matrix multiplication mixup I was watching a youtube video on linear algebra and it gave the following equality.
$\frac{1}{3}\left(\begin{matrix}1&2\\-1&1\\\end{matrix}\right)\left(\begin{matrix}4\\1\\\end{matrix}\right)=\left(\begin{matrix}2\\-1\\\end{matrix}\right)$
I must be mixing up my basics of matrix multiplication, but when I try it I do not get the same answer. Is it possible to break down the steps of this multiplication to help with my understanding?
 A: $$\frac 13 \begin{bmatrix} 1\cdot 4 + 2\cdot 1 \\ -1\cdot 4 + 1\cdot 1\end{bmatrix} $$ $$= \frac 13 \begin{bmatrix} 6\\ -3\end{bmatrix}$$
$$= \begin{bmatrix} \frac 63\\ \frac {-3}{3}\end{bmatrix}$$
$$= \begin{bmatrix}2\\-1\end{bmatrix}$$
A: $$
\left(
\begin{matrix}
1 & 2 \\
-1 & 1
\end{matrix}
\right)
\left(\begin{matrix}
4 \\ 1
\end{matrix}\right)
=
\left(\begin{matrix}
1·4+2·1 \\
(-1)·4+1·1
\end{matrix}\right)
=
\left(\begin{matrix}
6\\
-3
\end{matrix}\right)
$$
A: To get the $(i,j)$ th element of the product matrix , multiply the $i$th  rows of the first matrix with the $j$th column of the second matrix just as in usual dot product of two vectors.
A: We have $\left(\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\\\end{matrix}\right)\left(\begin{matrix}c\\d\\\end{matrix}\right)=\left(\begin{matrix}a_{11}c+a_{12}d\\a_{21}c+a_{22}d\\\end{matrix}\right)$  .
A: We can first expand out $$\frac13 \left(\begin{matrix}1&2\\-1&1\\\end{matrix}\right)$$ as $$\left(\begin{matrix}\frac13&\frac23\\-\frac13&\frac13\\\end{matrix}\right).$$ We can than multiply the two matrices to get $$\left(\begin{matrix}2\\-1\\\end{matrix}\right),$$ as desired.
