11.4.1 Example of Matrix-Matrix Multiplication via Dot Products

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Matrix-Matrix Multiplication via Dot Products

X = S_{21} (- 3) = [\begin{matrix} 1 & 0 \\ - 3 & 1 \end{matrix}] and A = [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 4 \end{matrix}]

B = X A

\begin{proof}[Solution.]
Since we are doing matrix-matrix multiplication via dot products, we will be finding $B$ using an entry-by-entry approach.

B = {[\begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{matrix}]}_{2 \times 3}

Let's compute the individual entries of $B$ .

\begin{aligned} b_{11} & = X (1, :) \cdot A (:, 1) = [\begin{array}{c} 1 & 0 \end{array}] [\begin{array}{c} 2 \\ 6 \end{array}] = 1 (2) + 0 (6) = 2, \\ b_{12} & = X (1, :) \cdot A (:, 2) = [\begin{array}{c} 1 & 0 \end{array}] [\begin{array}{c} 1 \\ 2 \end{array}] = 1 (1) + 0 (2) = 1, \\ b_{13} & = X (1, :) \cdot A (:, 3) = [\begin{array}{c} 1 & 0 \end{array}] [\begin{array}{c} 2 \\ 4 \end{array}] 1 (2) + 0 (4) = 2, \\ b_{21} & = X (2, :) \cdot A (:, 1) = [\begin{array}{c} - 3 & 1 \end{array}] [\begin{array}{c} 2 \\ 6 \end{array}] = - 3 (2) + 1 (6) = 0, \\ b_{22} & = X (2, :) \cdot A (:, 2) = [\begin{array}{c} - 3 & 1 \end{array}] [\begin{array}{c} 1 \\ 2 \end{array}] = - 3 (1) + 1 (2) = - 1, \\ b_{23} & = X (2, :) \cdot A (:, 3) = [\begin{array}{c} - 3 & 1 \end{array}] [\begin{array}{c} 2 \\ 4 \end{array}] = - 3 (2) + 1 (4) = - 2 \end{aligned}

Putting this all together, we get

⟹ [\begin{matrix} 1 & 0 \\ - 3 & 1 \end{matrix}] [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 4 \end{matrix}] = [\begin{matrix} 2 & 1 & 2 \\ 0 & - 1 & 2 \end{matrix}]

\end{proof}