Emmy Combs

2022-01-29

Matrix transformation
$f:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$
$\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)$
Establish x,y and z such that,
$f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$
Do I just need to multiply the values of f for the $3×3$ matrix? What does this mean overall?

dodato0n

Step 1 You need to solve the non-homogeneous system $\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4\\ 5\\ -1\end{array}\right)$ Form then the augmented coefficients matrix and reduce it by rows: $\left(\begin{array}{ccccc}4& 1& 3& :& 4\\ 2& -1& 3& :& 5\\ 2& 2& 0& :& -1\end{array}\right)\stackrel{{R}_{1}↔{R}_{3}\cdot \left(\frac{1}{2}\right)}{⟶}\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 2& -1& 3& :& 5\\ 4& 1& 3& :& 4\end{array}\right)\stackrel{{R}_{2}-2{R}_{1}\phantom{\rule{0ex}{0ex}}{R}_{3}-4{R}_{1}}{⟶}$ $\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 0& -3& 3& :& 6\\ 0& -3& 3& :& 6\end{array}\right)\stackrel{{R}_{3}-{R}_{2}}{⟶}\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 0& -3& 3& :& 6\\ 0& 0& 0& :& 0\end{array}\right)$ Well, there are infinite solutions to the above system: begin with row 3 , and remember the columns represent x,y,z from left to right, so: ${R}_{2}:\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}-3y+3z=6\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}y=z-2\phantom{\rule{0ex}{0ex}}{R}_{1}:\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}x+y=-\frac{1}{2}\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}x=-\frac{1}{2}-y=-\frac{1}{2}-z+2=\frac{3}{2}-z$ Now just choose a nice z , say $z=0\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}y=-2\phantom{\rule{0.167em}{0ex}},\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}x=\frac{3}{2}$, and one solution to your problem is $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3/2\\ -2\\ 0\end{array}\right)$ as you can easily check.

kumewekwah0

Step 1
You are given that
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$
Of course,
$\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4x+y+3z\\ 2x-y+3z\\ 2x+2y\end{array}\right)$
So we have
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4x+y+3z\\ 2x-y+3z\\ 2x+2y\end{array}\right)$
That gives us three equations:
$x=4x+y+3z\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}3x+y+3z=0.$
$y=2x-y+3z\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}2x-2y+3z=0.$
$z=2x+2y\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}2x+2y-z=0.$
An obvious solution to that is $x=y=z=0$. Notice that the determinant is $\begin{array}{rl}|\begin{array}{ccc}3& 1& 3\\ 2& -2& 3\\ 2& 2& -1\end{array}|& =3|\begin{array}{cc}-2& 3\\ 2& -1\end{array}|-|\begin{array}{cc}2& 3\\ 2& -1\end{array}|+3|\begin{array}{cc}2& -2\\ 2& 2\end{array}|\\ & =3\left(2-6\right)-\left(-2-6\right)+3\left(4+4\right)\\ & =-12+8+24\\ & =20\end{array}$
Since that is not $0,x=y=z=0$ is the only solution.

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