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)=(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)=(1)/((\cos x)(\cos x)-(-\sin x)(\sin x))\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}`

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

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

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

`(A^(-1))^(-1)=(1)/(1)A`

`(A^(-1))^(-1)=A`

Hence proved.