91Ó°ÊÓ

Let$A=\left[\begin{array}{cc}−3& 0\\ 0& −3\end{array}\right]$be the given matrix.

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

Therefore,

$\begin{array}{c}\left|A\right|=\left|\begin{array}{cc}−3& 0\\ 0& −3\end{array}\right|\\ =9−0\\ =9\\ ≠0\end{array}$

Then:

$∴\left|A\right|≠0$.

Thus, the inverse of matrix exists.

Let us find the inverse of matrix.

$\begin{array}{c}{A}^{−1}=\frac{1}{\left|A\right|}adjA\\ =\frac{1}{9}\left[\begin{array}{cc}−3& 0\\ 0& −3\end{array}\right]\\ =\left[\begin{array}{cc}−\frac{3}{9}& 0\\ 0& −\frac{3}{9}\end{array}\right]\\ =\left[\begin{array}{cc}−\frac{1}{3}& 0\\ 0& −\frac{1}{3}\end{array}\right]\end{array}$

Geometrically,$\left[\begin{array}{cc}−3& 0\\ 0& −3\end{array}\right]$ represents a rotation of 180 degrees and scaling of 3 while the inverse$\left[\begin{array}{cc}−\frac{1}{3}& 0\\ 0& −\frac{1}{3}\end{array}\right]$ represents a rotation of 180 degrees and scaling of 1/3.

Now, let us interpret the determinant of Ageometrically.

Consider the matrix $A=\left[\begin{array}{cc}−3& 0\\ 0& −3\end{array}\right]$.

Let $\stackrel{→}{v}=\left[\begin{array}{c}−3\\ 0\end{array}\right],\text{ â¶Ä‰}\stackrel{→}{w}=\left[\begin{array}{c}0\\ −3\end{array}\right]$.

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 $\stackrel{→}{w}$.