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Question

Mathematics Question on Determinants

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

A

det

B

1det(A)\frac{1}{det(A)}

C

1

D

0

Answer

1det(A)\frac{1}{det(A)}

Explanation

Solution

Since A is an invertible matrix,A-1 exists and A-1=1∣A∣adjA.\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=[d−b−ca]\begin{bmatrix}d&-b\\\\-c&a\end{bmatrix}.

Now,A-1=1∣A∣adjA\frac{1}{\mid A\mid} adj A=[d∣A∣−b∣A∣ −c∣A∣a∣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=∣d∣A∣−b∣A∣ −c∣A∣a∣A∣∣\begin{vmatrix}\frac{d}{\mid A\mid}&\frac{-b}{\mid A \mid}\\\ \frac{-c}{\mid A \mid}&\frac{a}{\mid A\mid}\end{vmatrix}
=1∣A∣2[d−b−ca]\frac{1}{\mid A \mid^2}\begin{bmatrix}d&-b\\\\-c&a\end{bmatrix}
=1∣A∣2(ad−bc)=1∣A∣2.∣A∣=1∣A∣\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)=1det(A)\frac{1}{det(A)}

Hence, the correct answer is B.