StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q21E

Expert-verifiedFound in: Page 289

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. ${\mathit{T}}{\mathbf{\left(}}{\mathit{f}}{\mathbf{\right(}}{\mathit{t}}{\mathbf{\left)}}{\mathbf{\right)}}{\mathbf{=}}{\mathit{f}}{\mathbf{(}}{\mathbf{-}}{\mathit{t}}{\mathbf{)}}{\mathbf{}}{\mathit{f}}{\mathit{r}}{\mathit{o}}{\mathit{m}}{\mathbf{}}{{\mathit{P}}}_{{\mathbf{3}}}{\mathbf{}}{\mathit{t}}{\mathit{o}}{\mathbf{}}{{\mathit{P}}}_{{\mathbf{3}}}$**

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

det *T* = det *B* = 1

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.

Given linear transformation,

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

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

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

We have that

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

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

94% of StudySmarter users get better grades.

Sign up for free