Matrices: Operations, Identity and Inverses

A table of numbers with rules

A snack stand at a high school football field sells hot dogs, pretzels and lemonade. Its sales for two games fit in a table, and a table of numbers is a matrix:

$$S = \begin{bmatrix} 40 & 25 & 60 \\ 55 & 30 & 80 \end{bmatrix}$$

Row one is Friday’s game and row two is Saturday’s. The columns are hot dogs, pretzels and lemonade. A matrix with $2$ rows and $3$ columns is a $2 \times 3$ matrix.

The point of writing data this way is that one calculation then does many jobs at once. With prices of $3$, $2$ and $1.50$ dollars in a column, a single multiplication gives both nights’ revenue:

$$\begin{bmatrix} 40 & 25 & 60 \\ 55 & 30 & 80 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\ 1.5 \end{bmatrix} = \begin{bmatrix} 260 \\ 345 \end{bmatrix}$$

Friday brought in $260$ dollars and Saturday $345$. How that multiplication works is the core of this lesson.

Adding, subtracting and scaling

Matrices of the same size add and subtract entry by entry:

$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 8 & 6 \end{bmatrix}$$

Multiplying by a number, a scalar, multiplies every entry. If every sale doubled, the new table would be $2S$:

$$2S = \begin{bmatrix} 80 & 50 & 120 \\ 110 & 60 & 160 \end{bmatrix}$$

Multiplying matrices

Each entry of a product pairs a row of the first matrix with a column of the second. Multiply matching entries and add.

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \qquad B = \begin{bmatrix} 0 & 1 \\ 5 & 2 \end{bmatrix}$$

The top-left entry of $AB$ uses row $1$ of $A$ and column $1$ of $B$: $1 \cdot 0 + 2 \cdot 5 = 10$. Doing all four:

$$AB = \begin{bmatrix} 1\cdot0 + 2\cdot5 & 1\cdot1 + 2\cdot2 \\ 3\cdot0 + 4\cdot5 & 3\cdot1 + 4\cdot2 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 20 & 11 \end{bmatrix}$$

Sizes have to fit. A row of the first matrix must be as long as a column of the second, so an $m \times n$ matrix can multiply an $n \times p$ matrix, and the product is $m \times p$. The sales example was $(2 \times 3)(3 \times 1)$, giving $2 \times 1$.

Why the order of multiplication matters

Multiply the same two matrices the other way round:

$$BA = \begin{bmatrix} 0\cdot1 + 1\cdot3 & 0\cdot2 + 1\cdot4 \\ 5\cdot1 + 2\cdot3 & 5\cdot2 + 2\cdot4 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 11 & 18 \end{bmatrix}$$

$AB$ and $BA$ are different matrices. The reason is built into the rule: $AB$ pairs the rows of $A$ with the columns of $B$, while $BA$ pairs the rows of $B$ with the columns of $A$. Different pairs of numbers meet, so different products come out.

For numbers, $3 \times 5 = 5 \times 3$ without exception. For matrices, the order is part of the instruction, and $AB = BA$ is the rare exception.

Two familiar laws do survive. With $C = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$:

Property	Holds?	Check with $A$, $B$, $C$
commutative, $AB = BA$	no	$\begin{bmatrix} 10 & 5 \\ 20 & 11 \end{bmatrix} \ne \begin{bmatrix} 3 & 4 \\ 11 & 18 \end{bmatrix}$
associative, $(AB)C = A(BC)$	yes	both give $\begin{bmatrix} 25 & 5 \\ 51 & 11 \end{bmatrix}$
distributive, $A(B + C) = AB + AC$	yes	both give $\begin{bmatrix} 14 & 7 \\ 30 & 15 \end{bmatrix}$

The zero and identity matrices

Two matrices act like $0$ and $1$.

The zero matrix $O$ has every entry $0$. Adding it changes nothing: $A + O = A$.

The identity matrix $I$ has $1$s down the diagonal and $0$s elsewhere. Multiplying by it changes nothing, on either side:

$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad IA = AI = A$$

Inverses and the determinant

A number’s reciprocal undoes it: $5 \cdot \tfrac{1}{5} = 1$. A matrix’s inverse $A^{-1}$ does the same job with $I$ in place of $1$: $AA^{-1} = A^{-1}A = I$.

For a $2 \times 2$ matrix the inverse has a formula. Its denominator is the determinant:

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \qquad \det A = ad - bc \qquad A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

For $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, the determinant is $4 - 6 = -2$, so

$$A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}$$

Multiplying back confirms it: $AA^{-1} = I$.

A matrix has an inverse exactly when its determinant is not zero. The formula divides by the determinant, so a zero determinant leaves no inverse. For $\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$ the determinant is $4 - 4 = 0$: the first row is twice the second, and a matrix whose rows repeat each other loses information that nothing can recover.

Worked examples

Common mistakes

Practice problems