What is the inverse of a matrix?

The inverse of a square matrix A is the matrix A⁻¹ such that A times A⁻¹ and A⁻¹ times A both equal the identity matrix. It undoes the transformation A performs. An inverse exists exactly when the determinant of A is not zero, and when it exists it is unique.

How do you find the inverse of a matrix using cofactors?

Compute the cofactor of every entry: strike its row and column, take the determinant of what remains, and attach the sign (−1) to the power i + j. Arrange the cofactors in a matrix and transpose it to get the adjugate. Compute the determinant of A, for example by multiplying the first row by its cofactors and adding. Then divide every entry of the adjugate by the determinant. The result is the inverse.

What is the formula for the inverse of a 2×2 matrix?

For a matrix with rows (a, b) and (c, d), the inverse is 1 over (ad − bc) times the matrix with rows (d, −b) and (−c, a). Swap the two diagonal entries, negate the two off-diagonal entries, and divide by the determinant ad − bc. If ad − bc is zero the matrix has no inverse.

Why does a matrix with zero determinant have no inverse?

The adjugate formula divides by the determinant, and division by zero is undefined. More fundamentally, a zero determinant means the matrix collapses space onto a lower-dimensional set, sending different inputs to the same output, and such a collapse cannot be reversed. The adjugate still exists in that case, but A times its adjugate equals the zero matrix rather than the identity.

Why is the cofactor matrix transposed to get the inverse?

For A times the adjugate to equal the determinant times the identity, row i of A must meet the cofactors of row i, and in a matrix product row i of the left factor meets column i of the right factor. So the cofactors of row i must sit in column i of the adjugate, which is what transposing the cofactor matrix achieves. Without the transpose the product would not be a multiple of the identity unless A were symmetric.

Matrix Inverse

Symbolic visualization of A⁻¹ = adj(A) / det A — cofactors, transpose, determinant, divide.

Size of A (square)?The inverse of a square matrix A is the matrix A⁻¹ with A·A⁻¹ = A⁻¹·A = I. It exists exactly when det A ≠ 0. The adjugate formula builds it in four moves: compute every cofactor (a signed minor), transpose the cofactor matrix to get the adjugate, compute det A by expanding along the first row, then divide every entry of the adjugate by det A. Row reduction of [A | I] is the other standard route and scales better; this one shows the structure.

A3×32×2 or 3×3 — the cofactors are written out in full

A3×3

a1,1

a1,2

a1,3

a2,1

a2,2

a2,3

a3,1

a3,2

a3,3

→

C3×3

Step 1 / 22

Step explanations

1Inverse of A by the adjugate formula

A is 3×3. Its inverse, when it exists, is A⁻¹ = adj(A) / det A. The run computes every cofactor of A (a signed minor), transposes the cofactor matrix to get the adjugate, computes det A from the first-row cofactors, and finally divides every entry of the adjugate by det A. The last step is the one that can fail: if det A = 0 there is no inverse.

Nine small determinants, a transpose, one big determinant, a division. Learn more about the opening scene · what it is

Key Terms

Getting Started with the Visualizer

The Four Phases

The Opening Scene: A Matrix and an Empty Cofactor Grid

Phase 1: Cofactors, One Entry at a Time

Phase 2: Transposing into the Adjugate

Phase 3: The Determinant from the First Row

Phase 4: Dividing by the Determinant

Reading the Scene Player

Key Terms

Inverse — the matrix

A^{-1}

with

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

. It exists exactly when

\det A \neq 0

, and then it is unique.

Minor —

M_{i,j}

, the determinant of the submatrix left after striking row

i

and column

j

A

.

Cofactor —

C_{i,j} = (-1)^{i+j} M_{i,j}

, the minor with the checkerboard sign attached.

Cofactor matrix — the matrix

C

whose

(i,j)

entry is

C_{i,j}

.

Adjugate —

\operatorname{adj} A = C^T

, the transpose of the cofactor matrix; also called the classical adjoint.

Adjugate formula —

A^{-1} = \dfrac{\operatorname{adj} A}{\det A}

.

Singular matrix — a square matrix with

\det A = 0

; it has no inverse.

Row reduction — the other standard route: reduce

[A \mid I]

[I \mid A^{-1}]

by row operations. It scales to any size; the adjugate formula shows the structure.

