Invertible Matrix — Theorem, Properties, Examples

TL;DR

An invertible matrix is a square matrix AAA for which a matrix A−1 exists with AA−1 = I — equivalently, a square matrix whose determinant is non-zero. This article covers the definition, the invertible matrix theorem, the key properties, how to check invertibility, and six worked examples.

What Is an Invertible Matrix?

An invertible matrix is a square matrix AAA for which there exists another matrix A−1 of the same order such that AA−1 = I, where III is the identity matrix. The matrix A−1 is the inverse of the matrix, and it is unique — an invertible matrix has exactly one inverse. An invertible matrix is also called non-singular, non-degenerate, or regular.

The single cleanest test is the determinant: a square matrix is invertible if and only if its determinant is non-zero. If det A = 0, the matrix is a singular matrix and no inverse exists. This makes "invertible" the same idea as "the inverse exists" — the two are not separate properties but two names for the same condition. Only square matrices are ever invertible; a rectangular matrix cannot be invertible at all.

What Is the Invertible Matrix Theorem?

The invertible matrix theorem states that for a square n×n matrix AAA, a long list of conditions are all equivalent — if any one holds, every one holds, and if any one fails, all fail. The most-used conditions in the chain:

AAA is invertible (an A−1 exists).
det A ≠ 0.
AAA is row-equivalent to the identity matrix In (row reduction produces III).
AAA has n pivot positions.
The equation Ax=0 has only the trivial solution x=0.
The columns of AAA are linearly independent.
The columns of AAA span Rn (and form a basis).
Ax=b has a unique solution for every b.
The rank of AAA equals n (full rank).
0 is not an eigenvalue of AAA.
ATA^T is also invertible.

The power of the theorem is the shortcut it gives: to prove a matrix invertible, you do not have to compute the inverse — you check whichever condition is easiest. Often that is just the determinant.

What Are the Properties of an Invertible Matrix?

Invertible matrices follow a short, dependable set of rules:

Unique inverse. An invertible matrix has exactly one inverse.
Inverse of the inverse: (A−1)−1 = A.
Product is invertible (reversed order): if AAA and BBB are invertible, so is AB, and (AB)−1 = B−1A−1.
Transpose is invertible: (A^T)−1 = (A−1)^T.
Determinant of the inverse: det(A−1) = 1/det A — which is why det A can never be zero for an invertible matrix.
Scalar multiple: (kA)−1 = (1/k)A−1 for non-zero k.
Powers stay invertible: (A^n)−1 = (A−1)^n for any positive integer n.

The reversed order in the product rule is the property most worth committing to memory — it is the one that trips people up under pressure.

How Do You Check If a Matrix Is Invertible?

Three checks, fastest first:

Is it square? A non-square matrix is never invertible — stop immediately if rows ≠ columns.
Compute the determinant. Non-zero means invertible; zero means singular. This is the single most efficient test for small matrices.
Row-reduce (for larger matrices). If AAA reduces to the identity, it is invertible; if a row of zeros appears, it is singular. Row reduction also hands you the inverse via the A|I method at the same time.

You rarely need to compute the full inverse just to know a matrix is invertible — that is exactly what the invertible matrix theorem buys you.

Examples of Invertible Matrix

Example 1

Is A= [\begin{bmatrix} 3 & 1 \ 2 & 4 \end{bmatrix}] invertible?

Square, so check the determinant: det A = (3)(4)−(1)(2) = 10.

Final answer: det A = 10 ≠ 0, so AAA is invertible. No need to find A−1 to answer the question.

Example 2

Is A= [\begin{bmatrix} 2 & 0 \ 0 & 0 \end{bmatrix}] invertible? Decide using eigenvalues.

Correct. Because 0 is an eigenvalue, AAA is singular. Confirm with the determinant: det A = (2)(0)−0 = 0.

Final answer: AAA is not invertible. One zero eigenvalue is enough to make a matrix singular, regardless of the others.

Example 3

Is A= [\begin{bmatrix} 1 & 2 & 3 \ 2 & 4 & 6 \ 0 & 1 & 1 \end{bmatrix}] invertible?

The second row is exactly twice the first, so the rows are linearly dependent. By the theorem, dependent columns (or rows) force a zero determinant: det A = 0.

Final answer: not invertible. Linear dependence and a zero determinant are the same verdict seen from two angles.

Example 4

If AAA and BBB are invertible 2×2 matrices, show AB is invertible and find (AB)−1 for A= [\begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}], B= [\begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}].

Both have determinant 1, so det(AB) = det A ⋅ det B = 1 ≠ 0 — invertible. Using (AB)−1 = B−1A−1: (AB)−1 = \begin{bmatrix} 1 & 0 \ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \ -3 & 7 \end{bmatrix}.

Final answer: AB is invertible with (AB)−1 = [\begin{bmatrix} 1 & -2 \ -3 & 7 \end{bmatrix}]. The product of invertibles is always invertible.

Example 5

Use the determinant of the inverse to find det(A−1) when det A = −5.

det(A−1) = \frac{1}{det A} = -\frac{1}{5}.

Final answer: det(A−1) = -\frac{1}{5}. Because the determinant of an invertible matrix is non-zero, this reciprocal is always defined.

Example 6

Given A−1= [\begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}], find AAA.

Since (A−1)−1 = A, invert the given matrix. Its determinant is (2)(1)−(1)(1)=1, so A= [\begin{bmatrix} 1 & -1 \ -1 & 2 \end{bmatrix}].

Final answer: A= [\begin{bmatrix} 1 & -1 \ -1 & 2 \end{bmatrix}]. Inverting the inverse recovers the original matrix — a useful sanity check whenever a problem hands you A−1.

Key Takeaways

An invertible matrix is a square matrix with a unique inverse — equivalently, one whose determinant is non-zero.
"Invertible," "non-singular," "non-degenerate," and "regular" all mean the same thing.
The invertible matrix theorem ties together a chain of equivalent conditions; check whichever is easiest, usually the determinant.
Key properties include: (A−1)−1 = A, (AB)−1 = B−1A−1 (reversed order), and det(A−1) = 1/det A.
Zero being an eigenvalue makes a matrix singular — a single zero eigenvalue is fatal to invertibility.
Invertibility decides whether Ax=b has a unique solution — the basis of well-posed linear systems, cryptography keys, and eigenvalue analysis.