KCET · Maths · Matrices
If \(A\) is a \(3 \times 3\) non-singular matrix and if \(|\mathrm{A}|=3\), then \(\left|(2 \mathrm{~A})^{-1}\right|\) is
- A 24
- B 3
- C \(\frac{1}{3}\)
- D \(\frac{1}{24}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{24}\)
Step-by-step Solution
Detailed explanation
Given, \(|A|_{3 \times 3} \neq 0\) and \(|A|=3\)
Then,
\[
\begin{aligned}
&\left|(2 \mathrm{~A})^{-1}\right|=\left|\frac{1}{2 \mathrm{~A}}\right|=\frac{1}{|2 \mathrm{~A}|} \\
&=\frac{1}{(2)^{3}} \cdot \frac{1}{|\mathrm{~A}|} \quad\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right) \\
&=\frac{1}{8} \cdot \frac{1}{3}=\frac{1}{24}
\end{aligned}
\]
Then,
\[
\begin{aligned}
&\left|(2 \mathrm{~A})^{-1}\right|=\left|\frac{1}{2 \mathrm{~A}}\right|=\frac{1}{|2 \mathrm{~A}|} \\
&=\frac{1}{(2)^{3}} \cdot \frac{1}{|\mathrm{~A}|} \quad\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right) \\
&=\frac{1}{8} \cdot \frac{1}{3}=\frac{1}{24}
\end{aligned}
\]
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