Course 8.1Lesson 2 of 3

Multiplying matrices

The product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$ . Each entry of $AB$ is a row of $A$ multiplied entry by entry with a column of $B$

about 30 min1 simulation1 practice question1 Python program5 questions to check yourself

This lesson is free to read. Enroll free to keep your place and take the tests.

Lesson 2 of 3Show all lessons

Try the code and the simulations beside the lesson.

What you will be able to do

decide whether a product exists and find the order of the answer;
multiply two matrices, entry by entry, by the row-by-column rule;
state which rules of number arithmetic hold for matrices and which fail;
use a product to follow two moves of a shape, or to count routes in a network.

When can two matrices be multiplied?

Multiplying matrices is not entry by entry. It combines a row of the first matrix with a column of the second. For that, the row and the column must have the same length.

The product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$ . If $A$ has order $m \times n$ and $B$ has order $n \times p$ , then $AB$ has order $m \times p$ . The two inner numbers must match, and the two outer numbers give the order.

The row-by-column rule

The entry of $AB$ in row $i$ and column $j$ is row $i$ of $A$ times column $j$ of $B$ . Multiply the first entries, the second entries and so on, then add:

\displaystyle (AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj}.

Take $A =

\displaystyle \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

$and$ B =

\displaystyle \begin{bmatrix} 0 & 1 \\ 5 & 2 \end{bmatrix}

\displaystyle AB = \begin{bmatrix} 1 \cdot 0 + 2 \cdot 5 & 1 \cdot 1 + 2 \cdot 2 \\ 3 \cdot 0 + 4 \cdot 5 & 3 \cdot 1 + 4 \cdot 2 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 20 & 11 \end{bmatrix}.

Now multiply in the other order.

\displaystyle BA = \begin{bmatrix} 0 \cdot 1 + 1 \cdot 3 & 0 \cdot 2 + 1 \cdot 4 \\ 5 \cdot 1 + 2 \cdot 3 & 5 \cdot 2 + 2 \cdot 4 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 11 & 18 \end{bmatrix}.

So $AB \ne BA$ . For numbers, $3 \times 5 = 5 \times 3$ always. For matrices the order of the factors matters.

Rectangular matrices

Let $C$ be $2 \times 3$ and $D$ be $3 \times 2$ :

\displaystyle C = \begin{bmatrix} 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix}, \qquad D = \begin{bmatrix} 2 & 1 \\ 1 & 0 \\ 0 & 3 \end{bmatrix}.

The inner numbers are both $3$ , so $CD$ exists and has order $2 \times 2$ . It is

\displaystyle CD = \begin{bmatrix} 2 & 7 \\ 7 & 6 \end{bmatrix}.

The product $DC$ also exists, but it has order $3 \times 3$ . Here the two products do not even have the same order. If $A$ is $2 \times 3$ and $B$ is $4 \times 2$ , then $BA$ exists but $AB$ does not.

What stays the same

These rules of arithmetic still hold, whenever the products and sums exist:

$(AB)C = A(BC)$ , so we may write $ABC$ ;
$A(B + C) = AB + AC$ and $(A + B)C = AC + BC$ ;
$k(AB) = (kA)B = A(kB)$ ;
$AI = IA = A$ for the identity matrix $I$ of the right order, and $AO = OA = O$ .

What changes

No commutative law. In general $AB \ne BA$ .
Zero products. $AB = O$ can happen when neither $A$ nor $B$ is $O$ . For example $

\displaystyle \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}

\displaystyle \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}

= O$.

No cancelling. $AB = AC$ does not mean $B = C$ .
Identities with squares change. $(A + B)^2 = A^2 + AB + BA + B^2$ , which equals $A^2 + 2AB + B^2$ only when $AB = BA$ .

Why multiplication is defined this way

Think of a matrix as a move of the plane. The matrix $B$ takes a point $v$ to $Bv$ . Then $A$ takes that point to $A(Bv)$ . The product $AB$ is the single matrix that does both moves, right factor first. If you turn and then stretch, you do not get what you get by stretching and then turning, so $AB \ne BA$ .

Two panels show the same arrow shape, a right-angled triangle with corners (0, 0), (2, 0) and (0, 1). In the left panel it is first reflected in the x-axis and then rotated 90 degrees anticlockwise, and the final triangle has corners (0, 0), (0, 2) and (1, 0), pointing up and to the right. In the right panel the order is reversed: rotate first, then reflect. The final triangle has corners (0, 0), (0, -2) and (-1, 0), pointing down and to the left. The titles say that R times F and F times R give different pictures.

