• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q26E

Expert-verified
Found in: Page 160

### Linear Algebra With Applications

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

# In Exercises 25 through 30, find the matrix B of the linear transformation ${\mathbit{T}}\left(\stackrel{\to }{x}\right){\mathbf{=}}{\mathbit{A}}\left(\stackrel{\to }{x}\right)$ with respect to the basis ${{J}}{\mathbf{=}}\left({\stackrel{\to }{v}}_{1},\dots ..,{\stackrel{\to }{v}}_{m}\right)$. ${\mathbit{A}}{\mathbf{=}}\left(\begin{array}{cc}0& 1\\ 2& 3\end{array}\right){\mathbf{;}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{=}}\left[\begin{array}{c}1\\ 2\end{array}\right]{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{=}}\left[\begin{array}{c}1\\ 1\end{array}\right]$

The matrix is, $B=\left(\begin{array}{cc}6& 4\\ -4& 3\end{array}\right)$.

See the step by step solution

## Step 1: Consider the vectors.

The matrix and vectors are,

$A=\left(\begin{array}{cc}0& 1\\ 2& 3\end{array}\right);{\stackrel{\to }{v}}_{1}=\left[\begin{array}{c}1\\ 2\end{array}\right],{\stackrel{\to }{v}}_{2}=\left[\begin{array}{c}1\\ 1\end{array}\right]$

## Step 2: Compute the matrix using formula.

The formula is, $B={S}^{-1}AS$ .

Compute the matrix S

The inverse of the matrix is,

$S=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\phantom{\rule{0ex}{0ex}}⇒{S}^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

For S put the values.

role="math" $S=\left({\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)\phantom{\rule{0ex}{0ex}}⇒{S}^{-1}=\frac{1}{-1}\left(\begin{array}{cc}1& -1\\ -2& 1\end{array}\right)$

Substitute these values in the formula.

$B={S}^{-1}AS\phantom{\rule{0ex}{0ex}}⇒B=\frac{1}{-1}\left(\begin{array}{cc}1& -1\\ -2& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 2& 3\end{array}\right)\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)\phantom{\rule{0ex}{0ex}}\therefore B=\left(\begin{array}{cc}6& 4\\ -4& 3\end{array}\right)$

The matrix is, $B=\left(\begin{array}{cc}6& 4\\ -4& 3\end{array}\right)$ .