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Q18E

Expert-verifiedFound in: Page 289

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

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

$detT=detB=8$.

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.

Given linear transformation,

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

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

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

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.

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