veneciasp

Answered

2022-06-29

If the transformation is from ${\mathbb{R}}^{3}\to \mathbb{R}$ is

$T\{a,b,c\}={\int}_{0}^{\pi}2a{e}^{t}+2b\mathrm{sin}(t)+3c\mathrm{cos}(t)\phantom{\rule{thinmathspace}{0ex}}dt$

How to find the standard matrix?

$T\{a,b,c\}={\int}_{0}^{\pi}2a{e}^{t}+2b\mathrm{sin}(t)+3c\mathrm{cos}(t)\phantom{\rule{thinmathspace}{0ex}}dt$

How to find the standard matrix?

Answer & Explanation

Alexia Hart

Expert

2022-06-30Added 19 answers

By "standard matrix" maybe you mean the matrix with respect to the basis

$B=\{(1,0,0),(0,1,0),(0,0,1)\}$ for ${\mathbb{R}}^{3}$

We could rewrite $T$ this way:

$T:{\mathbb{R}}^{3}\to \mathbb{R}$

$T(v)=T((x,y,z))={\int}_{0}^{\pi}2x{e}^{t}+2ysin(t)+3zcos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

$=2x{\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt+2y{\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt+3z{\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

To write the matrix of a transformation with respect to some basis we should first apply the transformation on the basis vectors, so for example:

$T((1,0,0))=2(1){\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt+2(0){\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt+3(0){\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

$=2{\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt$

$=2({e}^{\pi}-{e}^{0})$

$=2({e}^{\pi}-1)$

To get the matrix you should apply $T$ to the other base vectors and form the matrix with 3 rows and 1 column.

$T((0,1,0))=2{\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt=4$

$T((0,0,1))=3{\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt=0$

So the matrix of $T$ with respect to $B$ is: $\left(\begin{array}{c}2({e}^{\pi}-1)\\ 4\\ 0\end{array}\right)$

To apply $T$ to any $(x,y,z)\in {\mathbb{R}}^{3}$ simply multiply:

$T((x,y,z))=\left(\begin{array}{ccc}x& y& z\end{array}\right)\left(\begin{array}{c}2({e}^{\pi}-1)\\ 4\\ 0\end{array}\right)$

$B=\{(1,0,0),(0,1,0),(0,0,1)\}$ for ${\mathbb{R}}^{3}$

We could rewrite $T$ this way:

$T:{\mathbb{R}}^{3}\to \mathbb{R}$

$T(v)=T((x,y,z))={\int}_{0}^{\pi}2x{e}^{t}+2ysin(t)+3zcos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

$=2x{\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt+2y{\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt+3z{\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

To write the matrix of a transformation with respect to some basis we should first apply the transformation on the basis vectors, so for example:

$T((1,0,0))=2(1){\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt+2(0){\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt+3(0){\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt$

$=2{\int}_{0}^{\pi}{e}^{t}\phantom{\rule{thinmathspace}{0ex}}dt$

$=2({e}^{\pi}-{e}^{0})$

$=2({e}^{\pi}-1)$

To get the matrix you should apply $T$ to the other base vectors and form the matrix with 3 rows and 1 column.

$T((0,1,0))=2{\int}_{0}^{\pi}sin(t)\phantom{\rule{thinmathspace}{0ex}}dt=4$

$T((0,0,1))=3{\int}_{0}^{\pi}cos(t)\phantom{\rule{thinmathspace}{0ex}}dt=0$

So the matrix of $T$ with respect to $B$ is: $\left(\begin{array}{c}2({e}^{\pi}-1)\\ 4\\ 0\end{array}\right)$

To apply $T$ to any $(x,y,z)\in {\mathbb{R}}^{3}$ simply multiply:

$T((x,y,z))=\left(\begin{array}{ccc}x& y& z\end{array}\right)\left(\begin{array}{c}2({e}^{\pi}-1)\\ 4\\ 0\end{array}\right)$

gaiaecologicaq2

Expert

2022-07-01Added 6 answers

You find the matrix of a transformation by calculating what the transformation does to a basis of the domain of the transformation.

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