91Ó°ÊÓ

Let be$A=\left[\begin{array}{cc}1& 1\\ −1& 1\end{array}\right]$the given matrix.

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

Therefore,

$$\begin{gathered} \begin{array}{c}\left|A\right|=\left|\begin{array}{cc}1& 1\\ −1& 1\end{array}\right|\\ =1+1\\ =2\\ ≠0\end{array} \\ ∴\left|A\right|≠0 \end{gathered}$$

.

Thus, the inverse of matrix exists.

Let us find the inverse of matrix.

$\begin{array}{c}{A}^{−1}=\frac{1}{\left|A\right|}adjA\\ =\frac{1}{2}\left[\begin{array}{cc}1& −1\\ 1& 1\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{2}& −\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]\end{array}$

Geometrically,$\left[\begin{array}{cc}1& 1\\ −1& 1\end{array}\right]$ represents a scaling of$\sqrt{2}$ and rotation clockwise through an angle$\frac{\mathrm{π}}{4}$ and its inverse$\left[\begin{array}{cc}\frac{1}{2}& −\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]$ represents a scaling ofdata-custom-editor="chemistry" $\frac{\sqrt{2}}{2}$ and rotation counter-clockwise through an angle $\frac{\mathrm{π}}{4}$.

Now, let us interpret the determinant of Ageometrically.

Consider the matrix $A=\left[\begin{array}{cc}1& 1\\ −1& 1\end{array}\right]$.

Let $\stackrel{→}{v}=\left[\begin{array}{c}1\\ −1\end{array}\right],\text{ â¶Ä‰}\stackrel{→}{w}=\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Then $\det A=||\stackrel{→}{v}||\sin \mathrm{θ}||\stackrel{→}{w}||$, where θ is the oriented angle from $\stackrel{→}{v}$to$\stackrel{→}{w}$ and $\mathrm{θ}=\frac{\mathrm{π}}{4}$.

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 $\stackrel{→}{w}$.