Let $R =

\displaystyle \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

$turn the plane$ 90^\circ $anticlockwise and let$ F =

\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

$reflect it in the$ x $-axis. Then$ RF =

\displaystyle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

$is the reflection in the line$ y = x $, and$ FR =

\displaystyle \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

$is the reflection in the line$ y = -x$. Computer graphics, A-Level Further Maths and IB all use matrices like this to move points.

Counting routes

A matrix of $0$ s and $1$ s can describe a network. Let the entry in row $i$ and column $j$ be $1$ if there is a direct flight from city $i$ to city $j$ . For three cities with flights $P \to Q$ , $P \to R$ , $Q \to R$ and $R \to P$ ,

\displaystyle N = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, \qquad N^2 = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}.

The entry of $N^2$ in row $i$ and column $j$ counts the routes from city $i$ to city $j$ with exactly two flights. The entry of $N^k$ counts the routes with exactly $k$ flights.

A worked example

Find $AB$ and $BA$ for $A =

\displaystyle \begin{bmatrix} 2 & -1 \\ 3 & 0 \end{bmatrix}

$and$ B =

\displaystyle \begin{bmatrix} 1 & 4 \\ -2 & 5 \end{bmatrix}

$by looking only at the entry in row$ 1 $, column$ 2$ of each.

For $AB$ . Row $1$ of $A$ is $(2, -1)$ . Column $2$ of $B$ is $(4, 5)$ . The entry is $2 \times 4 + (-1) \times 5 = 3$ .
For $BA$ . Row $1$ of $B$ is $(1, 4)$ . Column $2$ of $A$ is $(-1, 0)$ . The entry is $1 \times (-1) + 4 \times 0 = -1$ .
Compare. The entries are $3$ and $-1$ , so $AB \ne BA$ .

Practise with new numbers

Try it in Python

The program multiplies two matrices with two loops, using the row-by-column rule. Then it checks the answer against the @ operator and compares $AB$ with $BA$ .

Common mistakes

Multiplying entry by entry. The product is row by column, not place by place.
Assuming $AB = BA$ . Always keep the factors in the order given.
Using $(A + B)^2 = A^2 + 2AB + B^2$ . It needs $AB = BA$ .
Cancelling a matrix. From $AB = AC$ you may not conclude $B = C$ .
Forgetting to check the orders. Check that the inner numbers match before you start.

Remember these

$A$ is $m \times n$ and $B$ is $n \times p$ : then $AB$ is $m \times p$ .
$(AB)_{ij}$ is row $i$ of $A$ times column $j$ of $B$ .
Associative and distributive laws hold. The commutative law does not.
$AB = O$ does not force $A = O$ or $B = O$ .

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

$A$ is $2 \times 3$ and $B$ is $3 \times 4$ . Does $AB$ exist? Does $BA$ exist?

Find $

\displaystyle \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}

\displaystyle \begin{bmatrix} 3 \\ 4 \end{bmatrix}

\displaystyle \begin{bmatrix} 3 + 8 \\ 0 + 4 \end{bmatrix}

\displaystyle \begin{bmatrix} 11 \\ 4 \end{bmatrix}

Is $(A - B)(A + B) = A^2 - B^2$ for every pair of square matrices?

Let $A =

\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

$. Find$ A² $and$ A³ $, and guess$ A^n$.

\displaystyle \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}

$and$ A³ =

\displaystyle \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}

$. The pattern is$ A^n =

\displaystyle \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}

What is $AI$ if $A$ is any $3 \times 3$ matrix and $I$ is the identity of order $3$ ?

References

NCERT, Mathematics Textbook for Class XII, Part I, Chapter 3 (Matrices), the section on multiplication of matrices.
OpenStax, Precalculus 2e, Section 9.5 (Matrices and Matrix Operations).

Enroll free in Matrices and Matrix Algebra

Every lesson is open to read. Enrolling also lets you:

Mark lessons done and carry on where you stopped.
Take the tests of this lesson and its course.
Count the course towards the certificate for its chapter.

No payment. You can leave at any time.

Problems (5)

Let A = 2, -1; 3, 0 and B = 1, 4; -2, 5…Easy · High School (11–12) · by StemPy Lab

Let A = 1, -1; 2, -1. Find A², then use it to find A²⁰²⁵…Medium · High School (11–12) · by StemPy Lab

Four cities A, B, C, D have these one-way direct flights: A → B, A → C, B → A, B → C, B → D, C → D…Hard · High School (11–12) · by StemPy Lab

Next: lesson 3Transpose, symmetric and skew-symmetric matrices

← Matrices and Matrix Algebra

← Matrices, their types and sums Transpose, symmetric and skew-symmetric matrices →

Discussion (0)

Loading…