Question

If $A$ is an invertible matrix, then ${\left(A^{-1}\right)}^{\prime }$ is equal to

• A. $-\left(A^{-1}\right)$
• B. ${\left(A^{-1}\right)}^2$
• C. ${\left(A^{\prime }\right)}^{-1}$
• D. Does not exist

Answer 1

Answer: (C)

Since, $A$ is an invertible matrix, so it is non-singular.
We know that, $|A|=\left|A^{\prime }\right|$. But $|A|\ne 0$. So, $\left|A^{\prime }\right|\ne 0$ i.e., $A^{\prime }$ is invertible matrix.
Also, we know that, $A{A}^{-1}=A^{-1}A=1$.
Taking transpose on both sides, we get
$\left(A^{-1}\right)A^{\prime }=A^{\prime }{\left(A^{-1}\right)}^{\prime }=(I{)}^{\prime }=I$
Hence, $\left(A^{-1}\right)$ is inverse of $A^{\prime }$, i.e., ${\left(A^{\prime }\right)}^{-1}=\left(A^{-1}\right)$.