KCET · Maths · Determinants
If \(A=\left[\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right]\) and \(\left|A^{3}\right|=125\), then \(\alpha\) is equal to
- A \(\pm 1\)
- B \(\pm 2\)
- C \(\pm 3\)
- D \(\pm 5\)
Answer & Solution
Correct Answer
(C) \(\pm 3\)
Step-by-step Solution
Detailed explanation
Given, \(\quad A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]\)
Now, \(\quad|A|=\left|\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right|=\alpha^{2}-4...(i)\)
Also, given,
\(\begin{array}{ll}
& \left|A^{3}\right|=125 \\
\Rightarrow \quad & |A|^{3}=125=(5)^{3} \quad\left(\because\left|A^{n}\right|=|A|^{n}\right) \\
\therefore & |A|=5
\end{array}\)
On putting this value in Eq. (i), we get
\(\begin{aligned}
5 &=\alpha^{2}-4 \\
\Rightarrow \quad \alpha^{2} &=9=(3)^{2} \\
\Rightarrow \quad \alpha &=\pm 3
\end{aligned}\)
Now, \(\quad|A|=\left|\begin{array}{cc}\alpha & 2 \\ 2 & \alpha\end{array}\right|=\alpha^{2}-4...(i)\)
Also, given,
\(\begin{array}{ll}
& \left|A^{3}\right|=125 \\
\Rightarrow \quad & |A|^{3}=125=(5)^{3} \quad\left(\because\left|A^{n}\right|=|A|^{n}\right) \\
\therefore & |A|=5
\end{array}\)
On putting this value in Eq. (i), we get
\(\begin{aligned}
5 &=\alpha^{2}-4 \\
\Rightarrow \quad \alpha^{2} &=9=(3)^{2} \\
\Rightarrow \quad \alpha &=\pm 3
\end{aligned}\)
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