Question

which of the following is true, if A is an invertible square matrix :-

• A. ${\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T$
• B. ${\left(A^T\right)}^{-1}=\left(A^{-1}\right)\left(A^T\right)$
• C. $\left(A^T\right)={\left(A^{-1}\right)}^T$
• D. ${\left(A^T\right)}^{-1}=\left(A^T\right)\left(A^{-1}\right)$

Answer 1

The correct option is A ${\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T$

$\mathrm{Since}A\mathrm{is}\mathrm{invertible}\mathrm{matrix}.\left(A\right|\ne 0\to \left(A^T\right|\ne 0[\because (A^T|=(A\left|\right]A^T\mathrm{is}\mathrm{also}\mathrm{invertible}\mathrm{Now},A.A^{-1}=I=A^{-1}.A\Rightarrow {\left(A.A^{-1}\right)}^T={\left(I\right)}^T={\left(A^{-1}.A\right)}^T\left(\mathrm{by}\mathrm{reversal}\mathrm{law}\mathrm{for}\mathrm{transpose}\right]\Rightarrow {\left(A^{-1}\right)}^T\left(A^T\right)=I=A^T{\left(A^{-1}\right)}^T\Rightarrow {\left(A^T\right)}^{-1}={\left(A^{-1}\right)}^T$