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Q27E

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Found in: Page 287

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the determinants of the linear transformations in Exercises 17 through 28.27. ${\mathbit{T}}\left(f\right){\mathbf{=}}{\mathbit{a}}{\mathbit{f}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbit{b}}{\mathbit{f}}{\mathbf{"}}$, where a and b are arbitrary constants, from the space V spanned by ${\mathbit{c}}{\mathbit{o}}{\mathbit{s}}\left(x\right)$ and ${\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\left(x\right)$to V

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

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

See the step by step solution

## Step 1: Definition.

A determinant is a unique number associated with 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.

## Step 2: Given.

Given linear transformation,

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

## Step 3: To find determinant.

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

So we compute

$T\left(f\right)\left(x\right)=af\text{'}\left(x\right)+bf"\left(x\right)\phantom{\rule{0ex}{0ex}}T\left(f\right)\left(x\right)=a\left(-\alpha \mathrm{sin}\left(x\right)+\beta \mathrm{cos}\left(x\right)\right)+b\left(-\alpha \mathrm{cos}\left(x\right)-\beta \mathrm{sin}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}T\left(f\right)\left(x\right)=\left(-b\alpha +a\beta \right)\mathrm{cos}\left(x\right)+\left(-a\alpha -b\beta \right)\mathrm{sin}\left(x\right)$

Obviously $B=\left\{\mathrm{cos}\left(x\right),\mathrm{sin}\left(x\right)\right\}$ 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,

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