What Is an Invertible Matrix?
An invertible matrix is a square matrix $A$ for which there exists another matrix $A^{-1}$ of the same order such that
$$A A^{-1} = A^{-1} A = I,$$
where $I$ 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 is the central result of the topic. It says that for a square $n \times n$ matrix $A$, a long list of conditions are all equivalent — if any one holds, every one holds, and if any one fails, all fail. The matrix is either invertible in every sense at once, or in none.
The most-used conditions in the chain:
$A$ is invertible (an $A^{-1}$ exists).
$\det A \neq 0$.
$A$ is row-equivalent to the identity matrix $I_n$ (row reduction produces $I$).
$A$ has $n$ pivot positions.
The equation $A\mathbf{x} = \mathbf{0}$ has only the trivial solution $\mathbf{x} = \mathbf{0}$.
The columns of $A$ are linearly independent.
The columns of $A$ span $\mathbb{R}^n$ (and form a basis).
$A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$.
The rank of $A$ equals $n$ (full rank).
$0$ is not an eigenvalue of $A$.
$A^T$ 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 $A$ and $B$ are invertible, so is $AB$, and $(AB)^{-1} = B^{-1}A^{-1}$.
Transpose is invertible: $(A^T)^{-1} = (A^{-1})^T$.
Determinant of the inverse: $\det(A^{-1}) = \dfrac{1}{\det A}$ — which is why $\det A$ can never be zero for an invertible matrix.
Scalar multiple: $(kA)^{-1} = \dfrac{1}{k}A^{-1}$ for non-zero $k$.
Powers stay invertible: $A^n$ is invertible for any positive integer $n$, with $(A^n)^{-1} = (A^{-1})^n$.
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 $A$ 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 $[A \mid I]$ 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
The set runs from a quick determinant check, through the most common eigenvalue trap, to a singular case, the product property, an eigenvalue test, and recovering a matrix from its inverse.
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: $\det A = 10 \neq 0$, so $A$ 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.
Wrong attempt. A student reads the diagonal entries $2$ and $0$ as eigenvalues, sees a non-zero one, and concludes "it has a non-zero eigenvalue, so it's invertible."
Recheck the rule. Invertibility requires that zero is not an eigenvalue — having some non-zero eigenvalue is not enough if zero is also present. Here the eigenvalues are $2$ and $0$, and that $0$ is fatal.
Correct. Because $0$ is an eigenvalue, $A$ is singular. Confirm with the determinant: $\det A = (2)(0) - 0 = 0$.
Final answer: $A$ 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 $A$ and $B$ are invertible $2 \times 2$ 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 $\det(AB) = \det A \cdot \det B = 1 \neq 0$ — invertible. Using $(AB)^{-1} = B^{-1}A^{-1}$ with $A^{-1} = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix}$, $B^{-1} = \begin{bmatrix} 1 & 0 \ -3 & 1 \end{bmatrix}$:
$$(AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 1 & 0 \ -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \ -3 & 7 \end{bmatrix}.$$
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 $\det A = -5$.
The property $\det(A^{-1}) = \frac{1}{\det A}$ applies directly:
$$\det(A^{-1}) = \frac{1}{-5} = -\frac{1}{5}.$$
Final answer: $\det(A^{-1}) = -\tfrac{1}{5}$. 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 $A$.
Since $(A^{-1})^{-1} = A$, invert the given matrix. Its determinant is $(2)(1) - (1)(1) = 1$, so
$$A = \frac{1}{1}\begin{bmatrix} 1 & -1 \ -1 & 2 \end{bmatrix}.$$
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 $A^{-1}$.
Why the Invertible Matrix Theorem Earns Its Place
"Can this transformation be undone, or is the information gone?"
The web of equivalent conditions was assembled as linear algebra matured through the late 1800s and early 1900s, building on Arthur Cayley (1821–1895, England), whose 1858 matrix algebra gave "invertible" its modern meaning. The theorem's value is that it ties geometry, equations, determinants, and eigenvalues into one statement.
Where invertibility decides the outcome:
Unique solutions to systems. $A\mathbf{x} = \mathbf{b}$ has exactly one solution for every $\mathbf{b}$ precisely when $A$ is invertible — the foundation of every well-posed linear model in engineering and economics.
Stability of numerical methods. A nearly-singular matrix (determinant close to zero) makes computations wildly sensitive to rounding — knowing invertibility, and how close to losing it, matters in real software.
Eigenvalue analysis. "Zero is not an eigenvalue" links invertibility to the modes of a physical system; a zero eigenvalue signals a direction the system can't push back on.
Cryptography. A cipher key matrix must be invertible, or messages encoded with it could never be decoded.
Where Students Trip Up on the Invertible Matrix
Mistake 1: Thinking a non-square matrix can be invertible
Where it slips in: A problem gives a $2 \times 3$ matrix and asks "is it invertible?"
Don't do this: Try to compute a determinant or an inverse for a rectangular matrix.
The correct way: Only square matrices can be invertible. If rows ≠ columns, the answer is "not invertible" before any computation.
Mistake 2: Treating "has a non-zero eigenvalue" as enough
Where it slips in: Judging invertibility from eigenvalues.
Don't do this: Conclude a matrix is invertible because it has some non-zero eigenvalue.
The correct way: Invertibility requires that zero is not an eigenvalue at all. A single zero eigenvalue makes the matrix singular, no matter how many non-zero ones sit beside it. The second-guesser who has just learned the eigenvalue condition often flips it — anchoring on "zero kills invertibility" keeps it straight.
Mistake 3: Confusing the determinant being small with being zero
Where it slips in: A matrix with a tiny determinant, like $0.0001$.
Don't do this: Round it to zero and call the matrix singular.
The correct way: A non-zero determinant means invertible — even a tiny one. (Such matrices are ill-conditioned, which is a numerical caution, not a verdict of singularity.) The memorizer who learned "small determinant = bad" overgeneralises here.
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: $(A^{-1})^{-1} = A$, $(AB)^{-1} = B^{-1}A^{-1}$ (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 $A\mathbf{x} = \mathbf{b}$ has a unique solution — the basis of well-posed linear systems, cryptography keys, and eigenvalue analysis.
Practice These Before Moving On
Is $\begin{bmatrix} 5 & 2 \ 10 & 4 \end{bmatrix}$ invertible? Justify with the determinant.
If $\det A = 7$ for a $3 \times 3$ matrix, find $\det(A^{-1})$.
The eigenvalues of a matrix are $3, -2, 0$. Is the matrix invertible?
Answer to Question 1: $\det = 0$, so not invertible (the rows are dependent — row 2 is twice row 1). Answer to Question 2: $\det(A^{-1}) = \tfrac{1}{7}$. Answer to Question 3: not invertible — the eigenvalue $0$ makes it singular. If Question 3 tempted you toward "yes" because of the $3$ and $-2$, return to Mistake 2.
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