Question

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

17.$\mathit{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{f}\mathbf{+}\mathbf{3}\mathit{f}\mathbf{\text{'}}\mathbf{}\mathbf{from}\mathbf{}{\mathit{P}}_{\mathbf{2}}\mathbf{}\mathbf{to}\mathbf{}{\mathit{P}}_{\mathbf{2}}$

Short answer

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

det T = det B = 8 .

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,

$T\left(\mathrm{f}\right)=2f+3f\text{'}\mathrm{from}{P}_2\mathrm{to}{P}_2$

To find determinant

For$f\in P_2$we have$f\left(t\right)=a{t}^2+bt+c$. So we compute

$$\begin{gathered} T\left(f\right)\left(t\right)=2f\left(t\right)+3f\text{'}\left(t\right) \\ T\left(f\right)\left(t\right)=2\left(a{t}^2+bt+c\right)+3\left(2at+b\right) \\ T\left(f\right)\left(t\right)=2a{t}^2+(6a+2b)t+(3b+2c) \end{gathered}$$

Since $\mathrm{B}=\left\{{\mathrm{t}}^2,\mathrm{t},1\right\}$is a basis for $P_2$, the matrix of T corresponding to B is

$B=\left[\begin{array}{ccc}2& 0& 0\\ 6& 2& 0\\ 0& 3& 2\end{array}\right]$

Therefore,

det T =det B = 8.