Question

Find the determinants of the linear transformations in Exercises 17 through 28.

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

Short answer

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

det T = det B = 1

Step-by-step solution

Definition. 

A determinant is a unique number associatedwith 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. 

Given linear transformation,

$T\left(f\right(t\left)\right)=f(-t)from{P}_{3}to{P}_3$

To find determinant.

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

$$\begin{gathered} T\left(1\right)=1 \\ T\left(1\right)=1\left(1\right)+0t+0t^2+0t^3 \\ T\left(t\right)=-t \\ T\left(t\right)=0\left(1\right)+(-1)t+0t^2+0t^3 \\ T\left(t^2\right)={(-t)}^{2}T\left(t^2\right)=t^2=0\left(1\right)+0t+1t^2+0t^{3}and \\ T\left(t^3\right)={(-t)}^{3}T\left(t^3\right)=-t^3 \\ T\left(t^3\right)=0\left(1\right)+0t+0t^2+(-1)t^3 \end{gathered}$$

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

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

Therefore,

det T = det B =1