Question

If $A$ is an invertible matrix of order $2$, then $det{\left(A\right)}^{-1}$ is equal to

• A. $det\left(A\right)$
• B. $\frac{1}{det\left(A\right)}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{det\left(A\right)}$

We know that

$A^{-1}=\frac{adjA}{|A|}$

$\Rightarrow adjA=|A|A^{-1}$

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

$\Rightarrow |A{|}^{n-1}=|A{|}^n|A^{-1}|$ (If A is a non-singular matrix of order n, then $|adjA|=|A{|}^{n-1}$ and $|kA|=k^n|A|$)

$\Rightarrow |A^{-1}|=|A{|}^{-1}$ for any square matrix

So, $det\left(A^{-1}\right)=\frac{1}{detA}$