Question

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

26. $\mathit{T}\left(M\right)\mathbf{=}\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\mathit{M}\mathbf{+}\mathit{M}\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$from the space V of symmetric 2 × 2 matrices to V

Short answer

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

$detT=detB=-16$

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,

$T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]M+M\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

To find determinant.

For$M\in V$, we have

$M=\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]$

So we compute

$$\begin{gathered} T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]M+M\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right] \\ T\left(M\right)=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]+\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right] \\ T\left(M\right)=\left[\begin{array}{cc}2a+4b& 2a+4b+2d\\ 2a+4b+2d& 4b+6d\end{array}\right] \end{gathered}$$

Since,

$B=\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}$is a basis for$V$, this means that the matrix of$T$corresponding to$B$is

$B=\left[\begin{array}{ccc}2& 4& 0\\ 2& 4& 2\\ 0& 4& 6\end{array}\right]$

Therefore,

$detT=detB=-16$