Question

If $A$ is a non-singular matrix, then

• A. $A^{-1}$ is symmetric if $A$ is symmetric
• B. $A^{-1}$ is skew-symmetric if $A$ is symmetric
• C. $\left|A^{-1}\right|=\left|A\right|$
• D. $\left|A^{-1}\right|={\left|A\right|}^{-1}$

Answer 1

Answer: (A), (D)

The correct options are
A $A^{-1}$ is symmetric if $A$ is symmetric
D $\left|A^{-1}\right|={\left|A\right|}^{-1}$

Since $\left|A\right|\ne 0$, therefore $A^{-1}$ exists.

Now, $A{A}^{-1}=I=A^{-1}A$

$\Rightarrow {\left(A{A}^{-1}\right)}^{\prime }=I^{\prime }={\left(A^{-1}A\right)}^{\prime }\Rightarrow {\left(A^{-1}\right)}^{\prime }{A}^{\prime }=I=A^{\prime }{\left(A^{-1}\right)}^{\prime }(\because A^{\prime }=A)$

$\Rightarrow {\left(A^{-1}\right)}^{\prime }A=I=A{\left(A^{-1}\right)}^{\prime }\Rightarrow A^{-1}={\left(A^{-1}\right)}^{\prime }\Rightarrow A^{-1}$ is symmetric

Also, since $\left|A\right|\ne 0,\therefore A^{-1}$ exists such that

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

$\Rightarrow \left|A\right|\left|A^{-1}\right|=1(\because \left|AB\right|=\left|A\right|\left|B\right|)$

$\Rightarrow \left|A^{-1}\right|=\frac{1}{\left|A\right|}$