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Q18E

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Found in: Page 289

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

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

$detT=detB=8$.

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,

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

## Step 3: To find determinant.

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

$T\left(f\right)\left(t\right)=2f\left(t\right)+3f\text{'}\left(t\right)\phantom{\rule{0ex}{0ex}}T\left(f\right)\left(t\right)=2\left(a{t}^{2}+bt+c\right)+3\left(2at+b\right)\phantom{\rule{0ex}{0ex}}T\left(f\right)\left(t\right)=2a{t}^{2}+\left(6a+2b\right)t+\left(3b+2c\right)$

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

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

Therefore,

det T =det B = 8.