AP EAMCET · Maths · Determinants
If \(k>1\) and the determinant of the matrix \(A^2\), where \(A=\left[\begin{array}{ccc}k & k \alpha & \alpha \\ 0 & \alpha & k \alpha \\ 0 & 0 & k\end{array}\right]\), is \(k^2\), then \(|\alpha|\) equal to
- A \(\frac{1}{k^2}\)
- B \(k\)
- C \(k^2\)
- D \(\frac{1}{k}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{k}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given } A=\left[\begin{array}{rrr} k & k \alpha & \alpha \\ 0 & \alpha & k \alpha \\ 0 & 0 & k \end{array}\right] \\ & \therefore \quad|A|=\alpha k^2 \\ & \text { Now } \quad \begin{aligned} \left|A^2\right| & =|A|^2 \\ & =\left(\alpha k^2\right)^2 \\ &…
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