MHT CET · Maths · Matrices
If \(B=\left[\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right]\) is the adjoint of a \(3 \times 3\) matrix \(\mathrm{A}\) and \(|\mathrm{A}|=5\), then \(\alpha\) is equal to
- A 25
- B 27
- C \(3 \sqrt{3}\)
- D 5
Answer & Solution
Correct Answer
(B) 27
Step-by-step Solution
Detailed explanation
Using \(|\operatorname{Adj} A|=|A|^{n-1}\)
But \(\mathrm{B}=\operatorname{Adj}(\mathrm{A})\)...['Given]
\(\therefore \quad|\mathrm{B}|=|\mathrm{A}|^2\)
\(\begin{aligned} & \Rightarrow\left|\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow \alpha-2=5^2 \\ & \Rightarrow \alpha-2=25 \\ & \Rightarrow \alpha=27\end{aligned}\)
But \(\mathrm{B}=\operatorname{Adj}(\mathrm{A})\)...['Given]
\(\therefore \quad|\mathrm{B}|=|\mathrm{A}|^2\)
\(\begin{aligned} & \Rightarrow\left|\begin{array}{lll}1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3\end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow \alpha-2=5^2 \\ & \Rightarrow \alpha-2=25 \\ & \Rightarrow \alpha=27\end{aligned}\)
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