Invertible Matrix Theorem: Examples & Tests

The invertible matrix theorem says that for a square n × n matrix A, many statements are all true or all false together: A is invertible, A has n pivots, Ax = 0 has only the trivial solution, the columns of A are independent and span ℝⁿ, and det A ≠ 0. One check settles them all.

Before this lesson: Elementary Matrices, Linear Independence, One-to-One and Onto Transformations and Determinant of a Matrix

Why so many tests agree

For a general m × n matrix, the questions “does Ax = b always have a solution?” and “is the solution unique?” are separate. The first needs a pivot in every row; the second needs a pivot in every column. A matrix can pass one test and fail the other.

A square matrix is different. It has as many rows as columns, so n pivots means a pivot in every row and every column at once, and fewer than n pivots means both tests fail. That single count, the number of pivot positions, is what ties every statement in the theorem together.

A diagram with — For a square matrix, n pivots gives a pivot in every row and every column, so all of these statements stand or fall together.

The invertible matrix theorem

Let A be an n × n matrix. The following statements are equivalent: for a given A, they are either all true or all false.

(a) A is invertible.
(b) A is row equivalent to the identity matrix Iₙ (its reduced row echelon form is Iₙ).
(c) A has n pivot positions.
(d) The equation Ax = 0 has only the trivial solution x = 0.
(e) The columns of A are linearly independent.
(f) The linear transformation x ↦ Ax is one-to-one.
(g) The equation Ax = b has at least one solution for every b in ℝⁿ.
(h) The columns of A span ℝⁿ.
(i) The linear transformation x ↦ Ax maps ℝⁿ onto ℝⁿ.
(j) There is an n × n matrix C with CA = I.
(k) There is an n × n matrix D with AD = I.
(l) Aᵀ is invertible.
(m) det A ≠ 0.
(n) A is a product of elementary matrices.

When A is invertible, the solution of Ax = b is also unique: it is x = A⁻¹b. Later lessons add more statements to the list, such as “0 is not an eigenvalue of A” and “the rank of A is n.”

A square matrix that is not invertible is called singular. For a singular matrix, every statement on the list is false: it has fewer than n pivots, Ax = 0 has nontrivial solutions, its columns are dependent and do not span ℝⁿ, and det A = 0.

How to use the invertible matrix theorem

Make sure A is square. The theorem says nothing about m × n matrices with m ≠ n.
Check the statement that is easiest for this matrix: a determinant for a 2 × 2 or 3 × 3 matrix, a pivot count after row reduction, or an obvious relation between the columns.
If that statement is true, every statement is true. If it is false, every statement is false.
Answer the question asked by reading off the matching statement from the list.

Invertible matrix theorem examples

Example 1: Count pivots, then read off the rest

Let A = [1 2 −1; 2 5 0; −1 −1 4] (rows separated by semicolons). Use row reduction to decide whether A is invertible. Then say what this means for Ax = 0, for the columns of A and for Ax = b.

\begin{bmatrix} 1 & 2 & -1 \\ 2 & 5 & 0 \\ -1 & -1 & 4 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & -1 \\ 0 & 1 & 2 \\ -1 & -1 & 4 \end{bmatrix}

R₂ − 2R₁ → R₂. Row 2: (2 − 2, 5 − 4, 0 + 2) = (0, 1, 2) (see Gaussian elimination).

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

R₃ + R₁ → R₃. Row 3: (−1 + 1, −1 + 2, 4 − 1) = (0, 1, 3).

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

R₃ − R₂ → R₃. Row 3: (0, 1 − 1, 3 − 2) = (0, 0, 1).

\text{pivots in columns } 1, 2, 3

The echelon form has 3 pivots, and A is 3 × 3.

Answer: A has n = 3 pivots, so A is invertible. By the theorem, Ax = 0 has only the trivial solution, the columns of A are linearly independent and span ℝ³, and Ax = b has exactly one solution for every b in ℝ³.

Example 2: Spot a column relation

Let A = [1 2 3; 0 1 1; 2 −1 1]. Notice that column 3 equals column 1 plus column 2. What does the theorem say about A? Find a nontrivial solution of Ax = 0.

\mathbf{a}_1 + \mathbf{a}_2 = (1 + 2,\ 0 + 1,\ 2 - 1) = (3, 1, 1) = \mathbf{a}_3

Add columns 1 and 2: the sum is column 3.

\mathbf{a}_1 + \mathbf{a}_2 - \mathbf{a}_3 = \mathbf{0}

Subtract a₃ from both sides.

\mathbf{x} = (1, 1, -1)

The weights 1, 1, −1 give a solution of Ax = 0.

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

Check by multiplying.

Answer: the columns are linearly dependent, so A is singular. By the theorem, A has fewer than 3 pivots, det A = 0, the columns do not span ℝ³, and some equations Ax = b have no solution. One nontrivial solution of Ax = 0 is x = (1, 1, −1).

Example 3: When is a 2 × 2 matrix invertible?

For which values of k is A = [k 4; 3 k + 1] invertible?

