91Ó°ÊÓ

Let the matrix be,

$T\left(\stackrel{→}{x}\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\stackrel{→}{x}$

The letter $\mathit{L}$is made up of vectors $\left[\begin{array}{c}1\\ 0\end{array}\right]$and $\left[\begin{array}{c}0\\ 2\end{array}\right]$

Let the matrix be,

$T\left(\stackrel{→}{x}\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\stackrel{→}{x}$

Consider the vector $\left[\begin{array}{c}1\\ 0\end{array}\right]$

role="math" localid="1659697428926" $$\begin{gathered} T\left(\stackrel{→}{x}\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\stackrel{→}{x} \\ ⇒T\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right] \\ ∴T\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ -2\end{array}\right] \end{gathered}$$

Consider the vector $\left[\begin{array}{c}0\\ 2\end{array}\right]$

role="math" localid="1659697437236" $$\begin{gathered} T\left(\stackrel{→}{x}\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\stackrel{→}{x} \\ ⇒T\left[\begin{array}{c}0\\ 2\end{array}\right]=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\left[\begin{array}{c}0\\ 2\end{array}\right] \\ ∴T\left[\begin{array}{c}0\\ 2\end{array}\right]=\left[\begin{array}{l}4\\ 0\end{array}\right] \end{gathered}$$

Now, graph the original vectors and the obtained vectors as follow:

$T\stackrel{→}{\left(x\right)}$is obtained by rotating the vector$\stackrel{→}{x}$ through an angle of${90}^{∘}$ in the clockwise direction.

The matrix $\mathit{\left[}\begin{array}{cc}\mathit{a}& \mathit{b}\\ \mathit{c}& \mathit{d}\end{array}\mathit{\right]}$is invertible if and only if$\mathit{a}\mathit{d}\mathbf{-}\mathit{b}\mathit{c}\mathbf{≠}\mathbf{0}$.

The inverse of the matrix$\mathit{\left[}\begin{array}{cc}\mathit{a}& \mathit{b}\\ \mathit{c}& \mathit{d}\end{array}\mathit{\right]}$is, ${\mathit{\left[}\begin{array}{cc}\mathit{a}& \mathit{b}\\ \mathit{c}& \mathit{d}\end{array}\mathit{\right]}}^{\mathit{-}\mathit{1}}\mathit{=}\frac{1}{ad-bc}\mathit{\left[}\begin{array}{cc}\mathit{d}& \mathit{-}\mathit{b}\\ \mathit{-}\mathit{c}& \mathit{a}\end{array}\mathit{\right]}$

Consider the matrix,

$\begin{array}{rcl}T& =& \left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\\ & =& (0×0)-(2×-2)\\ & =& 4≠0\end{array}$

The inverse of the matrix is,

$\begin{array}{rcl}{\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]}^{-1}& =& \frac{1}{0×0-2×(-2)}\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]\\ & =& \frac{1}{4}\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]\\ & =& \left[\begin{array}{cc}0& -\frac{1}{2}\\ \frac{1}{2}& 0\end{array}\right]\end{array}$

Therefore, the matrix $\begin{array}{rcl}& & \left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]\end{array}$is invertible and it’s inverse is $\begin{array}{rcl}& & \left[\begin{array}{cc}0& -\frac{1}{2}\\ \frac{1}{2}& 0\end{array}\right]\end{array}$, and the shape of L gets transformed in terms of scalability of 2 and also gets rotated by ${90}^{∘}$.