Question

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

Answer 1

We know that $A{A}^{-1}=I$
$\therefore |A{A}^{-1}|=|I|\Rightarrow |A||A^{-1}|=1$ (Using $A{A}^{-1}|=A||A^{-1}and|I|=1$)
$\Rightarrow |A^{-1}|=\frac{1}{|A|}=\frac{1}{det\left(A\right)}$. Hence, the correct options is (b).

Alternate method
Since, A is an invertible matrix, $A^{-1}$ exists and $A^{-1}=\frac{1}{|A|}adj\left(A\right)$
As matrix A is of order 2.

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ then, $|A|=ad-bc$ and $adj\left(A\right)=A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|}adj\left(A\right)=\frac{1}{|A|}A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]=\left[\begin{array}{cc}\frac{d}{|A|}& -\frac{d}{|A|}\\ -\frac{c}{|A|}& \frac{a}{|A|}\end{array}\right]$

$\therefore \left[\begin{array}{cc}\frac{d}{|A|}& -\frac{b}{|A|}\\ -\frac{c}{|A|}& \frac{a}{|A|}\end{array}\right]=\frac{1}{|A|}\times \frac{1}{|A|}\left|\begin{array}{cc}d& -b\\ -c& a\end{array}\right|=\frac{1}{|A{|}^2}(ad-bc)$
$={\frac{1}{|A{|}^2}}^{\prime }|A|(\because |A|=ad-bc)$
$=\frac{1}{|A|}$
$det\left(A^{-1}\right)=\frac{1}{det\left(A\right)}$. Hence, the correct option is (b).