Getting Started with the Visualizer

Set the size of

A

, then watch

A^{-1}

build in four phases.

• Use the Size stepper to choose

2 \times 2

3 \times 3

; the cofactors are written out in full, which is why larger sizes are not offered
• Hover the ? icon for a reminder of what the inverse is and how the adjugate formula relates to row reduction
• Press play or step manually through the scene player; the speed selector and step log let you control pace and review
• The run computes every cofactor, transposes them into the adjugate, expands the determinant along the first row, and divides — the same four moves at either size
• Everything is symbolic: the entries of

A

stay as

a_{i,j}

, so what you see is the formula, not a numerical example

The Four Phases

The visualizer separates the adjugate formula into four phases.

• Phase 1 — cofactors: one scene per entry in row-major order. Row

i

and column

j

A

are struck, the minor

M_{i,j}

is read from what remains, and the cofactor

C_{i,j} = (-1)^{i+j} M_{i,j}

is written into the cofactor matrix
• Phase 2 — transpose: the cofactor matrix is transposed in one scene, producing

\operatorname{adj} A

• Phase 3 — determinant:

\det A = a_{1,1} C_{1,1} + a_{1,2} C_{1,2} + \cdots

, the expansion along the first row using the cofactors already in hand, in one scene
• Phase 4 — divide: one scene per entry;

(A^{-1})_{i,j} = (\operatorname{adj} A)_{i,j} / \det A

3 \times 3

that is

9 + 1 + 1 + 9

steps plus the intro and outro; at

2 \times 2

4 + 1 + 1 + 4

. Phase 3 is where the construction can fail: if

\det A = 0

there is nothing to divide by and

A

has no inverse.

The Opening Scene: A Matrix and an Empty Cofactor Grid

The player opens with

A

on the left and an empty grid of the same shape on the right, waiting to hold the cofactors. At the default size

A

3 \times 3

.

Nothing is computed yet. What the scene establishes is the plan: the inverse will not be attacked directly, but assembled from nine smaller determinants, one per entry of

A

Opening scene, frozen

A beside an empty grid of the same shape, waiting for its nine cofactors. The inverse is assembled from smaller determinants, one per entry, not attacked directly.

The shape is worth a moment. The inverse of an

n \times n

matrix is

n \times n

, and so are the cofactor matrix and the adjugate that lead to it. Nothing changes size along the way; only a single number,

\det A

, is extracted.

Only square matrices have inverses. A rectangular

A

cannot satisfy

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

, because the two products would have different shapes.

Phase 1: Cofactors, One Entry at a Time

The first sweep visits every entry of

A

in row-major order. For each, the entry's own row and column are struck, the determinant of what remains is the minor, and the sign

(-1)^{i+j}

turns the minor into the cofactor.

The frozen picture below is the centre entry at

3 \times 3

: row

2

and column

2

struck, the four corner entries forming the minor, and

C_{2,2} = a_{1,1} a_{3,3} - a_{1,3} a_{3,1}

being written with four cofactors already in place.

Phase 1, the centre cofactor

Row 2 and column 2 struck through, the four corners forming the minor, and C_2,2 being written with four cofactors already in place. The sign is + here: the centre of the checkerboard.

The sign follows a checkerboard:

+

at the corners and centre,

-

at the four edge-centres. The tool folds the sign into the written cofactor, so a negated minor appears with its two terms swapped rather than with a leading minus; the caption states which happened.

This phase is nine

2 \times 2

determinants, and that is the whole cost of the method at this size. It is also why the method does not scale: at

4 \times 4

it would be sixteen

3 \times 3

determinants.

Phase 2: Transposing into the Adjugate

The second phase is a single scene: the cofactor matrix is transposed, so the cofactor of

a_{i,j}

moves to row

j

, column

i

. The result is the adjugate.

The frozen picture below shows the cofactor matrix on the left and the adjugate on the right, every entry in its mirrored position.

Phase 2, the transpose

Every cofactor moving to its mirrored position. The diagonal stays, the off-diagonal entries swap - this is the step that makes A · adj A come out as a multiple of I.

This is the step most often skipped, and skipping it is fatal unless

A

happens to be symmetric. The reason for the transpose is in the product

A \cdot \operatorname{adj} A

: for the

(i,i)

entry to be the cofactor expansion of

\det A

along row

i

, row

i

A

must meet column

i

of the adjugate, and that column has to hold the cofactors of row

i

. Only the transpose puts them there.

