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Question:
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If $A$ is an invertible matrix of order 2 , then det. $\left(A^{-1}\right)$ is equal to :
(a) det. (A)
(b) $\frac{1}{\text { det.(A) }}$
(c) 1
(d) 0
(a) det. (A)
(b) $\frac{1}{\text { det.(A) }}$
(c) 1
(d) 0
Solution:
1767 Upvotes
Verified Answer
$\begin{aligned}
&|\mathrm{A}| \neq 0 \\
&\Rightarrow \quad \mathrm{A}^{-1} \text { exists } \Rightarrow \mathrm{AA}^{-1}=\mathrm{I} \\
&\Rightarrow \quad\left|\mathrm{AA}^{-1}\right|=|\mathrm{I}|=1 \\
&\Rightarrow \quad|\mathrm{A}|\left|\mathrm{A}^{-1}\right|=1 \quad\left|\mathrm{~A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}
\end{aligned}$
Hence option (b) is correct.
&|\mathrm{A}| \neq 0 \\
&\Rightarrow \quad \mathrm{A}^{-1} \text { exists } \Rightarrow \mathrm{AA}^{-1}=\mathrm{I} \\
&\Rightarrow \quad\left|\mathrm{AA}^{-1}\right|=|\mathrm{I}|=1 \\
&\Rightarrow \quad|\mathrm{A}|\left|\mathrm{A}^{-1}\right|=1 \quad\left|\mathrm{~A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}
\end{aligned}$
Hence option (b) is correct.
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