Multiple Choice

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

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

Solution

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

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

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

Now $$Q=\dfrac{k}{|P|}(adjP)I$$ $$\Rightarrow Q=\dfrac{k}{(20+12\alpha)}\begin{bmatrix} 5\alpha&10&-\alpha \\3\alpha&0&(-3\alpha -4)\\-10&-12&0 \end{bmatrix} \begin{bmatrix} 1&0&0 \\0&1&0\\0&0&1 \end{bmatrix}$$

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

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

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

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

(A) incorrect

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

(C) $$|P(adjQ)| = |P| |adjQ| = P| |Q|^2 = 2^3(2^3)^2 = 2^9$$ correct

(D) $$|Q(adjP)| = |Q| |adjP| = |Q| |P|^2 = 2^3(2^3)^2 = 2^9$$ incorrect

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