\end{bmatrix}$
$| Q |=\frac{ k ^{2}}{2}=\frac{ k ^{3}}{8^{3}}(-5(12+12)-10(-6-10)+1(-36+60))$
$\frac{1}{2}=\frac{ k }{8.8 .8}(-120+160+24)$
$\frac{1}{2}=\frac{ k }{8}$
$k =4$
$| P \text{adj} Q |$
$=\left| P\, \text{adj}\left(4 P ^{-1}\right)\right|$
$=\left|16 P\, \text{adj} P ^{-1}\right|$
$=\left|16 P \,\cdot \frac{ P }{\mid P \|}\right|=\frac{16^{3}}{| P |^{3}}\left| P ^{2}\right|$
$=\frac{16^{3}}{| P |}=\frac{16^{3}}{8}=2^{9}$
$\mid Q\,\text{adj} P \mid$
$=\mid Q \,\text{adj} \left( kQ ^{-1} \mid\right.$
$=\left| k ^{2} Q \cdot \frac{ Q }{| Q |}\right|$
$=2^{3}| Q |^{2}=2^{3} \cdot 2^{6}=2^{9}$