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Inverse of a Matrix — Definition & Formula

Algebra

TL;DR

The inverse of a matrix AAA is the square matrix A−1A^{-1}A−1 satisfying AA−1=A−1A=IAA^{-1} = A^{-1}A = IAA−1=A−1A=I, and it exists only when AAA is square with a non-zero determinant. This article covers what an inverse really is, the fast 2x2 formula, the adjugate-over-determinant method for any size, the invertibility condition, and six worked examples.

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Bhanzu Team Last updated on June 10, 2026 9 min read

What Is the Inverse of a Matrix?

The inverse of a matrix AAA is the unique matrix A−1A^{-1}A−1 such that

AA−1=A−1A=I,AA^{-1} = A^{-1} A = I,AA−1=A−1A=I,

where III is the identity matrix of the same order. Multiplying AAA by A−1A^{-1}A−1 in either order returns the identity — the matrix that leaves every vector unchanged, exactly as multiplying a number by 1 leaves it unchanged.

Two conditions must hold for an inverse to exist. First, AAA must be square — same number of rows and columns — because only square matrices can map a space back onto itself. Second, its determinant must be non-zero. A square matrix with a non-zero determinant is invertible (non-singular); one whose determinant is zero is a singular matrix and has no inverse. When the inverse exists, it is unique — a matrix never has two different inverses.

What Is the Inverse of a Matrix Formula?

There is one general formula, and one fast shortcut for the 2x2 case.

The general (adjugate) formula works for any invertible square matrix:

A−1=1det⁡A,adj(A),A^{-1} = \frac{1}{\det A},\text{adj}(A),A−1=detA1​,adj(A),

where det⁡A\det AdetA is the determinant and adj(A)\text{adj}(A)adj(A) is the adjugate — the transpose of the cofactor matrix. The cofactor of each entry is the determinant of the smaller matrix left after deleting that entry's row and column, signed by the checkerboard pattern (−1)i+j(-1)^{i+j}(−1)i+j.

The 2x2 shortcut is worth memorising because it appears everywhere. For

A=[abcd], A−1=1ad−bc[d−b−ca].A = \begin{bmatrix} a & b \ c & d \end{bmatrix}, \qquad A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}.A=[a​bc​d​],A−1=ad−bc1​[d​−b−c​a​].

Swap the diagonal entries, negate the off-diagonal entries, and divide by the determinant ad−bcad - bcad−bc. The full 2x2 derivation and more practice lives in its own article; for the 3x3 case, where the cofactor work gets heavier, see the inverse of a 3x3 matrix walkthrough.

How Do You Find the Inverse of a Matrix?

Two methods dominate, and which one you choose depends on the size.

Adjugate method — best for 2x2 and 3x3 by hand:

1. Compute det⁡A\det AdetA. If it is zero, stop — no inverse.
2. Build the cofactor matrix, signing each minor by (−1)i+j(-1)^{i+j}(−1)i+j.
3. Transpose it to get the adjugate.
4. Divide every entry by det⁡A\det AdetA.

Elementary row operations (Gauss-Jordan) — best for larger matrices: augment AAA with the identity to form [A∣I][A \mid I][A∣I], then row-reduce until the left block becomes III. Whatever the right block becomes is A−1A^{-1}A−1:

[,A∣I,];⟶;[,I∣A−1,].[,A \mid I,] ;\longrightarrow; [,I \mid A^{-1},].[,A∣I,];⟶;[,I∣A−1,].

If the left block can never reach III, the matrix is singular. Both methods give the same answer when an inverse exists — they are different routes to the same unique matrix.

What Are the Properties of the Inverse of a Matrix?

A short list of rules lets you rearrange inverse expressions without recomputing from scratch:

• The inverse is unique. A matrix has at most one inverse.
• Inverse of an inverse:(A−1)−1=A(A^{-1})^{-1} = A(A^{-1})^{-1}=A. Undoing the undo returns the original.
• Product rule (note the reversed order):(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B−1A^{-1}. The order flips — like taking off shoes then socks in reverse.
• Transpose rule:(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T(AT)^{-1}=(A^{-1})^T.
• Determinant of the inverse: det⁡(A−1)=1det⁡A\det(A^{-1}) = \dfrac{1}{\det A}det(A−1)=detA1​.
• Scalar multiple:(kA)−1=1kA−1(kA)^{-1} = \dfrac{1}{k}A^{-1}(kA)^{-1}=k1​A^{-1} for any non-zero scalar kkk.

The reversed order in the product rule is the one most worth remembering — it trips up nearly everyone the first time.

