Matrix Equation Ax = b: Steps and Examples

The matrix equation Ax = b writes a whole linear system with one matrix A and one vector of unknowns x. The product Ax is the linear combination of the columns of A that uses the entries of x as weights. So Ax = b has a solution exactly when b is in the span of the columns of A.

Before this lesson: Span of Vectors and Linear Combinations

What the product Ax means

Let A be an m × n matrix with columns a₁, …, aₙ, and let x be a vector in ℝⁿ. The product Ax is defined as a linear combination of the columns, with the entries of x as the weights:

A\mathbf{x} = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n

The product only makes sense when x has one entry for each column of A. The answer is a vector in ℝᵐ, with one entry for each row of A.

For A with columns a₁ = (2, 1) and a₂ = (−1, 1) and x = (2, 1), Ax walks 2 copies of a₁ and then 1 copy of a₂, ending at (3, 3).

There is also a faster way to compute by hand, the row-vector rule: entry i of Ax is the dot product of row i of A with x. Both ways give the same vector, and this is the same product you know from matrix multiplication.

Three ways to write one linear system

A linear system, a vector equation and a matrix equation are three ways of writing the same problem. They all have the same solutions, and all three are solved by row reducing the augmented matrix [A | b].

Form	What it looks like	Use it to
Linear system	x₁ + 2x₂ = 5 and 3x₁ + 4x₂ = 11	Solve by elimination or row reduction
Vector equation	x₁(1, 3) + x₂(2, 4) = (5, 11)	Ask whether b is a linear combination of the columns
Matrix equation	Ax = b, where A has rows (1, 2) and (3, 4) and b = (5, 11)	Write the system compactly and use matrix tools

When does Ax = b have a solution?

For one vector b, row reduce [A | b]. The equation has a solution exactly when no row reads [0 ⋯ 0 | c] with c ≠ 0. In other words, b must be in the span of the columns of A.

For every b at once, there is a single test. If A is an m × n matrix, these statements are either all true or all false:

Ax = b has a solution for every b in ℝᵐ.
Every b in ℝᵐ is a linear combination of the columns of A.
The columns of A span ℝᵐ.
A has a pivot position in every row.

Use the last statement to check, since it only needs the coefficient matrix A. A pivot in every column is a different condition and does not answer this question.

The product also follows two rules that later lessons use often. For vectors u and v in ℝⁿ and any scalar c:

A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} \qquad A(c\mathbf{u}) = c(A\mathbf{u})

Matrix equation Ax = b examples

Example 1: Compute Ax two ways

Find Ax for the matrix and vector below.

A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 4 \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}

A\mathbf{x} = 3\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1\begin{bmatrix} 2 \\ -3 \end{bmatrix} + 2\begin{bmatrix} -1 \\ 4 \end{bmatrix}

Column way: A is 2 × 3 and x has 3 entries, so use x₁, x₂, x₃ as the weights.

A\mathbf{x} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ -3 \end{bmatrix} + \begin{bmatrix} -2 \\ 8 \end{bmatrix}

Multiply each column by its weight.

A\mathbf{x} = \begin{bmatrix} 3 + 2 - 2 \\ 0 - 3 + 8 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix}

Add entry by entry.

(1)(3) + (2)(1) + (-1)(2) = 3

Row way, entry 1: row 1 of A dotted with x.

(0)(3) + (-3)(1) + (4)(2) = 5

Row way, entry 2: row 2 of A dotted with x.

Answer: Ax = (3, 5). Both ways agree.

Example 2: Write a system as a matrix equation

Write the system 2x₁ − x₂ + 4x₃ = 7 and x₁ + 3x₃ = −2 as a vector equation and as a matrix equation. Then check that x = (4, −7, −2) is a solution.

x_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} -1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} 4 \\ 3 \end{bmatrix} = \begin{bmatrix} 7 \\ -2 \end{bmatrix}

Vector equation: the coefficients of each variable form a column. The second equation has no x₂, so that entry is 0.

\begin{bmatrix} 2 & -1 & 4 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7 \\ -2 \end{bmatrix}

Matrix equation Ax = b: the same columns make A.

(2)(4) + (-1)(-7) + (4)(-2) = 8 + 7 - 8

Row 1 of A dotted with x = (4, −7, −2).

8 + 7 - 8 = 7

Add: entry 1 matches b₁ = 7.

(1)(4) + (0)(-7) + (3)(-2) = 4 + 0 - 6

Row 2 of A dotted with x.

