Problem on matrix multiplication

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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
    The matrix product of an $n \times m$ matrix with an $m \times \ell$ matrix is an $n \times \ell$ matrix.

    The $(i,j)$ entry of the matrix product $\bfA \mathbf{B}$ is the dot product of the $i$th row of $\bfA$ with the $j$th column of $\mathbf{B}$.
    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, $$\begin{pmatrix}1 & 0 & 2 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix} = \begin{pmatrix}11 & 14 \\ 17 & 20 \end{pmatrix}.$$

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