Matrices And Determinants Ques 9
- If $P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A|=4$, then $\alpha$ is equal to
(2013 Main)
(a) $4$
(b) $11$
(c) $5$
(d) $0$
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Answer:
Correct Answer: 9.(b)
Solution: (b) Given, $P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$
$ \begin{aligned} & \therefore \quad|P|=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6 \\ & \because \quad P=\operatorname{adj}(A) \quad \text { [given] } \\ & \therefore \quad|P|=|\operatorname{adj} A|=|A|^2=16 \quad\left[\because|\operatorname{adj} A|=|A|^{n-1}\right] \\ & \therefore \quad 2 \alpha-6=16 \\ & \Rightarrow \quad 2 \alpha=22 \\ & \Rightarrow \quad \alpha=11 \\ \end{aligned} $
BETA
