Question

If $\mathrm{A}$ is an invertible matrix of order $2$, then $\left|{\mathrm{A}}^{-1}\right|$ is equal to?

• A. $\left|\mathrm{A}\right|$
• B. $\frac{1}{\left|\mathrm{A}\right|}$
• C. $1$
• D. $0$

Answer 1

The correct option is B

$\frac{1}{\left|\mathrm{A}\right|}$

Explanation of correct answer:

Since,

$\mathrm{A}{\mathrm{A}}^{-1}=\mathrm{I}$

Now, taking determinant on both sides

$\left|\mathrm{A}{\mathrm{A}}^{-1}\right|=\left|\mathrm{I}\right|$

$\Rightarrow \left|\mathrm{A}\right|\left|{\mathrm{A}}^{-1}\right|=\left|\mathrm{I}\right|$ [$\left|\mathrm{A}\mathrm{B}\right|=\left|\mathrm{A}\right|\left|\mathrm{B}\right|$]

$\Rightarrow$$\left|\mathrm{A}\right|\left|{\mathrm{A}}^{-1}\right|=1$ [ $\left|\mathrm{I}\right|=1$]

$\Rightarrow$ $\left|{\mathrm{A}}^{-1}\right|=\frac{1}{\left|\mathrm{A}\right|}$ [$\left|\mathrm{A}\right|\ne 0$]

Thus, $\left|{\mathrm{A}}^{-1}\right|$ is equal to $\frac{1}{\left|A\right|}$.

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.