Matrix Multiplication - Express a row as a linear combination $$ Let \ A \ = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 3 & 6 \\ \end{bmatrix} and 
\ let \ B = \begin{bmatrix} 0 & 1 & -3 \\ -3 & 1 & 4 \end{bmatrix}  $$
Express the third row of AB as a linear combination of the rows of B
$$ AB \ = \ \begin{bmatrix} -6 & 3 & 5 \\ -15 & 9 & 8 \\ -18 & 9 & 15 \end{bmatrix} $$
3rd row of AB would be
$$ \begin{bmatrix} 0 \ (-18) & 1 \ (9) & -3 \ (15) \\ -3 \ (-18) & 1 \ (9) & 4 \ (15) \end{bmatrix} $$
ANS: So the third row represented as a linear combination of the rows of B is given by:
$$ -18 \ \begin{bmatrix} \ \ \ 0 \\ -3  \end{bmatrix} \ \  
+ \ \ 9 \ \begin{bmatrix} \ \  1 \  \\ \ \ 1 \  \end{bmatrix}  \ \ 
+ \ \ 15 \ \begin{bmatrix} -3 \\ \ \ 4  \end{bmatrix} \ \ $$
Is my answer correct? 
If not any suggestion or help would be appreciated. 
Thanks for your time and cooperation from now.
 A: From the definition of matrix multiplication it follows that:
$$\underbrace{\begin{bmatrix}-18&9&15\end{bmatrix}}_{\text{third row of }AB} = 3\cdot \underbrace{\begin{bmatrix}0&1&-3\end{bmatrix}}_{\text{first row of }B}+6\underbrace{\begin{bmatrix}-3&1&4\end{bmatrix}}_{\text{second row of }B}$$
A: \begin{align} \textrm{third row of } AB 
&= \begin{bmatrix} 
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 0 \\ -3 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 4 \end{bmatrix}
\end{bmatrix} \\
&= \begin{bmatrix} 
3 \cdot \color{red}0 + 6 \cdot \color{blue}{(-3)} &
3 \cdot \color{red}1 + 6 \cdot \color{blue}1 &
3 \cdot \color{red}{(-3)} + 6 \cdot \color{blue}4
\end{bmatrix} \\
&= 3 \begin{bmatrix} \color{red}0 & \color{red}1 & \color{red}{-3} \end{bmatrix} + 6 \begin{bmatrix} \color{blue}{-3} & \color{blue}1 & \color{blue}4 \end{bmatrix} \\
&= 3(\textrm{first row of } B) + 6(\textrm{second row of } B).
\end{align}
