Question

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

25. $\mathit{T}\left(M\right)\mathbf{=}\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]$M from the space V of upper triangular 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,

$T\left(M\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]$

To find determinant.

For $M\in V$, we have

$M=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]$

So we compute

$$\begin{gathered} T\left(M\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]M \\ T\left(M\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right] \\ T\left(M\right)=\left[\begin{array}{cc}2a& 2b+3d\\ 0& 4d\end{array}\right] \end{gathered}$$

Since$B=\left\{E_{1,1},E_{1,2},E_{2,2}\right\}$is the canonical basis for$V$, this means that the matrix of$T$corresponding to$B$is

$B=\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 3\\ 0& 0& 4\end{array}\right]$

Therefore,

$deT=detB=16.$