4 + 0 - 6 = -2

Add: entry 2 matches b₂ = −2.

Answer: Ax = (7, −2) = b, so x = (4, −7, −2) is a solution.

Example 3: Solve Ax = b

Solve Ax = b, where A has columns (1, 3, −1) and (2, 4, 1) and b = (5, 11, 1).

\left[\begin{array}{rr|r} 1 & 2 & 5 \\ 3 & 4 & 11 \\ -1 & 1 & 1 \end{array}\right]

Augmented matrix [A | b].

\left[\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & -2 & -4 \\ -1 & 1 & 1 \end{array}\right]

R₂ − 3R₁ → R₂: 4 − 6 = −2 and 11 − 15 = −4 (see Gaussian elimination).

\left[\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & -2 & -4 \\ 0 & 3 & 6 \end{array}\right]

R₃ + R₁ → R₃: 1 + 2 = 3 and 1 + 5 = 6.

\left[\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]

−(1/2)R₂ → R₂.

\left[\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]

R₃ − 3R₂ → R₃.

x_2 = 2

Row 2.

x_1 + 2(2) = 5

Row 1 with x₂ = 2.

x_1 = 1

Subtract 4 from both sides.

A\mathbf{x} = 1\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + 2\begin{bmatrix} 2 \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 + 4 \\ 3 + 8 \\ -1 + 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 11 \\ 1 \end{bmatrix}

Check: Ax equals b.

Answer: x = (1, 2).

Example 4: Is Ax = b consistent for every b?

Let A have rows (1, −2, 3), (2, −3, 8) and (−1, 4, 1). Is Ax = b consistent for every b in ℝ³? If not, describe the vectors b that work.

\left[\begin{array}{rrr|c} 1 & -2 & 3 & b_1 \\ 2 & -3 & 8 & b_2 \\ -1 & 4 & 1 & b_3 \end{array}\right]

Augmented matrix with a general b.

\left[\begin{array}{rrr|c} 1 & -2 & 3 & b_1 \\ 0 & 1 & 2 & b_2 - 2b_1 \\ -1 & 4 & 1 & b_3 \end{array}\right]

R₂ − 2R₁ → R₂: −3 + 4 = 1 and 8 − 6 = 2.

\left[\begin{array}{rrr|c} 1 & -2 & 3 & b_1 \\ 0 & 1 & 2 & b_2 - 2b_1 \\ 0 & 2 & 4 & b_3 + b_1 \end{array}\right]

R₃ + R₁ → R₃: 4 − 2 = 2 and 1 + 3 = 4.

\left[\begin{array}{rrr|c} 1 & -2 & 3 & b_1 \\ 0 & 1 & 2 & b_2 - 2b_1 \\ 0 & 0 & 0 & 5b_1 - 2b_2 + b_3 \end{array}\right]

R₃ − 2R₂ → R₃: b₃ + b₁ − 2(b₂ − 2b₁) = 5b₁ − 2b₂ + b₃.

A has no pivot in row 3, so Ax = b is not consistent for every b. It is consistent only when the last right side is 0.

5b_1 - 2b_2 + b_3 = 0

The condition on b.

5(1) - 2(2) + (-1) = 0

Check: column 1 of A, (1, 2, −1), satisfies it.

5(-2) - 2(-3) + 4 = 0

Check: column 2, (−2, −3, 4).

5(3) - 2(8) + 1 = 0

Check: column 3, (3, 8, 1).

Answer: no. The columns of A span only the plane 5x − 2y + z = 0, so Ax = b has a solution only for b on that plane. For example, b = (1, 0, 0) gives 5 ≠ 0, so it has no solution.

Example 5: How many batches did the bakery make?

A bakery makes muffins, scones and cookies. One batch of muffins uses 2 cups of flour, 1 cup of sugar and 1 cup of butter. A batch of scones uses 3, 1 and 2 cups, and a batch of cookies uses 2, 2 and 1 cups. Yesterday the bakery used 20 cups of flour, 12 cups of sugar and 11 cups of butter. How many batches of each did it make?

\begin{bmatrix} 2 & 3 & 2 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 20 \\ 12 \\ 11 \end{bmatrix}

Each column is one product’s recipe (flour, sugar, butter), and x holds the numbers of batches.

\left[\begin{array}{rrr|r} 2 & 3 & 2 & 20 \\ 1 & 1 & 2 & 12 \\ 1 & 2 & 1 & 11 \end{array}\right]

Augmented matrix [A | b].

