Matrix Multiplication

Combining Rows and Columns to Form a Product

Matrix multiplication combines each row of the first matrix with each column of the second by multiplying each paired entry and summing the products. Unlike matrix addition, it does not operate on entries in matching positions.

Condition for Matrix Multiplication

Two matrices, say matrix $A$ and matrix $B$ , can be multiplied ( $A \times B$ ) only if the number of columns in matrix $A$ is equal to the number of rows in matrix $B$ .

Suppose matrix $A$ has an order of $m \times n$ (meaning $m$ rows and $n$ columns) and matrix $B$ has an order of $n \times p$ (meaning $n$ rows and $p$ columns).

Since the number of columns in matrix $A$ ( $n$ ) is equal to the number of rows in matrix $B$ ( $n$ ), then matrix $A$ and $B$ can be multiplied.

The result of the multiplication, call it matrix $C = AB$ , will have an order of $m \times p$ .

The dimension check fits in one line: $A\in\mathbb{R}^{m\times n}$ and $B\in\mathbb{R}^{n\times p}$ imply $AB\in\mathbb{R}^{m\times p}$ . The shared inner dimension $n$ is consumed by each dot product, while the outer dimensions $m$ and $p$ determine the result.

How to Calculate the Elements of the Resultant Matrix

The element $c_{ij}$ in matrix $C$ (i.e., the element in the $i$ -th row and $j$ -th column) is calculated by multiplying each element in the $i$ -th row of matrix $A$ by the corresponding element in the $j$ -th column of matrix $B$ , and then summing all these products.

Mathematically, if $A = [a_{ik}]$ and $B = [b_{kj}]$ , then the element $c_{ij}$ of matrix $C = AB$ is:

\begin{aligned} c_{ij} &= \sum_{k=1}^{n} a_{ik}b_{kj} \\ &= a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj} \end{aligned}

The notation $\sum$ (sigma) means summation.

In the formula above, we sum the products $a_{ik}b_{kj}$ for all values of $k$ from $1$ to $n$ .

Steps to Multiply Matrices

In the following example, each entry of matrix $R$ is the dot product of one row of matrix $P$ and one column of matrix $Q$ .

The two matrices are:

P = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix}

Matrix $P$ has an order of $2 \times 2$ and matrix $Q$ also has an order of $2 \times 2$ . The number of columns in $P$ (which is $2$ ) is equal to the number of rows in $Q$ (which is $2$ ), so we can multiply them. The result, $R = PQ$ , will have an order of $2 \times 2$ .

R = \begin{bmatrix} r_{11} & r_{12} \\ r_{21} & r_{22} \end{bmatrix}

The elements of matrix $R$ are calculated as follows:

\begin{aligned} r_{11} &= (\text{row 1 of } P) \\ &\quad\cdot (\text{column 1 of } Q) \\ &= p_{11}q_{11} + p_{12}q_{21} \end{aligned}

Example of Multiplying Two Matrices

Given two matrices:

A = \begin{bmatrix} -7 & 2 & -2 \\ 1 & 0 & -1 \\ 2 & 3 & -1 \end{bmatrix}

Matrix $A$ has an order of $3 \times 3$ and matrix $B$ has an order of $3 \times 2$ .

The number of columns in matrix $A$ (which is $3$ ) is equal to the number of rows in matrix $B$ (which is $3$ ).

So, $AB$ can be calculated and will result in a matrix of order $3 \times 2$ .

Calculate $C = AB$ :

C = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ c_{31} & c_{32} \end{bmatrix}

So, the result of the matrix multiplication $AB$ is:

AB = \begin{bmatrix} -13 & -8 \\ -1 & 2 \\ -3 & 13 \end{bmatrix}

Now, what about $BA$ ?

Matrix $B$ has an order of $3 \times 2$ and matrix $A$ has an order of $3 \times 3$ .

The number of columns in matrix $B$ (which is $2$ ) is not equal to the number of rows in matrix $A$ (which is $3$ ).

$BA$ is undefined. Whether a matrix product exists can change when the factors are reversed.

Properties of Matrix Multiplication

Matrix multiplication satisfies the following rules:

Generally Not Commutative:

This means $AB \neq BA$ . We have already seen an example above where $AB$ is defined but $BA$ is not. Even if both are defined, the results are not necessarily the same.
Associative:

If the products on both sides are defined, then $(AB)C = A(BC)$ holds. This means the order of grouping the multiplication does not change the final result.
Distributive:

Matrix multiplication is distributive over matrix addition or subtraction:
$A(B + C) = AB + AC$
This holds if all involved addition and multiplication operations are defined.
Multiplication by Identity Matrix ( $I$ ):

If $A$ has dimensions $m \times n$ , then the left identity $I_m$ has dimensions $m \times m$ and the right identity $I_n$ has dimensions $n \times n$ :

$I_mA=A=AI_n$

The identity matrix acts like the number $1$ in ordinary number multiplication.
Multiplication by a Scalar ( $k$ ):

If $k$ is a scalar (real number), then:

$k(AB) = (kA)B = A(kB)$

Calculating Revenue

Matrix multiplication can combine a table of quantities with a compatible table of prices. The row and column labels determine which quantities are multiplied and added, so matching dimensions alone is not enough.

