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.

From the definition of matrix multiplication it follows that:
$$\underbrace{\begin{bmatrix}-18&9&15\end{bmatrix}}_{\text{third row of }AB} = 3\cdot \underbrace{\begin{bmatrix}0&1&-3\end{bmatrix}}_{\text{first row of }B}+6\underbrace{\begin{bmatrix}-3&1&4\end{bmatrix}}_{\text{second row of }B}$$
