Chapter 3 · Matrices

Example 12: Find $AB$

ExamplesMultiplication of matrices · Section 3.4.5

The question

Find AB, if

A = \begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix}

and

B = \begin{bmatrix} 2 & 6 & 0 \\ 7 & 9 & 8 \end{bmatrix}.

Exam answer

$A$ is $2\times2$ and $B$ is $2\times3$ ⁠, so $AB$ is defined and is of order $2\times3$ ⁠.

Let $AB = [c_{ij}]$ ⁠.

\begin{aligned} c_{11} &= 6{\cdot} 2 + 9{\cdot} 7 = 75 \\ c_{12} &= 6{\cdot} 6 + 9{\cdot} 9 = 117 \\ c_{13} &= 6{\cdot} 0 + 9{\cdot} 8 = 72 \\ c_{21} &= 2{\cdot} 2 + 3{\cdot} 7 = 25 \\ c_{22} &= 2{\cdot} 6 + 3{\cdot} 9 = 39 \\ c_{23} &= 2{\cdot} 0 + 3{\cdot} 8 = 24 \end{aligned}

$\therefore\ AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Each entry of a product is one row of the first matrix times one column of the second: pair the entries up in order, multiply each pair, and add. The entry in row $i$ ⁠, column $k$ of $AB$ uses row $i$ of $A$ and column $k$ of $B$ ⁠, and nothing else. For this to work, a row of $A$ and a column of $B$ must have the same length, which is why the number of columns of $A$ must equal the number of rows of $B$ ⁠.

Given, to find, and the plan

Given: $A$ of order $2\times2$ and $B$ of order $2\times3$ ⁠.
To find: The product $AB$ ⁠.
Plan: Check the product is defined and find its order. Then compute the six entries one at a time, each as a row-times-column sum, and assemble them. Finally note why the reverse product $BA$ does not exist.

Basics you need

1Order of a matrix

" $m\times n$ " means $m$ rows and $n$ columns, in that order. $A$ has 2 rows and 2 columns; $B$ has 2 rows and 3 columns. Basics: rows, columns and the address

2When a product is defined

$AB$ is defined exactly when (columns of $A$ ⁠) = (rows of $B$ ⁠). If $A$ is $m\times n$ and $B$ is $n\times p$ ⁠, then $AB$ is $m\times p$ ⁠: the inner sizes must match and the outer sizes give the answer's order.

3Row times column

If $A=[a_{ij}]$ is $m\times n$ and $B=[b_{jk}]$ is $n\times p$ ⁠, the $(i,k)$ entry of $AB$ is $c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}+\dots+a_{in}b_{nk}$ ⁠. Basics: pair up, multiply, add

Solution

Check that $AB$ is defined and find its order

$A$ is $2\times2$ and $B$ is $2\times3$ ⁠. Columns of $A$ = 2 and rows of $B$ = 2. These are equal, so $AB$ is defined, and its order is (rows of $A$ ⁠) $\times$ (columns of $B$ ⁠) $=2\times3$ ⁠.

Each row of $A$ has 2 entries and each column of $B$ has 2 entries, so they pair up exactly with nothing left over. The answer has one entry for every choice of a row of $A$ (2 choices) and a column of $B$ (3 choices): $2\times3=6$ entries.

Definition of the product, Section 3.4.5: $A=[a_{ij}]_{m\times n}$ ⁠, $B=[b_{jk}]_{n\times p}$ gives $AB=[c_{ik}]_{m\times p}$ ⁠. Lesson: matrix multiplication
Row 1 of $A$ with each column of $B$

Row 1 of $A$ is $[6\ \ 9]$ ⁠. The columns of $B$ are $(2,7)$ ⁠, $(6,9)$ and $(0,8)$ read from top to bottom. Pair, multiply, add:
$\begin{aligned} c_{11} &= 6\cdot 2 + 9\cdot 7 = 12 + 63 = 75\\ c_{12} &= 6\cdot 6 + 9\cdot 9 = 36 + 81 = 117\\ c_{13} &= 6\cdot 0 + 9\cdot 8 = 0 + 72 = 72 \end{aligned}$
The first entry of the row always pairs with the first (top) entry of the column, and the second with the second. That is why we read each column of $B$ from top to bottom.

