SATHEE: Matrices and Determinants 3 Question 11

Matrices and Determinants 3 Question 11

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13. Let $P = \begin{array}{cccc} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{array}$ , where $α \in R$ . Suppose $Q = [q_{i j}]$ 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 $\det (Q) = \frac{k^{2}}{2}$ , then

======= ####13. Let $P = [\begin{matrix} 3 & - 1 & - 2 \\ 2 & 0 & α \\ 3 & - 5 & 0 \end{matrix}]$ , where $α \in R$ . Suppose $Q = [q_{i j}]$ 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 $\det (Q) = \frac{k^{2}}{2}$ , then

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed

(2016 Adv.)

(a) $α = 0, k = 8$

(b) $4 α - k + 8 = 0$

(d) $\det (Q adj (P)) = 2^{13}$

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Answer:

Correct Answer: 13. (b,c)

Solution:

Here, $P = [\begin{matrix} 3 & - 1 & - 2 \\ 2 & 0 & α \\ 3 & - 5 & 0 \end{matrix}]$

Now, $| P | = 3 (5 α) + 1 (- 3 α) - 2 (- 10)$

$= 12 α + 20$

$∴ adj (P)$ = $[\begin{matrix} 5 α & 2 α & - 10 \\ - 10 & 6 & 12 \\ - α & - (3 α + 4) & 2 \end{matrix}]$

= $[\begin{matrix} 5 α & - 10 & - α \\ 2 α & 6 & - (3 α + 4) \\ - 10 & 12 & 2 \end{matrix}]$

$\begin{aligned} As, P Q = k I \\ \Rightarrow | P | | Q | = | k I | \\ \Rightarrow | P | | Q | = k^{3} \\ \Rightarrow | P | \frac{k^{2}}{2} = k^{3} given, | Q | = \frac{k^{2}}{2} \end{aligned}$

$\begin{aligned} \Rightarrow | P | = 2 k \\ ∵ P Q = k I \\ ∴ Q = k p^{- 1} I \\ = k \cdot \frac{adj P}{| P |} = \frac{k (adj P)}{2 k} [from Eq. (iii)] \\ = \frac{adj P}{2} = \frac{1}{2} [\begin{array}{c} 5 α & - 10 & - α \\ 2 α & 6 & - 3 α - 4 \\ - 10 & 12 & 2 \end{array}] \\ ∴ q_{23} = \frac{- 3 α - 4}{2} given, q_{23} = - \frac{k}{8} \\ \Rightarrow - \frac{(3 α + 4)}{2} = - \frac{k}{8} \\ \Rightarrow (3 α + 4) \times 4 = k \\ \Rightarrow 12 α + 16 = k \\ From Eq. (iii), | P | = 2 k \\ \Rightarrow 12 α + 20 = 2 k \end{aligned}$