\det A = k(k + 1) - 4(3)

For a 2 × 2 matrix, det A = ad − bc.

= k^2 + k - 12

Distribute and multiply.

= (k + 4)(k - 3)

Factor the trinomial (see factoring trinomials).

(k + 4)(k - 3) = 0 \Rightarrow k = -4 \text{ or } k = 3

Set each factor equal to 0.

Answer: A is invertible for every k except k = −4 and k = 3. Check k = 3: A = [3 4; 3 4] has two equal rows, so it is clearly singular.

Example 4: Reason from one fact

A is a 4 × 4 matrix, and the equation Ax = b has a solution for every b in ℝ⁴. Is the solution always unique? Can the columns of A be linearly dependent?

\text{(g) is true}

“Ax = b has a solution for every b” is statement (g) of the list.

\text{(a) is true: } A^{-1} \text{ exists}

A is square, so the theorem makes every statement true.

\mathbf{x} = A^{-1}\mathbf{b}

Multiply both sides of Ax = b on the left by A⁻¹.

\text{(e) is true}

Statement (e) says the columns are linearly independent.

Answer: yes, each solution is unique, namely x = A⁻¹b, and no, the columns cannot be dependent. The argument needs A to be square: a 4 × 5 matrix could have solutions for every b that are not unique.

Example 5: Can every grain order be filled?

A mill sells three blends of flour. One bag of blend 1 holds 2 kg wheat, 1 kg rye and 1 kg oats; blend 2 holds 1, 2 and 1 kg; blend 3 holds 1, 1 and 2 kg. A baker wants exact amounts of wheat, rye and oats. Can every order be filled by exactly one mix of bag counts? Then find the mix for 9 kg wheat, 8 kg rye and 7 kg oats.

A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}

Each column is one blend; each row is one grain.

\det A = 2(2 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 1) + 1(1 \cdot 1 - 2 \cdot 1)

Expand along row 1.

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

Work out each 2 × 2 determinant.

= 4

Multiply and add. Since 4 ≠ 0, A is invertible.

2x_1 + x_2 + x_3 = 9,\ \ x_1 + 2x_2 + x_3 = 8,\ \ x_1 + x_2 + 2x_3 = 7

Write Ax = b for b = (9, 8, 7).

4(x_1 + x_2 + x_3) = 24

Add the three equations.

x_1 + x_2 + x_3 = 6

Divide both sides by 4.

x_1 = 9 - 6 = 3,\ \ x_2 = 8 - 6 = 2,\ \ x_3 = 7 - 6 = 1

Each equation is (sum) + one variable, so subtract 6 from each right side.

Answer: det A = 4 ≠ 0, so every order (any b in ℝ³) has exactly one solution. For 9, 8 and 7 kg the mix is 3 bags of blend 1, 2 of blend 2 and 1 of blend 3. Check: 2(3) + 2 + 1 = 9, 3 + 2(2) + 1 = 8, 3 + 2 + 2(1) = 7. A real order also needs whole, nonnegative bag counts, which the theorem does not promise.

Common mistakes

Mistake	Example of the error	Fix
Using the theorem on a non-square matrix	Saying a 3 × 4 matrix with independent rows is invertible	Only square matrices can be invertible; check the size first.
Mixing up “every row” and “every column”	Thinking a 3 × 3 matrix can have a pivot in every row but not in every column	For a square matrix, a pivot in every row means n pivots, which is a pivot in every column too.
Concluding from a nonzero entry	Saying A is invertible because no entry is 0	A = [1 2; 2 4] has no zero entries, but det A = 0.
Forgetting that Ax = 0 always has x = 0	Calling A singular because Ax = 0 has a solution	Every matrix has the trivial solution. A is singular when there is a nontrivial one.
Reading “invertible” as “every solution is nice”	Expecting whole-number bag counts in Example 5 for every order	The theorem gives exactly one real solution; whole numbers are a separate question.

Related ideas

Inverse from one side. Statements (j) and (k) mean that for square matrices a one-sided inverse is automatically two-sided: if AD = I, then D = A⁻¹ and DA = I as well. This fails for non-square matrices. The inverse itself is found with the [A | I] method from the inverse matrices lesson.

What comes next. LU Decomposition factors A without finding A⁻¹. The determinant test (m) is developed in Properties of Determinants, and Cramer’s rule gives a formula for each entry of x = A⁻¹b when det A ≠ 0.

Practice problems

Is A = [3 6; 1 2] invertible? Explain with the theorem.
Is A = [1 2 0; 0 1 4; 0 0 5] invertible?
A is 3 × 3 and its columns are linearly dependent. Does Ax = b have a solution for every b in ℝ³?
A is 5 × 5 and x ↦ Ax is one-to-one. Is it onto ℝ⁵?
Find every value of h that makes A = [1 h; 2 6] singular.
Row reduce A = [1 −1 2; 2 1 1; 3 0 3]. Is A invertible? If not, find a nontrivial solution of Ax = 0.
A and D are n × n and AD = I. Explain why DA = I too.
A is 4 × 4 with det A = −2. Is Aᵀ invertible? Does Ax = 0 have a nontrivial solution?
Is T(x₁, x₂) = (x₁ + 3x₂, 2x₁ + 6x₂) one-to-one? If not, find a nonzero x with T(x) = 0.
Use a determinant to decide whether A = [2 1 0; 1 2 1; 0 1 2] is invertible.