Examples of Inverse of a Matrix

The set runs from a clean 2x2, through the most common sign mistake, to a singular matrix that has no inverse, a system solved by inversion, the product rule in action, and a verification check.

Example 1

Find the inverse of A=[4726]A = \begin{bmatrix} 4 & 7 \ 2 & 6 \end{bmatrix}A=[4​72​6​].

Determinant: det⁡A=(4)(6)−(7)(2)=24−14=10\det A = (4)(6) - (7)(2) = 24 - 14 = 10detA=(4)(6)−(7)(2)=24−14=10. Apply the 2x2 shortcut — swap the diagonal, negate the off-diagonal, divide by 10:

A−1=110[6−7−24]=[0.6−0.7−0.20.4].A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \ -0.2 & 0.4 \end{bmatrix}.A−1=101​[6​−7−2​4​]=[0.6​−0.7−0.2​0.4​].

Final answer: the matrix above. Check: AA−1A A^{-1}AA−1 should return the identity, and it does.

Example 2

A common slip — find the inverse of A=[3512]A = \begin{bmatrix} 3 & 5 \ 1 & 2 \end{bmatrix}A=[3​51​2​].

Wrong attempt. A student computes det⁡A=(3)(2)−(5)(1)=1\det A = (3)(2) - (5)(1) = 1detA=(3)(2)−(5)(1)=1, then writes the inverse by swapping the diagonal but forgetting to negate the off-diagonal entries: [2513]\begin{bmatrix} 2 & 5 \ 1 & 3 \end{bmatrix}[2​51​3​]. It looks like the right shape.

Test it. Multiply by AAA: the off-diagonal terms come out non-zero, so the product is not the identity. The shortcut is not just "swap" — the two off-diagonal entries must change sign.

Correct. Swap the diagonal and negate the off-diagonal:

A−1=11[2−5−13].A^{-1} = \frac{1}{1}\begin{bmatrix} 2 & -5 \ -1 & 3 \end{bmatrix}.A−1=11​[2​−5−1​3​].

Final answer:[2−5−13]\begin{bmatrix} 2 & -5 \ -1 & 3 \end{bmatrix}[2​−5−1​3​]. The negation is the half of the shortcut that gets skipped most.

Example 3

Does A=[2436]A = \begin{bmatrix} 2 & 4 \ 3 & 6 \end{bmatrix}A=[2​43​6​] have an inverse?

Compute the determinant: det⁡A=(2)(6)−(4)(3)=12−12=0\det A = (2)(6) - (4)(3) = 12 - 12 = 0detA=(2)(6)−(4)(3)=12−12=0. A zero determinant means the matrix is singular.

Final answer: no inverse exists. The second row is 1.51.51.5 times the first — the rows are linearly dependent, which is what a zero determinant detects.

Example 4

Solve ;2x+y=5,;x+3y=10;;2x + y = 5,; x + 3y = 10;;2x+y=5,;x+3y=10; using the inverse.

Write as Ax=bA\mathbf{x} = \mathbf{b}Ax=b with A=[2113]A = \begin{bmatrix} 2 & 1 \ 1 & 3 \end{bmatrix}A=[2​11​3​] and b=(5,10)T\mathbf{b} = (5, 10)^Tb=(5,10)T. Determinant =6−1=5= 6 - 1 = 5=6−1=5, so

A−1=15[3−1−12],x=A−1b=15[15−10−5+20]=[13].A^{-1} = \frac{1}{5}\begin{bmatrix} 3 & -1 \ -1 & 2 \end{bmatrix}, \qquad \mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{5}\begin{bmatrix} 15 - 10 \ -5 + 20 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \end{bmatrix}.A^{-1}=51​[3​−1−1​2​ ext{and}]=A=7.

Final answer: x=1,;y=3x = 1,; y = 3x=1,;y=3. The inverse turns "solve the system" into a single multiplication.

Example 5

Verify the product rule (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B−1A^{-1} for A=[1021]A = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix}A=[1​02​1​], B=[1301]B = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}B=[1​30​1​].

Both are triangular with determinant 1, so A−1=[10−21]A^{-1} = \begin{bmatrix} 1 & 0 \ -2 & 1 \end{bmatrix}A^{-1}=[1​0−2​1​] and B−1=[1−301]B^{-1} = \begin{bmatrix} 1 & -3 \ 0 & 1 \end{bmatrix}B^{-1}=[1​−30​1​]. Then

B−1A−1=[1−301][10−21]=[7−3−21].B^{-1}A^{-1} = \begin{bmatrix} 1 & -3 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 \ -2 & 1 \end{bmatrix}.B−1A−1=[1​−30​1​][1​0−2​1​]=[7​−3−2​1​].

