Chapter 3 · Matrices

Example 13: Find $AB$ and $BA$ ⁠; show $AB\ne BA$

ExamplesNon-commutativity of matrix multiplication · Section 3.4.5

The question

A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}

and

B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix},

then find AB, BA. Show that AB $\ne$ BA.

Exam answer

$A$ is $2\times3$ and $B$ is $3\times2$ ⁠, so $AB$ is $2\times2$ and $BA$ is $3\times3$ ⁠.

Let $AB = [c_{ij}]$ ⁠.

\begin{aligned} c_{11} &= 1{\cdot} 2 + (-2){\cdot} 4 + 3{\cdot} 2 = 0 \\ c_{12} &= 1{\cdot} 3 + (-2){\cdot} 5 + 3{\cdot} 1 = -4 \\ c_{21} &= (-4){\cdot} 2 + 2{\cdot} 4 + 5{\cdot} 2 = 10 \\ c_{22} &= (-4){\cdot} 3 + 2{\cdot} 5 + 5{\cdot} 1 = 3 \end{aligned}

\therefore\ AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}

Let $BA = [d_{ij}]$ ⁠.

\begin{aligned} d_{11} &= 2{\cdot} 1 + 3{\cdot} (-4) = -10 \\ d_{12} &= 2{\cdot} (-2) + 3{\cdot} 2 = 2 \\ d_{13} &= 2{\cdot} 3 + 3{\cdot} 5 = 21 \\ d_{21} &= 4{\cdot} 1 + 5{\cdot} (-4) = -16 \\ d_{22} &= 4{\cdot} (-2) + 5{\cdot} 2 = 2 \\ d_{23} &= 4{\cdot} 3 + 5{\cdot} 5 = 37 \\ d_{31} &= 2{\cdot} 1 + 1{\cdot} (-4) = -2 \\ d_{32} &= 2{\cdot} (-2) + 1{\cdot} 2 = -2 \\ d_{33} &= 2{\cdot} 3 + 1{\cdot} 5 = 11 \end{aligned}

\therefore\ BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}

$AB$ is $2\times2$ and $BA$ is $3\times3$ ⁠, so $AB \ne BA$ ⁠.

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

The one idea this problem is built on

Swapping the order of the factors changes which rows meet which columns, and it can even change the size of the answer. Here $A$ is $2\times3$ and $B$ is $3\times2$ ⁠, so $AB$ is $2\times2$ while $BA$ is $3\times3$ ⁠. Two matrices of different orders can never be equal, so $AB\ne BA$ follows the moment we know the orders. We still compute both products in full, as the question asks.

Given, to find, and the plan

Given: $A$ of order $2\times3$ and $B$ of order $3\times2$ ⁠.
To find / show: $AB$ and $BA$ ⁠, and that $AB\ne BA$ ⁠.
Plan: Check both products are defined and find their orders. Compute the 4 entries of $AB$ and the 9 entries of $BA$ as row-times-column sums. Conclude using the definition of equal matrices.

Basics you need

1When a product is defined, and its order

If $A$ is $m\times n$ and $B$ is $n\times p$ ⁠, then $AB$ is defined and is $m\times p$ ⁠. Basics: rows, columns and the address

2Row times column

The $(i,k)$ entry of $AB$ is row $i$ of $A$ times column $k$ of $B$ ⁠: pair the entries in order, multiply, add. Basics: pair up, multiply, add

3Equal matrices

Two matrices are equal only if they have the same order and every pair of corresponding entries is equal. Lesson: equality of matrices

4Signs in products

$(-2)\cdot 4=-8$ ⁠, $(-4)\cdot(-2)=8$ ⁠, $(-4)\cdot 3=-12$ ⁠: a product of two numbers of opposite sign is negative, of the same sign positive.

