<s>[INST] Find a vector whose image under $T$ is the vector $b$ 
Hey everyone, I'm having some trouble solving this problem. To find the vector can I multiply the matrix $A$ by the column vector $[a, b, c]$ and set that equal to the vector $b$? or do I have to row reduce to solve this problem? Thanks for the help
 [/INST] An alternative way, using advanced methods: Scalar product and cross product.
Setting the columns from $A$
$$u = \left( {\begin{array}{*{20}{c}}
  4 \\ 
  { - 3} \\ 
  5 
\end{array}} \right),v = \left( {\begin{array}{*{20}{c}}
  4 \\ 
  { - 6} \\ 
  { - 3} 
\end{array}} \right),w = \left( {\begin{array}{*{20}{c}}
  3 \\ 
  2 \\ 
  6 
\end{array}} \right)$$
then I do first check, if $\{ u,v,w\}$ is a set of linear independent vectors.
Because of
$$v \times w = \left( {\begin{array}{*{20}{c}}
  4 \\ 
  { - 6} \\ 
  { - 3} 
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  3 \\ 
  2 \\ 
  6 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 36 + 6} \\ 
  { - 9 - 24} \\ 
  {8 + 18} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 30} \\ 
  { - 33} \\ 
  {26} 
\end{array}} \right)$$
and
$$u \cdot (v \times w) = \left( {\begin{array}{*{20}{c}}
  4 \\ 
  { - 3} \\ 
  5 
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  { - 30} \\ 
  { - 33} \\ 
  {26} 
\end{array}} \right) =  - 120 + 99 + 130 = 109 \ne 0$$
it is the case: $\{ u,v,w\} $ is a set of linear independent vectors.
Can be used as basis. Therefore $\{ u,v,w,b\}$ is a set of linear dependent
vectors. There exists a unique linear combination:
$$xu + yv + zw = b$$
These numbers are computed by the formulae:
$$\begin{gathered}
  x = \frac{{b \cdot (v \times w)}}{{u \cdot (v \times w)}} \hfill \\
  y = \frac{{b \cdot (u \times w)}}{{v \cdot (u \times w)}} \hfill \\
  z = \frac{{b \cdot (u \times v)}}{{w \cdot (u \times v)}} \hfill \\ 
\end{gathered} $$
Known as Cramer's rule, written in an old fashion style. 
</s>