Question

If A is an invertible matrix of order 2,then det(A^-1) is equal to

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

Answer 1

The Correct Option is B

Since A is an invertible matrix,A^-1 exists and A^-1=\(\frac{1}{\mid A\mid} adj A.\)  

As matrix A is of order 2,let A=\(\begin{bmatrix}a&b\\c&d\end{bmatrix}\).        

Then, IAI=ad-bc and adjA=\(\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).

Now,A^-1=\(\frac{1}{\mid A\mid} adj A\)=\(\begin{bmatrix}\frac{d}{\mid A\mid}&\frac{-b}{\mid A \mid}\\ \frac{-c}{\mid A \mid}&\frac{a}{\mid A\mid}\end{bmatrix}\).

therefore IA^-1I=\(\begin{vmatrix}\frac{d}{\mid A\mid}&\frac{-b}{\mid A \mid}\\ \frac{-c}{\mid A \mid}&\frac{a}{\mid A\mid}\end{vmatrix}\)
=\(\frac{1}{\mid A \mid^2}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
=\(\frac{1}{\mid A \mid^2}(ad-bc)=\frac{1}{\mid A \mid^2}.\mid A \mid=\frac{1}{\mid A\mid}\)

\(\therefore\) det(A^-1)=\(\frac{1}{det(A)}\)

Hence, the correct answer is B.