Lesson 11.5: Matrices and Matrix Operations

New Concept

Matrices are rectangular arrays used to organize data. This lesson introduces the fundamental operations you can perform on them: addition, subtraction, scalar multiplication, and matrix multiplication, each following specific rules based on matrix dimensions.

What’s next

Next, you will apply these concepts through a series of interactive examples and practice cards covering each matrix operation.

Property

A matrix is a rectangular array of numbers that is usually named by a capital letter: $A$, $B$, $C$, and so on. Each entry in a matrix is referred to as $a_{ij}$, such that $i$ represents the row and $j$ represents the column. Matrices are often referred to by their dimensions: $m \times n$ indicating $m$ rows and $n$ columns.

Examples

• Given matrix $A = \begin{bmatrix} 5 & -1 & 4 \\ 9 & 0 & 2 \end{bmatrix}$, the dimensions are $2 \times 3$ (2 rows, 3 columns). The entry at $a_{13}$ is 4.

• A matrix with dimensions $3 \times 3$ like $B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ is called a square matrix because it has the same number of rows and columns.

Property

Given matrices $A$ and $B$ of like dimensions, addition and subtraction of $A$ and $B$ will produce matrix $C$ or matrix $D$ of the same dimension.

Matrix addition is commutative ($A + B = B + A$) and associative ($(A + B) + C = A + (B + C)$). Addition and subtraction are only possible when matrices have the same dimensions.

Examples

• Find the sum of $A = \begin{bmatrix} 3 & 8 \\ 1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 0 \\ 4 & -1 \end{bmatrix}$. The sum is $\begin{bmatrix} 3+2 & 8+0 \\ 1+4 & 5-1 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 5 & 4 \end{bmatrix}$.

• Find the difference of $A = \begin{bmatrix} 10 & 2 \\ 5 & 9 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 6 \\ 3 & 1 \end{bmatrix}$. The difference is $\begin{bmatrix} 10-4 & 2-6 \\ 5-3 & 9-1 \end{bmatrix} = \begin{bmatrix} 6 & -4 \\ 2 & 8 \end{bmatrix}$.

Property

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

the scalar multiple $cA$ is

Scalar multiplication is distributive. For the matrices $A$, $B$, and $C$ with scalars $a$ and $b$,

Examples

• Multiply matrix $A = \begin{bmatrix} 6 & 2 \\ 10 & 3 \end{bmatrix}$ by the scalar 4. The result is $4A = \begin{bmatrix} 4 \cdot 6 & 4 \cdot 2 \\ 4 \cdot 10 & 4 \cdot 3 \end{bmatrix} = \begin{bmatrix} 24 & 8 \\ 40 & 12 \end{bmatrix}$.

• Given $B = \begin{bmatrix} 5 & -1 \\ -4 & 0 \end{bmatrix}$, find $-3B$. The result is $-3B = \begin{bmatrix} -3 \cdot 5 & -3 \cdot (-1) \\ -3 \cdot (-4) & -3 \cdot 0 \end{bmatrix} = \begin{bmatrix} -15 & 3 \\ 12 & 0 \end{bmatrix}$.

Property

Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If $A$ is an $m \times r$ matrix and $B$ is an $r \times n$ matrix, then the product matrix $AB$ is an $m \times n$ matrix. To obtain the entry in row $i$, column $j$ of $AB$, multiply the entries in row $i$ of $A$ by the corresponding entries in column $j$ of $B$ and add the results.

Examples

• Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 0 & 1 \end{bmatrix}$, their product is $AB = \begin{bmatrix} 1(5)+2(0) & 1(6)+2(1) \\ 3(5)+4(0) & 3(6)+4(1) \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 15 & 22 \end{bmatrix}$.

• Given $A = \begin{bmatrix} 1 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix}$, the product $AB$ is defined because their dimensions are $1 \times 3$ and $3 \times 1$. The result is $\begin{bmatrix} 1(7)+0(8)+2(9) \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix}$.

Property

For the matrices $A$, $B$, and $C$ the following properties hold.

• Matrix multiplication is associative: $(AB)C = A(BC)$.
• Matrix multiplication is distributive: $C(A + B) = CA + CB$, $(A + B)C = AC + BC$.

Note that matrix multiplication is not commutative.

Examples

• Let $A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ 5 & 0 \end{bmatrix}$. Then $AB = \begin{bmatrix} 1(4)+2(5) & 1(1)+2(0) \\ 0(4)+3(5) & 0(1)+3(0) \end{bmatrix} = \begin{bmatrix} 14 & 1 \\ 15 & 0 \end{bmatrix}$.

• Using the same matrices, $BA = \begin{bmatrix} 4(1)+1(0) & 4(2)+1(3) \\ 5(1)+0(0) & 5(2)+0(3) \end{bmatrix} = \begin{bmatrix} 4 & 11 \\ 5 & 10 \end{bmatrix}$. Notice that $AB \neq BA$.