Matrices and Determinants 3 Question 11
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13. Let $P=\begin{array}{cccc}3 & -1 & -2 \ 2 & 0 & \alpha \ 3 & -5 & 0\end{array}$, 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
======= ####13. 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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
(2016 Adv.)
(a) $\alpha=0, k=8$
(b) $4 \alpha-k+8=0$
(c) $\operatorname{det}(P \operatorname{adj}(Q))=2^{9}$
(d) $\operatorname{det}(Q \operatorname{adj}(P))=2^{13}$
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Answer:
Correct Answer: 13. (b,c)
Solution:
- Here, $ P=\begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix} $
Now, $\quad|P|=3(5 \alpha)+1(-3 \alpha)-2(-10)$
$ =12 \alpha+20 $
$\therefore \quad \operatorname{adj}(P)$ =$ \begin{bmatrix} 5 \alpha & 2 \alpha & -10\\ -10 & 6 & 12 \\ -\alpha & -(3 \alpha+4) & 2 \end{bmatrix}$
=$ \begin{bmatrix} 5 \alpha & -10 & -\alpha \\ 2 \alpha & 6 & -(3 \alpha+4)\\ -10 & 12 & 2 \end{bmatrix}$
$ \begin{aligned} & \text { As, } \quad P Q=k I \\ & \Rightarrow \quad|P||Q|=|k I| \\ & \Rightarrow \quad|P||Q|=k^{3} \\ & \Rightarrow \quad|P| \frac{k^{2}}{2}=k^{3} \quad \text { given, }|Q|=\frac{k^{2}}{2} \end{aligned} $
$ \begin{aligned} & \Rightarrow \quad|P|=2 k \\ & \because \quad P Q=k I \\ & \therefore \quad Q=k p^{-1} I \\ & =k \cdot \frac{\operatorname{adj} P}{|P|}=\frac{k(\operatorname{adj} P)}{2 k} \quad \text { [from Eq. (iii)] } \\ & =\frac{\operatorname{adj} P}{2}=\frac{1}{2} \begin{bmatrix} 5 \alpha & -10 & -\alpha \\ 2 \alpha & 6 & -3 \alpha-4 \\ -10 & 12 & 2 \end{bmatrix} \\ & \therefore \quad q _{23}=\frac{-3 \alpha-4}{2} \quad \text { given, } q _{23}=-\frac{k}{8} \\ & \Rightarrow \quad-\frac{(3 \alpha+4)}{2}=-\frac{k}{8} \\ & \Rightarrow \quad(3 \alpha+4) \times 4=k \\ & \Rightarrow \quad 12 \alpha+16=k \\ & \text { From Eq. (iii), } \quad|P|=2 k \\ & \Rightarrow \quad 12 \alpha+20=2 k \end{aligned} $
On solving Eqs. (iv) and (v), we get
$ \alpha=-1 \text { and } k=4 $
$\therefore \quad 4 \alpha-k+8=-4-4+8=0$
$\therefore$ Option (b) is correct.
Now, $\quad|P \operatorname{adj}(Q)|=|P||\operatorname{adj} Q|$
$ =2 k{\frac{k^{2}}{2}} _2^{2}=\frac{k^{5}}{2}=\frac{2^{10}}{2}=2^{9} $
$\therefore$ Option (c) is correct.
