91Ó°ÊÓ

Let $A=\left[\begin{array}{cc}\cos \mathrm{Î\pm }& −\sin \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]$be the given matrix.

In order to find the inverse of matrix, we find the determinant of the matrix.

Therefore,

$\begin{array}{c}\left|A\right|=\left|\begin{array}{cc}\cos \mathrm{Î\pm }& −\sin \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right|\\ ={\cos }^2\mathrm{Î\pm }+{\sin }^2\mathrm{Î\pm }\\ =1\\ ≠0\end{array}$

$∴\left|A\right|≠0$.

Thus, the inverse of matrix exists.

Let us find the inverse of matrix.

$\begin{array}{c}{A}^{−1}=\frac{1}{\left|A\right|}adjA\\ =\frac{1}{1}\left[\begin{array}{cc}\cos \mathrm{Î\pm }& \sin \mathrm{Î\pm }\\ −\sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]\\ =\left[\begin{array}{cc}\cos \mathrm{Î\pm }& \sin \mathrm{Î\pm }\\ −\sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]\end{array}$

Now, we have to interpret the linear transformation $T\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$and the inverse transformation$T^{−1}\left(\stackrel{→}{y}\right)=A^{−1}\stackrel{→}{y}$ .

Then

$T\left(\stackrel{→}{x}\right)=\left[\begin{array}{cc}\cos \mathrm{Î\pm }& −\sin \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]\stackrel{→}{x}$

And

role="math" localid="1660892785267" $\begin{array}{c}{T}^{−1}\left(\stackrel{→}{y}\right)=A^{−1}\stackrel{→}{y}\\ =\left[\begin{array}{cc}\cos \mathrm{Î\pm }& \sin \mathrm{Î\pm }\\ −\sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]\stackrel{→}{y}\end{array}$

Geometrically, $\left[\begin{array}{cc}\cos \mathrm{Î\pm }& −\sin \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]$represents a rotation of in the counter-clockwise direction while the inverse $\left[\begin{array}{cc}\cos \mathrm{Î\pm }& \sin \mathrm{Î\pm }\\ −\sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]$represents a rotation of$\mathrm{Î\pm }$ in the clockwise direction.

Now, let us interpret the determinant of$A$geometrically.

Consider the matrix$A=\left[\begin{array}{cc}\cos \mathrm{Î\pm }& −\sin \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }& \cos \mathrm{Î\pm }\end{array}\right]$.

Let$\stackrel{→}{v}=\left[\begin{array}{c}\cos \mathrm{Î\pm }\\ \sin \mathrm{Î\pm }\end{array}\right],\text{ â¶Ä‰}\stackrel{→}{w}=\left[\begin{array}{c}−\sin \mathrm{Î\pm }\\ \cos \mathrm{Î\pm }\end{array}\right]$.

Then$\det A=‖\stackrel{→}{v}‖\sin \mathrm{θ}‖\stackrel{→}{w}‖$, where$\mathrm{θ}$is the oriented angle from $\stackrel{→}{v}$to $\stackrel{→}{w}$.

Thus$\left|\det A\right|=‖\stackrel{→}{v}‖\text{ â¶Ä‰}\left|\sin \mathrm{θ}\right|\text{ â¶Ä‰}‖\stackrel{→}{w}‖$, is the area of the parallelogram spanned by$\stackrel{→}{v}$and.$\stackrel{→}{w}$

The graph is shown a below: