# 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 $L(2,-3,1)$?

• You have to learn what matrix multiplication is. See Wikipedia, for instance. – Bernard Nov 19 '15 at 22:51
• I know it sounds stupid but I multiplied them but I I am still not getting the right answer. – user287967 Nov 19 '15 at 22:55
• If you know the dot product, you have to do the dot product of each row of $A$ with the column-vector $\;\begin{bmatrix}2\\-3\\1\end{bmatrix}$. – Bernard Nov 19 '15 at 22:58
• I know. I am not even that stupid :P – user287967 Nov 19 '15 at 22:59
• What do you obtain? – Bernard Nov 19 '15 at 23:00

## 2 Answers

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.

• I did the same thing but I am still not getting the answer. That's why I asked it. – user287967 Nov 19 '15 at 22:57
• for my interest what answer did you get? – Chris Nov 19 '15 at 22:59
• [6 -4 -7] is my answer. – user287967 Nov 19 '15 at 22:59
• That is what I get as well so it is likely your answer key has a typo or the person making it made a mistake. It happens from time to time. – Chris Nov 19 '15 at 23:01

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 $v=\begin{pmatrix}x \\y\\z \end{pmatrix}.$