$c_{1k}=a_{11}b_{1k}+a_{12}b_{2k}$ for $k=1,2,3$ ⁠. Basics: pair up, multiply, add
Row 2 of $A$ with each column of $B$

Row 2 of $A$ is $[2\ \ 3]$ ⁠:
$\begin{aligned} c_{21} &= 2\cdot 2 + 3\cdot 7 = 4 + 21 = 25\\ c_{22} &= 2\cdot 6 + 3\cdot 9 = 12 + 27 = 39\\ c_{23} &= 2\cdot 0 + 3\cdot 8 = 0 + 24 = 24 \end{aligned}$
Same columns of $B$ ⁠, new row of $A$ ⁠. The second row of $AB$ is built entirely from the second row of $A$ ⁠.
Assemble the product

Placing each $c_{ik}$ in row $i$ ⁠, column $k$ ⁠:
$AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}$
This is $2\times3$ ⁠, as Step 1 predicted.

Checking the order of the finished answer against the prediction is a quick test that no entry has been lost or misplaced.

The product laid out so that each entry sits where its row and column meet: row 1 of $A$ (gold) and column 2 of $B$ (blue) give the entry $c_{12}=117$ (green). Every other entry is found the same way.
Why $BA$ is not defined

For $BA$ the first factor is $B$ ( $2\times3$ ⁠) and the second is $A$ ( $2\times2$ ⁠). Columns of $B$ = 3 but rows of $A$ = 2. These differ, so $BA$ is not defined: a row of $B$ has 3 entries but a column of $A$ has only 2, so they cannot be paired.

This is the textbook's Remark after the example: if $AB$ is defined, $BA$ need not be. In general, for $A$ of order $m\times n$ and $B$ of order $k\times l$ ⁠, both products exist exactly when $n=k$ and $l=m$ ⁠. Here $n=k=2$ ⁠, but $l=3\ne m=2$ ⁠.

The order of the factors matters in matrix multiplication, unlike multiplication of numbers. Lesson: matrix multiplication

Result

AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}

a $2\times3$ matrix, matching the textbook's solution. The product $BA$ is not defined.

Check it with numbers

Row 1 of $A$ ⁠, $[6\ \ 9]$ ⁠, is exactly 3 times row 2, $[2\ \ 3]$ ⁠. Since row $i$ of $AB$ is made only from row $i$ of $A$ ⁠, row 1 of $AB$ must be 3 times row 2 of $AB$ ⁠. Indeed $3\times25=75$ ⁠, $3\times39=117$ and $3\times24=72$ ⁠.

Common mistakes

Multiplying entry by entry. $AB$ is not $[a_{ij}b_{ij}]$ ⁠; the two matrices here are not even the same shape. Each entry is a row-times-column sum.
Using a row of $B$ instead of a column. Always run along the row of the first matrix and down the column of the second.
Getting the order backwards. The order of $AB$ is (rows of $A$ ⁠) $\times$ (columns of $B$ ⁠), here $2\times3$ ⁠, not $3\times2$ ⁠.
Assuming $BA$ exists because $AB$ does. It does not here (Step 5).

What to take away

Before multiplying, write the orders side by side: $(2\times\underline{2})(\underline{2}\times3)$ ⁠. Inner sizes equal means defined; outer sizes give the order of the answer.
Entry $(i,k)$ of $AB$ = row $i$ of $A$ times column $k$ of $B$ ⁠. There is one such sum for every entry.
Row $i$ of $AB$ depends only on row $i$ of $A$ ⁠, which gives a quick check whenever rows of $A$ are related.

Every fact used, and where it comes from

Fact	Used in	Source
Order $m\times n$ ⁠: rows first, then columns	Step 1	Section 3.2.1 (Lesson: what a matrix is)
$AB$ defined when columns of $A$ = rows of $B$ ⁠; order $m\times p$	Steps 1 and 5	Section 3.4.5 (Lesson: matrix multiplication)
$c_{ik}=\sum_j a_{ij}b_{jk}$	Steps 2 and 3	Section 3.4.5, definition of the product
Both $AB$ and $BA$ defined iff $n=k$ and $l=m$	Step 5	Remark after Example 12, Section 3.4.5

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Example 12: Find ABABAB

Basics you need

Solution

Check that ABABAB is defined and find its order

Row 1 of AAA with each column of BBB

Row 2 of AAA with each column of BBB

Assemble the product

Why BABABA is not defined

Every fact used, and where it comes from

Example 12: Find $AB$

Check that $AB$ is defined and find its order

Row 1 of $A$ with each column of $B$

Row 2 of $A$ with each column of $B$

Why $BA$ is not defined