\left[\begin{array}{rrr|r} 1 & 1 & 2 & 12 \\ 2 & 3 & 2 & 20 \\ 1 & 2 & 1 & 11 \end{array}\right]

R₁ ↔ R₂, so the first pivot is 1.

\left[\begin{array}{rrr|r} 1 & 1 & 2 & 12 \\ 0 & 1 & -2 & -4 \\ 1 & 2 & 1 & 11 \end{array}\right]

R₂ − 2R₁ → R₂: 3 − 2 = 1, 2 − 4 = −2 and 20 − 24 = −4.

\left[\begin{array}{rrr|r} 1 & 1 & 2 & 12 \\ 0 & 1 & -2 & -4 \\ 0 & 1 & -1 & -1 \end{array}\right]

R₃ − R₁ → R₃: 2 − 1 = 1, 1 − 2 = −1 and 11 − 12 = −1.

\left[\begin{array}{rrr|r} 1 & 1 & 2 & 12 \\ 0 & 1 & -2 & -4 \\ 0 & 0 & 1 & 3 \end{array}\right]

R₃ − R₂ → R₃: −1 + 2 = 1 and −1 + 4 = 3.

x_3 = 3

Row 3.

x_2 - 2(3) = -4

Row 2 with x₃ = 3.

x_2 = 2

Add 6 to both sides.

x_1 + 2 + 2(3) = 12

Row 1 with x₂ = 2 and x₃ = 3.

x_1 + 8 = 12

Add 2 + 6.

x_1 = 4

Subtract 8 from both sides.

4\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} + 3\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 8 + 6 + 6 \\ 4 + 2 + 6 \\ 4 + 4 + 3 \end{bmatrix} = \begin{bmatrix} 20 \\ 12 \\ 11 \end{bmatrix}

Check: Ax equals b.

Answer: 4 batches of muffins, 2 batches of scones and 3 batches of cookies.

Common mistakes

Mistake	Example of the error	Fix
Multiplying when the sizes do not match	Multiplying a 2 × 3 matrix by a vector with 2 entries	x needs one entry for each column of A.
Using the rows as the vectors in the combination	Writing Ax as x₁(row 1) + x₂(row 2)	Ax is a combination of the columns, with the entries of x as weights.
Testing “every b” with a pivot in every column	“A has 2 pivots and 2 columns, so Ax = b always has a solution” for a 3 × 2 matrix A	Ax = b is consistent for every b only when A has a pivot in every row.
Writing the system’s rows as the columns of A	Using (2, −1, 4) as column 1 for 2x₁ − x₂ + 4x₃ = 7	Each equation is a row of A; the coefficients of one variable form a column.
Forgetting a missing variable	Writing row 2 of A as [1 3] for x₁ + 3x₃ = −2	Put a 0 for the missing x₂, so the row is [1 0 3].

Related ideas

Square invertible matrices. When A is square with an inverse, Ax = b has exactly one solution, x = A⁻¹b. See inverse matrices.

The case b = 0. The equation Ax = 0 always has the solution x = 0. When it has others is the topic of the next lesson, Homogeneous Systems.

Reading the weights quickly. Reducing [A | b] all the way to reduced row echelon form lets you read x without back-substitution.

Practice problems

Compute Ax for A with rows (2, −1) and (0, 3) and x = (4, 1).
Compute Ax for A with rows (1, 0, 2) and (3, −1, 1) and x = (2, 5, −1).
Compute Ax for A with rows (1, 4), (−2, 0) and (3, 1) and x = (−1, 2).
A is 3 × 2 and x is in ℝ³. Is Ax defined?
Write 3x₁ − x₂ = 4 and x₁ + 2x₂ + 5x₃ = 0 as a matrix equation Ax = b.
Write x₁(1, 2) + x₂(−3, 1) + x₃(0, 4) = (5, 6) as a matrix equation.
Solve Ax = b for A with rows (1, −1) and (2, 1) and b = (1, 8).
Solve Ax = b for A with rows (1, 2), (2, 4) and (0, 1) and b = (3, 6, 1).
For A with rows (1, 3) and (−2, −6), describe all b for which Ax = b is consistent.
Is Ax = b consistent for every b in ℝ³ when A has rows (1, 0, 2), (0, 1, −1) and (2, 1, 4)?

Show answers

4\begin{bmatrix} 2 \\ 0 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 8 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 3 \end{bmatrix}

Multiply each column by its weight.

\begin{bmatrix} 8 - 1 \\ 0 + 3 \end{bmatrix} = \begin{bmatrix} 7 \\ 3 \end{bmatrix}

Add entry by entry.

