KCET · Maths · Continuity and Differentiability
If \( \mathrm{A}=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right] \) and \( \left|\mathrm{A}^{3}\right|=27 \), then \( \alpha= \)
- A \( \pm 1 \)
- B \( \pm 2 \)
- C \( \pm \sqrt{7} \)
- D \( 0 \pm \sqrt{5} \)
Answer & Solution
Correct Answer
(C) \( \pm \sqrt{7} \)
Step-by-step Solution
Detailed explanation
Given that \(A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]\)
And \(\left|A^{3}\right|=27=|A|^{3}\)
\(\Rightarrow|A|=3\)
So, \(|A|=\alpha^{2}-4=3\)
\(\Rightarrow \alpha^{2}=7 \Rightarrow \alpha=\pm \sqrt{7}\)
And \(\left|A^{3}\right|=27=|A|^{3}\)
\(\Rightarrow|A|=3\)
So, \(|A|=\alpha^{2}-4=3\)
\(\Rightarrow \alpha^{2}=7 \Rightarrow \alpha=\pm \sqrt{7}\)
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