Show answers

\det A = 3(2) - 6(1)

Use det A = ad − bc.

= 0

Multiply and subtract.

Answer: no. det A = 0, so A is singular. Column 2 is 2 times column 1, so the columns are also dependent.

\text{pivots: } 1, 1, 5 \text{ on the diagonal}

A is already upper triangular with nonzero diagonal entries.

\det A = 1(1)(5) = 5

A triangular determinant is the product of the diagonal.

Answer: yes. A has 3 pivots and det A = 5 ≠ 0.

\text{(e) is false}

Dependent columns means statement (e) fails.

\text{(g) is false}

A is square, so every statement fails.

Answer: no. Some b in ℝ³ give an equation Ax = b with no solution.

\text{(f) is true}

One-to-one is statement (f), and A is square.

\text{(i) is true}

Every statement is then true, including onto.

Answer: yes, x ↦ Ax maps ℝ⁵ onto ℝ⁵ (and A is invertible).

\det A = 1(6) - h(2) = 6 - 2h

Use det A = ad − bc.

6 - 2h = 0

A is singular exactly when det A = 0.

-2h = -6

Subtract 6 from both sides.

h = 3

Divide both sides by −2.

Answer: h = 3, which makes A = [1 3; 2 6] with column 2 = 3 × column 1.

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

R₂ − 2R₁ → R₂. Row 2: (2 − 2, 1 + 2, 1 − 4) = (0, 3, −3).

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

R₃ − 3R₁ → R₃. Row 3: (3 − 3, 0 + 3, 3 − 6) = (0, 3, −3).

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

R₃ − R₂ → R₃. Only 2 pivots, so A is singular.

x_3 = t

x₃ is a free variable.

3x_2 - 3t = 0 \Rightarrow x_2 = t

Row 2: add 3t to both sides, then divide by 3.

x_1 - t + 2t = 0 \Rightarrow x_1 = -t

Row 1: substitute x₂ = t, combine like terms and subtract t.

\mathbf{x} = t(-1, 1, 1)

Write the solution in parametric vector form.

Answer: not invertible. With t = 1, x = (−1, 1, 1): A x = (−1 − 1 + 2, −2 + 1 + 1, −3 + 0 + 3) = (0, 0, 0).

\text{(k) is true}

AD = I is statement (k).

A^{-1} \text{ exists}

So statement (a) is true.

D = A^{-1}AD = A^{-1}I = A^{-1}

Multiply AD = I on the left by A⁻¹.

DA = A^{-1}A = I

Substitute D = A⁻¹.

Answer: D must be A⁻¹, so DA = I.

\det A = -2 \ne 0

Statement (m) is true.

Answer: Aᵀ is invertible (statement l), and Ax = 0 has only the trivial solution (statement d).

A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}

Read the standard matrix from the coefficients.

\det A = 1(6) - 3(2) = 0

Use det A = ad − bc.

T(3, -1) = (3 - 3,\ 6 - 6) = (0, 0)

Column 2 is 3 times column 1, so try x = (3, −1).

Answer: T is not one-to-one; T(3, −1) = (0, 0) = T(0, 0).

10.

\det A = 2(2 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 0) + 0

Expand along row 1.

= 2(3) - 1(2)

Work out each 2 × 2 determinant.

= 4

Multiply and subtract.

Answer: yes, det A = 4 ≠ 0, so A is invertible.

Invertible matrix theorem FAQ

What is the invertible matrix theorem?

It is a list of statements about a square matrix A, such as “A is invertible,” “det A ≠ 0” and “the columns of A are linearly independent,” that are either all true or all false for each A.

Does the invertible matrix theorem apply to non-square matrices?

No. It is only for n × n matrices. A non-square matrix is never invertible, and for it “solutions always exist” and “solutions are unique” are separate questions.

What does it mean if the determinant of a matrix is 0?

The matrix is singular: it has no inverse, its columns are linearly dependent, and Ax = 0 has nontrivial solutions.

What is the fastest way to tell if a matrix is invertible?

For a 2 × 2 or 3 × 3 matrix, compute the determinant. For larger matrices, row reduce to echelon form and count the pivots; n pivots means invertible.

Summary

The theorem applies only to square n × n matrices.
All its statements are equivalent: invertible, n pivots, Ax = 0 has only x = 0, independent columns, columns span ℝⁿ, one-to-one, onto, det A ≠ 0, Aᵀ invertible, a product of elementary matrices.
The key is the pivot count: for a square matrix, a pivot in every row is the same as a pivot in every column.
If A is invertible, Ax = b has exactly one solution, x = A⁻¹b.
A singular matrix fails every statement on the list.

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