Answer: Ax = (7, 3).

(1)(2) + (0)(5) + (2)(-1) = 2 + 0 - 2 = 0

Row 1 dotted with x.

(3)(2) + (-1)(5) + (1)(-1) = 6 - 5 - 1 = 0

Row 2 dotted with x.

Answer: Ax = (0, 0), the zero vector.

(1)(-1) + (4)(2) = -1 + 8 = 7

Row 1 dotted with x.

(-2)(-1) + (0)(2) = 2 + 0 = 2

Row 2 dotted with x.

(3)(-1) + (1)(2) = -3 + 2 = -1

Row 3 dotted with x.

Answer: Ax = (7, 2, −1).

No. A has 2 columns, so x must have 2 entries. A vector in ℝ³ has 3.

\begin{bmatrix} 3 & -1 & 0 \\ 1 & 2 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}

Each equation is a row of A; the first equation has no x₃, so its entry is 0.

\begin{bmatrix} 1 & -3 & 0 \\ 2 & 1 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}

The three vectors become the columns of A.

\left[\begin{array}{rr|r} 1 & -1 & 1 \\ 2 & 1 & 8 \end{array}\right]

Augmented matrix [A | b].

\left[\begin{array}{rr|r} 1 & -1 & 1 \\ 0 & 3 & 6 \end{array}\right]

R₂ − 2R₁ → R₂: 1 + 2 = 3 and 8 − 2 = 6.

x_2 = 2

Row 2: 3x₂ = 6, so divide both sides by 3.

x_1 - 2 = 1

Row 1 with x₂ = 2.

x_1 = 3

Add 2 to both sides.

Answer: x = (3, 2). Check: 3 − 2 = 1 and 6 + 2 = 8.

\left[\begin{array}{rr|r} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 0 & 1 & 1 \end{array}\right]

Augmented matrix [A | b].

\left[\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{array}\right]

R₂ − 2R₁ → R₂: 4 − 4 = 0 and 6 − 6 = 0.

\left[\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]

R₂ ↔ R₃.

x_2 = 1

Row 2.

x_1 + 2(1) = 3

Row 1 with x₂ = 1.

x_1 = 1

Subtract 2 from both sides.

Answer: x = (1, 1). Check: (1 + 2, 2 + 4, 0 + 1) = (3, 6, 1).

\left[\begin{array}{rr|c} 1 & 3 & b_1 \\ -2 & -6 & b_2 \end{array}\right]

Augmented matrix with a general b.

\left[\begin{array}{rr|c} 1 & 3 & b_1 \\ 0 & 0 & b_2 + 2b_1 \end{array}\right]

R₂ + 2R₁ → R₂: −2 + 2 = 0 and −6 + 6 = 0.

2b_1 + b_2 = 0

The last right side must be 0.

Answer: Ax = b is consistent exactly when 2b₁ + b₂ = 0, that is, when b lies on the line y = −2x.

10.

\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 2 & 1 & 4 \end{bmatrix}

The coefficient matrix A.

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

R₃ − 2R₁ → R₃: 1 − 0 = 1 and 4 − 4 = 0.

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

R₃ − R₂ → R₃: 0 + 1 = 1.

Answer: yes. A has a pivot in every row, so its columns span ℝ³ and Ax = b has a solution for every b.

Matrix equation Ax = b FAQ

What does Ax = b mean in linear algebra?

It is a linear system written with a coefficient matrix A, a vector of unknowns x and a vector of constants b. Solving it means finding the weights x that combine the columns of A into b.

How do you multiply a matrix by a vector?

Multiply each column of A by the matching entry of x and add the results. By hand, it is often faster to take the dot product of each row of A with x.

When does Ax = b have a solution?

When b is in the span of the columns of A. For every b at once, A must have a pivot in every row.

Is Ax = b the same as an augmented matrix?

They describe the same system. The augmented matrix [A | b] is the tool you row reduce to solve Ax = b.

Summary

Ax is the combination x₁a₁ + ⋯ + xₙaₙ of the columns of A, with the entries of x as weights.
Ax is defined only when x has one entry per column of A; by hand, use the row-vector rule.
A linear system, a vector equation and Ax = b have the same solutions, found by row reducing [A | b].
Ax = b has a solution exactly when b is in the span of the columns of A.
Ax = b is consistent for every b exactly when A has a pivot in every row.

Previous lesson: Span of Vectors

Next lesson: Homogeneous Systems

Back to the Linear Algebra Index