The diagonal survives the transpose unchanged, which is why the

2 \times 2

shortcut swaps the diagonal entries of

A

but negates the off-diagonal ones in place.

Phase 3: The Determinant from the First Row

The third phase is also a single scene. The determinant is expanded along the first row: each entry

a_{1,j}

is multiplied by its cofactor

C_{1,j}

, already computed in phase 1, and the products are added.

The frozen picture below shows the first row of

A

against the first row of the cofactor matrix, with the

\det A

slot filled.

Phase 3, the determinant

The first row of A against the first row of the cofactor matrix, det A filled. One row of products, reusing work already done - and the number that decides whether an inverse exists.

Reusing the cofactors is the efficiency of the method: the determinant costs one row of multiplications rather than a fresh calculation. Any row or column would do, since expanding along any of them gives the same number.

This is the decision point. If

\det A = 0

the run should stop: the adjugate still exists, but dividing by zero is impossible and

A

has no inverse. Geometrically, a zero determinant means

A

flattens space, and a flattening cannot be undone.

Phase 4: Dividing by the Determinant

The last sweep divides every entry of the adjugate by

\det A

, one entry per scene, filling

A^{-1}

.

The frozen picture below is the centre entry at

3 \times 3

C_{2,2} / |A|

being written, with four entries of the inverse already in place.

Phase 4, mid-sweep

The centre entry of A⁻¹ being formed from the adjugate and det A, four entries already in place. One common divisor for every entry.

Every entry is divided by the same number, so the inverse is the adjugate rescaled. At

2 \times 2

the entries are short enough to write out; at

3 \times 3

the tool names them

C_{j,i} / |A|

rather than expanding each cofactor a second time.

The completed matrix satisfies

A A^{-1} = I

. Row

i

A

against column

i

A^{-1}

is the expansion of

\det A

along row

i

, divided by

\det A

, which is

1

. Row

i

against column

j

for

i \neq j

is the expansion of a determinant with two equal rows, which is

0

. The whole construction, and the single number that governs it, is on the screen at once.

Reading the Scene Player

Each scene combines highlights, arrows, and a caption.

• In phase 1, the pivot entry

a_{i,j}

is highlighted primary, the rest of its row and column are greyed and lined through, the surviving block is secondary, and the destination cofactor is accent; one arrow runs from the pivot to its cofactor
• In phase 2, every cofactor is secondary and every entry of the adjugate accent, with the caption describing the mirror swap
• In phase 3, the first row of

A

is primary, the first row of the cofactor matrix secondary, and the

\det A

slot accent
• In phase 4, the active adjugate entry is primary, the

\det A

slot secondary, and the destination entry of

A^{-1}

accent, with arrows from both
• Cofactor cells show the full signed minor, for example

a_{2,2} a_{3,3} - a_{2,3} a_{3,2}

; at

3 \times 3

the entries of

A^{-1}

are written as

C_{j,i} / |A|

to stay readable, and at

2 \times 2

they are written out
• The step log on the right keeps a record of every completed step across all phases

Choosing the Size

The stepper offers

2 \times 2

and

3 \times 3

, and the limit is deliberate.

• At

2 \times 2

every minor is a single entry, so the whole formula is visible at once: swap the diagonal, negate the off-diagonal, divide by

a_{1,1} a_{2,2} - a_{1,2} a_{2,1}

• At

3 \times 3

every minor is a

2 \times 2

determinant, and the nine cofactors are the real work; this is the size at which the checkerboard of signs starts to matter
• At

4 \times 4

each cofactor would be a

3 \times 3

determinant with six terms, and sixteen of them; the expressions no longer fit in a cell, and the method itself becomes impractical by hand
• For larger matrices, row reduction of

[A \mid I]

is the method of choice; the Gaussian elimination tool animates that process on numerical matrices

What the Inverse Is

The inverse of a square matrix

A

is the matrix

A^{-1}

satisfying

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

It exists if and only if

\det A \neq 0

, and when it exists it is unique. For a

2 \times 2

matrix the formula is short enough to memorize:

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

The general version of that formula is the adjugate formula,

A^{-1} = \frac{1}{\det A}\,\operatorname{adj} A, \qquad (\operatorname{adj} A)_{i,j} = C_{j,i} = (-1)^{i+j} M_{j,i}

where

M_{j,i}

is the minor obtained by deleting row

j

and column

i

. The

2 \times 2

case is exactly this: the cofactor of

a

d

, of

b

-c

, of

c

-b

, of

d

a

, and transposing puts

-b

and

-c

where the formula shows them.

