Question

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

20.$\mathit{L}\mathit{\left(}\mathit{A}\mathit{\right)}\mathbf{=}{\mathit{A}}^{\mathit{T}}\mathbf{\hspace{0.33em}}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{\hspace{0.33em}}{\mathbf{ℝ}}^{\mathbf{2}\mathbf{\times }\mathbf{2}}\mathit{\hspace{0.33em}}\mathit{t}\mathit{o}\mathbf{\hspace{0.33em}}{\mathbf{ℝ}}^{\mathbf{2}\mathbf{\times }\mathbf{2}}$

Short answer

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

$detT=detB=-1$.

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,

$L\mathit{\left(}A\mathit{\right)}=A^T\hspace{0.33em}from\hspace{0.33em}{\mathrm{ℝ}}^{2\times 2}\mathit{\hspace{0.33em}}to\hspace{0.33em}{\mathrm{ℝ}}^{2\times 2}$

To find determinant.

Consider the linear transformation$T:R^{2\times 2}\to R^{2\times 2}$defined as $T\left(f\right)=A^T$.

Consider the basis $B=\left\{E_{11},E_{12},E_{21},E_{22}\right\}$for${\mathrm{ℝ}}^{2\times 2}$where

$$\begin{gathered} E_{11}=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right] \\ {E}_{12}=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right] \\ {E}_{21}=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]and \\ {E}_{22}=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \end{gathered}$$

We have that

$$\begin{gathered} \mathrm{T}\left({\mathrm{E}}_{11}\right)=({\mathrm{E}}_{11}{)}^{\mathrm{T}}={\mathrm{E}}_{11}=1{\mathrm{E}}_{11}+0{\mathrm{E}}_{12}+0{\mathrm{E}}_{21}+0{\mathrm{E}}_{21} \\ \mathrm{T}({\mathrm{E}}_{12})=({\mathrm{E}}_{12}{)}^{\mathrm{T}}={\mathrm{E}}_{21}=0{\mathrm{E}}_{11}+0{\mathrm{E}}_{12}+1{\mathrm{E}}_{21}+0{\mathrm{E}}_{22} \\ \mathrm{T}\left({\mathrm{E}}_{21}\right)=({\mathrm{E}}_{21}{)}^{\mathrm{T}}={\mathrm{E}}_{12}=0{\mathrm{E}}_{11}+1{\mathrm{E}}_{12}+0{\mathrm{E}}_{21}+0{\mathrm{E}}_{22}\mathrm{and} \\ \mathrm{T}({\mathrm{E}}_{22})=({\mathrm{E}}_{22}{)}^{\mathrm{T}}={\mathrm{E}}_{22}=0{\mathrm{E}}_{11}+0{\mathrm{E}}_{12}+0{\mathrm{E}}_{21}+1{\mathrm{E}}_{22} \end{gathered}$$

Thus the matrix ofT with respect to B is

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

Observe that

$$\begin{gathered} detB=\left|\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right| \\ det\left(B\right)=-1 \end{gathered}$$

Therefore,

$detT=detB=-1$.