# 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](/content/math/algebra/inverse-of-a-matrix/index.html), 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](/content/math/algebra/singular-matrix/index.html) 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: 1. **Is it square?** A non-square matrix is never invertible — stop immediately if rows ≠ columns. 2. **Compute the determinant.** Non-zero means invertible; zero means singular. This is the single most efficient test for small matrices. 3. **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](/content/math/algebra/inverse-of-3x3-matrix/index.html) 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.