Suppose a home business sells tempeh chips, banana chips, and potato chips at Place A, Place B, and Place C.

The sales matrix $P$ uses product types as rows and places as columns. In other words, each entry of $P$ records how many jars of one product were sold at one place.

From top to bottom, the rows represent tempeh chips, banana chips, and potato chips:

P = \begin{bmatrix} 15 & 12 & 20 \\ 25 & 10 & 15 \\ 15 & 15 & 20 \end{bmatrix}

The first row ( $\begin{bmatrix} 15 & 12 & 20 \end{bmatrix}$ ) means that $15$ jars of tempeh chips were sold at Place A, $12$ at Place B, and $20$ at Place C.

The price per jar, in rupiah, is stored in the column matrix $Q$ using the same product order:

Q = \begin{bmatrix} 20{,}000 \\ 15{,}000 \\ 30{,}000 \end{bmatrix}

From top to bottom, the entries give the prices of tempeh chips, banana chips, and potato chips.

The product $PQ$ is numerically defined because both inner dimensions are $3$ , but it is not meaningful here. The columns of $P$ represent places, while the rows of $Q$ represent products. Those labels do not match.

To calculate revenue by place, transpose the price vector and multiply $Q^T$ by $P$ . The product $S = Q^T P$ pairs each product price with the sales of that same product.

Matrix $Q^T$ has dimensions $1 \times 3$ , and $P$ has dimensions $3 \times 3$ . Therefore $S$ has dimensions $1 \times 3$ , with one entry for each place.

\begin{aligned} S &= \begin{bmatrix} 20{,}000 & 15{,}000 & 30{,}000 \end{bmatrix} \\ &\quad{}\cdot\begin{bmatrix} 15 & 12 & 20 \\ 25 & 10 & 15 \\ 15 & 15 & 20 \end{bmatrix} \end{aligned}

$S = \begin{bmatrix} 1{,}125{,}000 & 840{,}000 & 1{,}225{,}000 \end{bmatrix}$ .

This means the total revenue from Place A is $\text{Rp}1{,}125{,}000$ , from Place B is $\text{Rp}840{,}000$ , and from Place C is $\text{Rp}1{,}225{,}000$ .

Revenue by product requires a different multiplication. First use a vector of ones $U$ to total each row of $P$ , then multiply those totals by the diagonal price matrix derived from $Q$ .

U = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

Matrix $R$ therefore gives product revenue: $\text{Rp}940{,}000$ for tempeh chips, $\text{Rp}750{,}000$ for banana chips, and $\text{Rp}1{,}500{,}000$ for potato chips.

A meaningful matrix product needs matching dimensions and must pair quantities that describe the same category in the same order.

Here, $Q^T$ and $P$ produce revenue by place in $S$ , while the diagonal price matrix and product totals produce revenue by product in $R$ . Those are different questions, so they require different products.

Exercises

Given the following matrices:

C = \begin{bmatrix} 1 & -3 \\ 1 & 2 \end{bmatrix}

Determine the matrices $CD$ and $DC$ .

Is $CD = DC$ ?

Worked Solutions

Calculating $CD$ :

Matrix $C$ has an order of $2 \times 2$ and $D$ has an order of $2 \times 2$ . The result will have an order of $2 \times 2$ .

$CD = \begin{bmatrix} 1 & -3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & -4 \\ 8 & 2 \end{bmatrix}$

Compute each entry by multiplying a row by a column:
$\begin{aligned} &m_{11}=(1)(1) + (-3)(8) \\ &m_{12}=(1)(-4) + (-3)(2) \\ &m_{21}=(1)(1) + (2)(8) \\ &m_{22}=(1)(-4) + (2)(2) \end{aligned}$
Calculating $DC$ :

Matrix $D$ has an order of $2 \times 2$ and $C$ has an order of $2 \times 2$ . The result will have an order of $2 \times 2$ .

$DC = \begin{bmatrix} 1 & -4 \\ 8 & 2 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 1 & 2 \end{bmatrix}$

Compute each entry by multiplying a row by a column:
$\begin{aligned} &m_{11}=(1)(1) + (-4)(1) \\ &m_{12}=(1)(-3) + (-4)(2) \\ &m_{21}=(8)(1) + (2)(1) \\ &m_{22}=(8)(-3) + (2)(2) \end{aligned}$

The two products differ, so $CD \neq DC$ . Matrix multiplication is not commutative in this example.