The Process of Matrix Multiplication

Core Concepts

Walks through the 'row-by-column' method for calculating the elements of a product matrix. This stack provides step-by-step examples to build procedural fluency in matrix multiplication.

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The Process of Matrix Multiplication

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To find the element in row \(i\) and column \(j\) of the product matrix \(AB\), what two things do you multiply?
You multiply the \(i\)-th row of the first matrix (\(A\)) by the \(j\)-th column of the second matrix (\(B\)).
Describe the 'dot product' process of multiplying a row vector by a column vector.
You multiply the first element of the row by the first element of the column, the second by the second, and so on. Then, you sum all of these products to get a single number.
Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}\). What is the formula for the element in the 1st row, 1st column of \(AB\)?
The element is \((AB)_{11}\). You multiply the 1st row of \(A\) by the 1st column of \(B\): \(ae + bg\).
Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}\). What is the formula for the element in the 2nd row, 1st column of \(AB\)?
The element is \((AB)_{21}\). You multiply the 2nd row of \(A\) by the 1st column of \(B\): \(ce + dg\).
$$ \text{Calculate: } \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} $$
$$ \begin{pmatrix} (1)(5)+(2)(7) & (1)(6)+(2)(8) \\ (3)(5)+(4)(7) & (3)(6)+(4)(8) \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix} $$
$$ \text{Calculate: } \begin{pmatrix} 2 & 0 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 5 \\ 4 & -2 \end{pmatrix} $$
$$ \begin{pmatrix} (2)(1)+(0)(4) & (2)(5)+(0)(-2) \\ (-1)(1)+(3)(4) & (-1)(5)+(3)(-2) \end{pmatrix} = \begin{pmatrix} 2 & 10 \\ 11 & -11 \end{pmatrix} $$
$$ \text{Calculate the product: } \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} $$
$$ (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = [32] $$. The result is a \(1 \times 1\) matrix.
$$ \text{Calculate: } \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 2 & 1 \\ 1 & 0 \end{pmatrix} $$
$$ \begin{pmatrix} (1)(3)+(0)(2)+(2)(1) & (1)(1)+(0)(1)+(2)(0) \\ (-1)(3)+(3)(2)+(1)(1) & (-1)(1)+(3)(1)+(1)(0) \end{pmatrix} = \begin{pmatrix} 5 & 1 \\ 4 & 2 \end{pmatrix} $$
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