## Problem on matrix multiplication Compute the matrix multiplication $$\begin{pmatrix}1 & 0 & 2 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$$
• ## Solution For convenience, we give names to the two matrices. Let $\bfA = \begin{pmatrix}1 & 0 & 2 \\ -1 & 1 & 3 \end{pmatrix}$ and $\bfB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}$. Recall that Because $\bfA$ is $2 \times 3$ and $\bfB$ is $3 \times 2$, their product is a $2\times2$ matrix. The $1,1$ entry of $\bfA\bfB$ is the product of the first row of $\bfA$ with the first column of $\bfB$: $$(\bfA \bfB)_{11} = \begin{pmatrix} 1 & 0 & 2 \end{pmatrix} \begin{pmatrix}1\\3\\5\end{pmatrix} = 1\cdot 1 + 0 \cdot 3 + 2 \cdot 5 = 11$$ The $1,2$ entry of $\bfA\bfB$ is the product of the first row of $\bfA$ with the second column of $\bfB$: $$(\bfA \bfB)_{12} = \begin{pmatrix} 1 & 0 & 2 \end{pmatrix} \begin{pmatrix}2\\4\\6\end{pmatrix} = 1\cdot 2 + 0 \cdot 4 + 2 \cdot 6 = 14$$ The $2,1$ entry of $\bfA\bfB$ is the product of the second row of $\bfA$ with the first column of $\bfB$: $$(\bfA \bfB)_{21} = \begin{pmatrix} -1 & 1 & 3 \end{pmatrix} \begin{pmatrix}1\\3\\5\end{pmatrix} = -1\cdot 1 + 1 \cdot 3 + 3 \cdot 5 = 17$$ The $2,2$ entry of $\bfA\bfB$ is the product of the second row of $\bfA$ with the second column of $\bfB$: $$(\bfA \bfB)_{22} = \begin{pmatrix} -1 & 1 & 3 \end{pmatrix} \begin{pmatrix}2\\4\\6\end{pmatrix} = -1\cdot 2 + 1 \cdot 4 + 3 \cdot 6 = 20$$ Hence, $$\begin{pmatrix}1 & 0 & 2 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix} = \begin{pmatrix}11 & 14 \\ 17 & 20 \end{pmatrix}.$$