Solution

Check both products and find their orders

$AB$ ⁠: $A$ is $2\times\mathbf{3}$ ⁠, $B$ is $\mathbf{3}\times2$ ⁠. Inner sizes $3=3$ ⁠, so $AB$ is defined, of order $2\times2$ ⁠.

$BA$ ⁠: $B$ is $3\times\mathbf{2}$ ⁠, $A$ is $\mathbf{2}\times3$ ⁠. Inner sizes $2=2$ ⁠, so $BA$ is defined, of order $3\times3$ ⁠.

Write the orders side by side. The inner sizes must match for the product to exist; the outer sizes give the order of the answer. Swapping the factors changes the outer sizes, so $AB$ and $BA$ have different orders.

This is the textbook's condition for both products to exist: with $A$ of order $m\times n$ and $B$ of order $k\times l$ ⁠, we need $n=k$ and $l=m$ ⁠. Here $m=2$ ⁠, $n=3$ ⁠, $k=3$ ⁠, $l=2$ ⁠, so both hold.

Order of a product, Section 3.4.5. Lesson: matrix multiplication
Compute the four entries of $AB$

Rows of $A$ ⁠: $[1\ \ {-2}\ \ 3]$ and $[-4\ \ 2\ \ 5]$ ⁠. Columns of $B$ ⁠: $(2,4,2)$ and $(3,5,1)$ ⁠.
$\begin{aligned} c_{11} &= 1\cdot 2 + (-2)\cdot 4 + 3\cdot 2\\ &= 2 - 8 + 6 = 0\\ c_{12} &= 1\cdot 3 + (-2)\cdot 5 + 3\cdot 1\\ &= 3 - 10 + 3 = -4\\ c_{21} &= (-4)\cdot 2 + 2\cdot 4 + 5\cdot 2\\ &= -8 + 8 + 10 = 10\\ c_{22} &= (-4)\cdot 3 + 2\cdot 5 + 5\cdot 1\\ &= -12 + 10 + 5 = 3 \end{aligned}$ $AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$
Each row of $A$ has three entries and each column of $B$ has three, so every entry of $AB$ is a sum of three products. Writing the products with their signs in brackets, such as $(-2)\cdot4$ ⁠, before adding is the surest way to keep the signs right.

$c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}+a_{i3}b_{3k}$ ⁠. Basics: pair up, multiply, add
Compute the nine entries of $BA$

Now the rows come from $B$ ⁠: $[2\ \ 3]$ ⁠, $[4\ \ 5]$ ⁠, $[2\ \ 1]$ ⁠. The columns come from $A$ ⁠: $(1,-4)$ ⁠, $(-2,2)$ ⁠, $(3,5)$ ⁠. Each entry is a sum of two products. Write $BA=[d_{ik}]$ ⁠:
$\begin{aligned} d_{11} &= 2\cdot 1 + 3\cdot(-4) = 2 - 12 = -10\\ d_{12} &= 2\cdot(-2) + 3\cdot 2 = -4 + 6 = 2\\ d_{13} &= 2\cdot 3 + 3\cdot 5 = 6 + 15 = 21\\ d_{21} &= 4\cdot 1 + 5\cdot(-4) = 4 - 20 = -16\\ d_{22} &= 4\cdot(-2) + 5\cdot 2 = -8 + 10 = 2\\ d_{23} &= 4\cdot 3 + 5\cdot 5 = 12 + 25 = 37\\ d_{31} &= 2\cdot 1 + 1\cdot(-4) = 2 - 4 = -2\\ d_{32} &= 2\cdot(-2) + 1\cdot 2 = -4 + 2 = -2\\ d_{33} &= 2\cdot 3 + 1\cdot 5 = 6 + 5 = 11 \end{aligned}$ $BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$
In $BA$ the roles swap: rows of $B$ (length 2) meet columns of $A$ (length 2). That is why these sums have two terms, while those for $AB$ had three. The two products are built from entirely different pairings, so there is no reason to expect them to agree.
Conclude that $AB\ne BA$

$AB$ has order $2\times2$ and $BA$ has order $3\times3$ ⁠. Equal matrices must have the same order, so $AB\ne BA$ .