Computing (AB)−1(AB)^{-1}(AB)−1 directly gives the same matrix.

Final answer:(AB)−1=[7−3−21](AB)^{-1} = \begin{bmatrix} 7 & -3 \ -2 & 1 \end{bmatrix}(AB)−1=[7​−3−2​1​] — and crucially, A−1B−1A^{-1}B^{-1}A^{-1}B^{-1} in the wrong order does not match. The order matters.

Example 6

Confirm that [1−1−12]\begin{bmatrix} 1 & -1 \ -1 & 2 \end{bmatrix}[1​−1−1​2​] is the inverse of [2111]\begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}[2​11​1​].

The fastest verification is to multiply them:

[2111][1−1−12]=[2−1−2+21−1−1+2]=[1001]=I.\begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \end{bmatrix} = \begin{bmatrix} 2-1 & -2+2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I.

Final answer: yes — the product is the identity, so the two matrices are inverses. Verifying by multiplication is faster than recomputing the inverse from scratch.

Why the Inverse of a Matrix Earns Its Place

"If you cannot divide by a matrix, how do you ever solve for one?"

The matrix inverse arrived with Arthur Cayley (1821–1895, England), whose 1858 A Memoir on the Theory of Matrices defined matrix multiplication, the identity, and the inverse as a single algebraic system. That paper is the reason "solve AX=BAX = BAX=B" became a one-line operation rather than a fresh round of elimination.

Where the inverse does real work today:

• Solving linear systems. Any system Ax=bA\mathbf{x} = \mathbf{b}Ax=b collapses to x=A−1b\mathbf{x} = A^{-1}\mathbf{b}x=A−1b, and the same inverse reuses for every new b\mathbf{b}b — circuit analysis, economics input-output models, and structural engineering all run on this.
• Computer graphics. Undoing a transformation — moving a camera back, un-rotating an object — is applying the inverse matrix.
• Cryptography. The Hill cipher decrypts by multiplying by the inverse of the key matrix, which exists only when the key is invertible.
• Statistics and machine learning. Least-squares regression solves β^=(XTX)−1XTy\hat{\beta} = (X^T X)^{-1} X^T yβ^​=(XTX)−1XTy — the inverse sits at the heart of fitting a line to data.

Where Students Trip Up on the Inverse of a Matrix

Mistake 1: Trying to invert a non-square matrix

Where it slips in: Reaching for the inverse of a rectangular matrix, like a 2×32 \times 32×3.

Don't do this: Apply the formula to a matrix that is not square — there is no A−1A^{-1}A−1 to find.

The correct way: Only square matrices have inverses. A rectangular matrix can have a one-sided pseudo-inverse, but that is a different object entirely.

Mistake 2: Flipping the product-rule order

Where it slips in: Inverting a product ABABAB.

Don't do this: Write (AB)−1=A−1B−1(AB)^{-1} = A^{-1}B^{-1}(AB)−1=A^{-1}B^{-1}. The order is wrong.

The correct way:(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B^{-1}A^{-1} — the order reverses. The second-guesser who knows the rule still flips it back under time pressure; the "shoes-then-socks, reversed" image keeps it anchored.

Mistake 3: Forgetting to divide by the determinant

Where it slips in: Building the adjugate (or applying the 2x2 swap-and-negate) and stopping there.

Don't do this: Hand in the adjugate as the inverse. The adjugate is only the inverse up to a scale factor.

The correct way: Divide every entry by det⁡A\det AdetA. The rusher who skips this step gets an answer that is off by exactly the determinant — and fails the AA−1=IAA^{-1} = IAA−1=I check.

Key Takeaways

• The inverse of a matrix AAA is the unique A−1A^{-1}A−1 with AA−1=A−1A=IAA^{-1} = A^{-1}A = IAA−1=A−1A=I — the matrix version of a reciprocal.
• An inverse exists only when AAA is square and det⁡A≠0\det A \neq 0detA=0; otherwise the matrix is singular.
• The 2x2 shortcut: swap the diagonal, negate the off-diagonal, divide by ad−bcad - bcad−bc.
• The general method is A−1=1det⁡Aadj(A)A^{-1} = \frac{1}{\det A}\text{adj}(A)A^{-1}=detA1​adj(A), or row-reduce [A∣I][A \mid I] for larger matrices.
• The product rule reverses order: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B^{-1}A^{-1}.
• The inverse turns Ax=bA\mathbf{x} = \mathbf{b}Ax=b into x=A−1b\mathbf{x} = A^{-1}\mathbf{b}x=A−1b — the reason it powers linear systems, graphics, ciphers, and regression.