Matrix transformation $f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ I am self-learning Linear Algebra (mainly through Khan Academy and Youtube). I encountered this linear transformation, and was wondering how would this be solved.
$f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$
$f \begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} $  = $ \begin{pmatrix}
4 &amp; 1 &amp; 3\\ 
2 &amp;-1 &amp; 3\\ 
2 &amp; 2 &amp; 0
\end{pmatrix} $ $\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} $
Establish $x, y$ and $z$ such that,
$f\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} $ = $\begin{pmatrix}
4\\ 
5\\ 
-1
\end{pmatrix} $
Do I just need to multiply the values of $f$ for the $3\times3$ matrix? What does this mean overall?
You need to solve the non-homogeneous system
$$ \begin{pmatrix}
4 &amp; 1 &amp; 3\\ 
2 &amp;-1 &amp; 3\\ 
2 &amp; 2 &amp; 0
\end{pmatrix} \begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix} =  \begin{pmatrix}
4\\ 
5\\ 
-1
\end{pmatrix}$$
Form then the augmented coefficients matrix and reduce it by rows:
$$ \begin{pmatrix}
4 &amp; 1 &amp; 3&amp;:&amp;4\\ 
2 &amp;-1 &amp; 3&amp;:&amp;5\\ 
2 &amp; 2 &amp; 0&amp;:&amp;-1
\end{pmatrix}\stackrel{R_1\leftrightarrow R_3\cdot\left(\frac12\right)}\longrightarrow\begin{pmatrix}
1 &amp; 1 &amp; 0&amp;:&amp;-1/2\\
2 &amp;-1 &amp; 3&amp;:&amp;5\\
4 &amp; 1 &amp; 3&amp;:&amp;4\\ 
 \end{pmatrix}\stackrel{R_2-2R_1\\R_3-4R_1}\longrightarrow$$
$$\begin{pmatrix}
1 &amp; 1 &amp; 0&amp;:&amp;-1/2\\
0 &amp;-3 &amp; 3&amp;:&amp;6\\
0 &amp; -3 &amp; 3&amp;:&amp;6\\ 
 \end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1 &amp; 1 &amp; 0&amp;:&amp;-1/2\\
0 &amp;-3 &amp; 3&amp;:&amp;6\\
0 &amp; 0 &amp; 0&amp;:&amp;0\\ 
 \end{pmatrix}$$
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:
$$R_2:\;\;-3y+3z=6\implies y=z-2\\R_1:\;\;x+y=-\frac12\implies x=-\frac12-y=-\frac12-z+2=\frac32-z$$
Now just choose a nice $\;z\;$ , say $\;z=0\implies y=-2\,,\,\,x=\cfrac32\;$ , and one solution to your problem is
$$\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
3/2\\ 
-2\\ 
0
\end{pmatrix}$$
as you can easily check.
You are given that $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4 &amp; 1 &amp; 3 \\ 2 &amp; -1 &amp; 3 \\ 2 &amp; 2 &amp; 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}$.
Of course, $\begin{pmatrix}4 &amp; 1 &amp; 3 \\ 2 &amp; -1 &amp; 3 \\ 2 &amp; 2 &amp; 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.
So we have $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.
That gives us three equations:
$$x= 4x+ y+ 3z\quad\text{which reduces to}\quad 3x+ y+ 3z= 0.$$
$$y= 2x- y+ 3z\quad\text{which reduces to}\quad 2x- 2y+ 3z= 0.$$
$$z= 2x+ 2y\quad\text{which reduces to}\quad 2x+ 2y- z= 0.$$
An obvious solution to that is $x= y= z= 0.$
Notice that the determinant is
$$\begin{align*}
\left|\begin{array}{ccc} 3 &amp; 1 &amp; 3 \\ 2 &amp; -2 &amp; 3 \\ 2 &amp; 2 &amp; -1 \end{array}\right|&amp;= 3\left|\begin{array}{cc} -2 &amp; 3 \\ 2 &amp; -1\end{array}\right|- \left|\begin{array}{cc}2 &amp; 3 \\ 2 &amp; -1\end{array}\right|+ 3\left|\begin{array}{cc} 2 &amp; -2 \\ 2 &amp; 2 \end{array}\right|\\[5pt]
&amp;= 3(2- 6)- (-2- 6)+ 3(4+ 4)\\[5pt]
&amp;= -12+ 8+ 24\\[5pt]
&amp;= 20\\[5pt]
\end{align*}$$
Since that is not $0, x= y= z= 0$  is the only solution.