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Q21E

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Found in: Page 289

### Linear Algebra With Applications

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

# Find the determinants of the linear transformations in Exercises 17 through 28.21. ${\mathbit{T}}{\mathbf{\left(}}{\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{f}}{\mathbf{\left(}}{\mathbf{-}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{}}{\mathbit{f}}{\mathbit{r}}{\mathbit{o}}{\mathbit{m}}{\mathbf{}}{{\mathbit{P}}}_{{\mathbf{3}}}{\mathbf{}}{\mathbit{t}}{\mathbit{o}}{\mathbf{}}{{\mathbit{P}}}_{{\mathbf{3}}}$

Therefore, the determinant of the linear transformations is given by,

det T = det B = 1

See the step by step solution

## Step 1: Definition.

A determinant is a unique number associated with a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

## Step 2: Given.

Given linear transformation,

$T\left(f\left(t\right)\right)=f\left(-t\right)from{P}_{3}to{P}_{3}$

## Step 3: To find determinant.

Consider the linear transformation $T:{P}_{3}\to {P}_{3}$defined as $T\left(f\right)=f\left(-t\right)$.

Consider the basis $B=\left\{1,t,{t}^{2},{t}^{3}\right\}$ for ${P}_{3}$.

We have that

$T\left(1\right)=1\phantom{\rule{0ex}{0ex}}T\left(1\right)=1\left(1\right)+0t+0{t}^{2}+0{t}^{3}\phantom{\rule{0ex}{0ex}}T\left(t\right)=-t\phantom{\rule{0ex}{0ex}}T\left(t\right)=0\left(1\right)+\left(-1\right)t+0{t}^{2}+0{t}^{3}\phantom{\rule{0ex}{0ex}}T\left({t}^{2}\right)={\left(-t\right)}^{2}T\left({t}^{2}\right)={t}^{2}=0\left(1\right)+0t+1{t}^{2}+0{t}^{3}and\phantom{\rule{0ex}{0ex}}T\left({t}^{3}\right)={\left(-t\right)}^{3}T\left({t}^{3}\right)=-{t}^{3}\phantom{\rule{0ex}{0ex}}T\left({t}^{3}\right)=0\left(1\right)+0t+0{t}^{2}+\left(-1\right){t}^{3}$

Thus the matrix of T with respect tois

$B=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]$

Observe that

$detB=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]\phantom{\rule{0ex}{0ex}}detB=1$

Therefore,

det T = det B =1