We do not need to compare entries at all: there is no entry $(3,3)$ in $AB$ to compare with the 11 in $BA$ ⁠. Equality fails at the first requirement, the order.

Definition 2, Section 3.3.1: $A=[a_{ij}]$ and $B=[b_{ij}]$ are equal if they are of the same order and $a_{ij}=b_{ij}$ for all $i,j$ ⁠. Lesson: equality of matrices
What this shows about matrix multiplication

Even when both $AB$ and $BA$ exist, they need not be equal. The textbook goes on (Example 14) to show that this can happen even when $AB$ and $BA$ have the same order, so the failure is not only about sizes. Matrix multiplication is not commutative.

For numbers $ab=ba$ always. For matrices, the order of the factors decides which rows meet which columns, so in general you must keep the order exactly as written. See Example 14 for two $2\times2$ matrices with $AB\ne BA$ ⁠.

Result

AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}

BA = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}

$AB$ is $2\times2$ and $BA$ is $3\times3$ ⁠, so $AB\ne BA$ ⁠. This matches the textbook's solution.

Check it with numbers

A fact that is not in the textbook (a property of the trace) gives a neat test: the sum of the diagonal entries of $AB$ always equals that of $BA$ ⁠, whenever both products exist. Here $0+3=3$ and $-10+2+11=3$ ⁠. They agree, which is good evidence that no diagonal entry was miscalculated.

A second spot check: entry $(2,3)$ of $BA$ is row 2 of $B$ ⁠, $[4\ \ 5]$ ⁠, times column 3 of $A$ ⁠, $(3,5)$ ⁠, giving $12+25=37$ ⁠, as in Step 3.

Common mistakes

Thinking $BA$ is " $AB$ turned round". It is a different matrix, here not even the same size.
Sign slips. $(-4)\cdot 3=-12$ and $3\cdot(-4)=-12$ ⁠; write the sign of every product before adding.
Mixing up which matrix gives the rows. In $BA$ the rows come from $B$ ⁠, the first factor, and the columns from $A$ ⁠.
Comparing entries before orders. Different orders already settle $AB\ne BA$ ⁠.

What to take away

For $A$ of order $m\times n$ and $B$ of order $n\times m$ ⁠, $AB$ is $m\times m$ and $BA$ is $n\times n$ ⁠. If $m\ne n$ they cannot be equal.
In a product the first factor supplies the rows and the second supplies the columns.
Matrix multiplication is not commutative: never swap factors unless you have a reason.

Every fact used, and where it comes from

Fact	Used in	Source
$AB$ defined iff columns of $A$ = rows of $B$ ⁠; order $m\times p$	Step 1	Section 3.4.5 and the Remark after Example 12 (Lesson: matrix multiplication)
$c_{ik}=\sum_j a_{ij}b_{jk}$	Steps 2 and 3	Section 3.4.5, definition of the product
Equal matrices have the same order and equal entries	Step 4	Definition 2, Section 3.3.1 (Lesson: equality)
Matrix multiplication is not commutative	Step 5	Section 3.4.5, "Non-commutativity of multiplication of matrices"
Diagonal sums of $AB$ and $BA$ are equal	Check box only	Beyond the syllabus (a property of the trace); not needed for the solution

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Example 13: Find ABABAB and BABABA⁠; show AB≠BAAB\ne BAAB=BA

Basics you need

Solution

Check both products and find their orders

Compute the four entries of ABABAB

Compute the nine entries of BABABA

Conclude that AB≠BAAB\ne BAAB=BA

What this shows about matrix multiplication

Every fact used, and where it comes from

Example 13: Find $AB$ and $BA$ ⁠; show $AB\ne BA$

Compute the four entries of $AB$

Compute the nine entries of $BA$

Conclude that $AB\ne BA$