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Q20E

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Found in: Page 54

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Give a geometric interpretation of the linear transformations defined by the matrices in Exercises ${\mathbf{16}}$ through ${\mathbf{23}}$. Show the effect of these transformations on the letter ${\mathbit{L}}$ considered in Example ${\mathbf{5}}$. In each case, decide whether the transformation is invertible. Find the inverse if it exists, and interpret it geometrically. See Exercise ${\mathbf{13}}$.20. $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

The matrix $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ has scalable change in shape of L and is invertible with inverse .

$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

The geometrical interpretation is:

See the step by step solution

## Step by Step Explanation: Step 1: Consider the matrix.

Let the matrix be,

$T\left(\stackrel{\to }{x}\right)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\stackrel{\to }{x}$

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

## Step 2: Compute the vectors.

Let the matrix be,

$T\left(\stackrel{\to }{x}\right)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\stackrel{\to }{x}$

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

role="math" localid="1659689477902" $T\left(\stackrel{\to }{x}\right)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\stackrel{\to }{x}\phantom{\rule{0ex}{0ex}}⇒T\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{lc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\therefore T\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]$

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

$T\left(\stackrel{\to }{x}\right)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\stackrel{\to }{x}\phantom{\rule{0ex}{0ex}}⇒T\left[\begin{array}{c}0\\ 2\end{array}\right]=\left[\begin{array}{lc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}0\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}\therefore T\left[\begin{array}{c}0\\ 2\end{array}\right]=\left[\begin{array}{c}2\\ 0\end{array}\right]$

## Step 3: Graph the letter using matrix.

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

$T\left(\stackrel{\to }{x}\right)$ is obtained by scaling the vector $\stackrel{\to }{x}$.

## Step 4: Check for the invertibility of the matrix and find the inverse if exists.

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

The inverse of the matrix ${\mathbf{\left[}}\begin{array}{cc}\mathbf{a}& \mathbf{b}\\ \mathbf{c}& \mathbf{d}\end{array}{\mathbf{\right]}}$ is role="math" localid="1659690085262" ${\mathbf{\left[}\begin{array}{cc}\mathbf{a}& \mathbf{b}\\ \mathbf{c}& \mathbf{d}\end{array}\mathbf{\right]}}^{\mathbf{-}\mathbf{1}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{a}\mathbf{d}\mathbf{-}\mathbf{b}\mathbf{c}}{\mathbf{\left[}}\begin{array}{cc}\mathbf{d}& \mathbf{-}\mathbf{b}\\ \mathbf{-}\mathbf{c}& \mathbf{a}\end{array}{\mathbf{\right]}}$ .

Consider the matrix,

$T=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left(0×0\right)-\left(1×1\right)=-1\ne 0$

Therefore, the matrix is invertible.

The inverse of the matrix is,

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

Therefore, the matrix $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ is invertible and its inverse is $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, and the shape of L gets transformed in terms of scalability