Question

If A = $\left[\begin{array}{ccc}cosx&-sinx\\sinx&cosx\end{array}\right]$, then show that $(A^{-1} )^{-1}$

Answer 1

Given:

The matrix is:

$A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}$

To show:

$(A^{-1})^{-1}=A$

Solution:

If a matrix is:

$M=\begin{bmatrix}a&b\\c&d\end{bmatrix}$

Then,

$M^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$

We have,

$A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}$

Using the above formula, we get

$A^{-1}=\dfrac{1}{(\cos x)(\cos x)-(-\sin x)(\sin x)}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}$

$A^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}$

$A^{-1}=\dfrac{1}{1}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}$

$A^{-1}=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}$

Now, the inverse of $A^{-1}$ is:

$(A^{-1})^{-1}=\dfrac{1}{(\cos x)(\cos x)-(\sin x)(-\sin x)}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}$

$(A^{-1})^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}$

$(A^{-1})^{-1}=\dfrac{1}{1}A$

$(A^{-1})^{-1}=A$

Hence proved.