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Exercises Systems of equations

Example 9 from Algebra and Trigonometry, 11.5 Matrices and Matrix Operations

Multiplying Two Matrices

Given $A$ and $B :$

ⓐ Find $A B .$
ⓑ Find $B A .$

A = [\begin{array}{l} \begin{matrix} −1 & 2 & 3 \end{matrix} \\ \begin{matrix} 4 & 0 & 5 \end{matrix} \end{array}] and B = [\begin{matrix} 5 \\ −4 \\ 2 \end{matrix} \begin{matrix} −1 \\ 0 \\ 3 \end{matrix}]

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐAs the dimensions of $A$ are $2 \times$ and the dimensions of $B$ are $3 \times$ these matrices can be multiplied together because the number of columns in $A$ matches the number of rows in $B .$ The resulting product will be a $2 \times$ matrix, the number of rows in $A$ by the number of columns in $B .$
  $\begin{array}{l} A B = [\begin{array}{r} −1 & 2 & 3 \\ 4 & 0 & 5 \end{array}] [\begin{array}{r} 5 & −1 \\ - 4 & 0 \\ 2 & 3 \end{array}] \\ = [\begin{array}{r} −1 (5) + 2 (−4) + 3 (2) & −1 (−1) + 2 (0) + 3 (3) \\ 4 (5) + 0 (−4) + 5 (2) & 4 (−1) + 0 (0) + 5 (3) \end{array}] \\ = [\begin{array}{r} −7 & 10 \\ 30 & 11 \end{array}] \end{array}$
2. ⓑThe dimensions of $B$ are $3 \times 2$ and the dimensions of $A$ are $2 \times 3.$ The inner dimensions match so the product is defined and will be a $3 \times 3$ matrix.
  $\begin{array}{l} B A = [\begin{array}{r} 5 & −1 \\ −4 & 0 \\ 2 & 3 \end{array}] [\begin{array}{r} −1 & 2 & 3 \\ 4 & 0 & 5 \end{array}] \\ = [\begin{array}{r} 5 (−1) + −1 (4) & 5 (2) + −1 (0) & 5 (3) + −1 (5) \\ −4 (−1) + 0 (4) & −4 (2) + 0 (0) & −4 (3) + 0 (5) \\ 2 (−1) + 3 (4) & 2 (2) + 3 (0) & 2 (3) + 3 (5) \end{array}] \\ = [\begin{array}{r} −9 & 10 & 10 \\ 4 & −8 & −12 \\ 10 & 4 & 21 \end{array}] \end{array}$

How did it go?

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From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.