Linear Transformation- Standard Matrix A standard matrix is given: $$A=\begin{bmatrix} 0 & -1 & 3 \\ 1 & 1 & -3 \\ 2 & 2 & -5 \end{bmatrix}$$ representing the linear transformation $L: \mathbb{R}^3 -> \mathbb{R}^3$.
How to find $L(2,-3,1)$?
 A: Assuming that this matrix is given with respect to the standard basis $(e_1,e_2,e_3)$, then the columns of the matrix are just $(L(e_1),L(e_2),L(e_3))$, respectively. Thus, $$L(e_1)=(0,1,2), \quad L(e_2)=(-1,1,2), \quad L(e_3) = (3,-3,-5).$$
Hence, $$L(2,-3,1)=L(2e_1-3e_2+e_3)=2L(e_1)-3L(e_2)+L(e_3) = (6,-4,-7).$$
Alternatively, for any vector $v \in \mathbb{R}^3$, the following is true: $$Lv=Av,$$ where $v=\begin{pmatrix}x \\y\\z \end{pmatrix}.$
A: The way I think of putting a vector through a matrix is you push it down from the top then add across the sides. I will show you this approach in a general way.
$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\Rightarrow\begin{bmatrix}ax&by&cz\\dx&ey&fz\\gx&hy&iz\end{bmatrix}\Rightarrow\begin{bmatrix}ax+by+cz\\dx+ey+fz\\gx+hy+iz\end{bmatrix}$. This gives you the anwser:
$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}ax+by+cz\\dx+ey+fz\\gx+hy+iz\end{bmatrix}$.
Now for your problem plug in your values and multiply then add.
