Question

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

27. $\mathit{T}\left(f\right)\mathbf{=}\mathit{a}\mathit{f}\mathbf{\text{'}}\mathbf{+}\mathit{b}\mathit{f}\mathbf{"}$, where a and b are arbitrary constants, from the space V spanned by $\mathit{c}\mathit{o}\mathit{s}\left(x\right)$and $\mathit{s}\mathit{i}\mathit{n}\left(x\right)$to V

Short answer

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

$detT=detB=a^2+b^2$

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,

$T\left(f\right)=af\text{'}+bf"$

To find determinant.

For$f\in V$, we have$f\left(x\right)=\alpha \cos \left(x\right)+\beta \sin \left(x\right)$.

So we compute

$$\begin{gathered} T\left(f\right)\left(x\right)=af\text{'}\left(x\right)+bf"\left(x\right) \\ T\left(f\right)\left(x\right)=a\left(-\alpha \sin \left(x\right)+\beta \cos \left(x\right)\right)+b\left(-\alpha \cos \left(x\right)-\beta \sin \left(x\right)\right) \\ T\left(f\right)\left(x\right)=\left(-b\alpha +a\beta \right)\cos \left(x\right)+\left(-a\alpha -b\beta \right)\sin \left(x\right) \end{gathered}$$

Obviously $B=\left\{\cos \left(x\right),\sin \left(x\right)\right\}$is a basis for $V$, so the matrix of $T$that corresponds to $B$is

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

Therefore,

$detT=detB=a^2+b^2$.