Why it works: the

(i,i)

entry of

A \cdot \operatorname{adj} A

\sum_k a_{i,k} C_{i,k}

, the cofactor expansion of

\det A

along row

i

. The

(i,j)

entry for

i \neq j

\sum_k a_{i,k} C_{j,k}

, which is the expansion of a matrix whose rows

i

and

j

are equal, and that determinant is zero. So

A \cdot \operatorname{adj} A = (\det A)\,I

.

For the full treatment, including properties of inverses and the row reduction method, see the matrix inverse theory page; for minors and cofactors in depth, see the cofactors theory page.

Key Properties

The inverse interacts cleanly with the other matrix operations, with one famous reversal.

• Existence:

A^{-1}

exists exactly when

\det A \neq 0

, equivalently when the columns of

A

are linearly independent, equivalently when

A\mathbf{x} = \mathbf{0}

has only the trivial solution
• Uniqueness: a matrix has at most one inverse
• Involution:

(A^{-1})^{-1} = A

• Product (order reverses):

(AB)^{-1} = B^{-1} A^{-1}

• Transpose:

(A^T)^{-1} = (A^{-1})^T

• Scalar:

(kA)^{-1} = \dfrac{1}{k} A^{-1}

for

k \neq 0

• Determinant:

\det(A^{-1}) = \dfrac{1}{\det A}

• Adjugate identity:

A \cdot \operatorname{adj} A = \operatorname{adj} A \cdot A = (\det A)\,I

, which holds even when

\det A = 0

• Solving systems: if

A

is invertible,

A\mathbf{x} = \mathbf{b}

has the unique solution

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

Why It Matters

The inverse is the matrix that undoes a linear transformation, and that idea appears everywhere.

• Solving linear systems:

A\mathbf{x} = \mathbf{b}

becomes

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

; in practice the system is solved directly, but the inverse is the concept behind it
• Change of basis: converting coordinates back requires the inverse of the change-of-basis matrix
• Undoing transformations: in graphics and robotics, the inverse of a rotation, scaling or shear returns an object to where it started
• Diagonalization:

A = P D P^{-1}

needs

P^{-1}

to move between the standard basis and the eigenbasis
• Cramer's rule: the adjugate formula, applied to

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

, gives each unknown as a ratio of determinants
• Statistics and least squares: the normal equations involve

(A^T A)^{-1}

, and covariance matrices are inverted to form precision matrices
• Invertibility as a test: whether

\det A

is zero decides whether a system has a unique solution, whether a transformation is reversible, and whether a set of vectors is a basis

Worked Example

2 \times 2

case first:

A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}, \quad \det A = 2 \cdot 3 - 1 \cdot 5 = 1

Cofactors:

C_{1,1} = 3

C_{1,2} = -5

C_{2,1} = -1

C_{2,2} = 2

. Transpose to get the adjugate and divide by

1

A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Now a

3 \times 3

case:

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}

Cofactors, row by row:

C_{1,1} = 0 - 24 = -24

C_{1,2} = -(0 - 20) = 20

C_{1,3} = 0 - 5 = -5

;

C_{2,1} = -(0 - 18) = 18

C_{2,2} = 0 - 15 = -15

C_{2,3} = -(6 - 10) = 4

;

C_{3,1} = 8 - 3 = 5

C_{3,2} = -(4 - 0) = -4

C_{3,3} = 1 - 0 = 1

.

Determinant along the first row:

\det A = 1 \cdot (-24) + 2 \cdot 20 + 3 \cdot (-5) = 1

.

Transpose the cofactor matrix and divide by

1

A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

Check the first row of

A A^{-1}

(1, 2, 3)

against the columns gives

-24 + 40 - 15 = 1

18 - 30 + 12 = 0

5 - 8 + 3 = 0

. Set the visualizer to

3 \times 3

and step through to see the same nine cofactors, the transpose, and the division assembled symbolically.

Common Mistakes

A few mistakes recur.

• Forgetting to transpose — the adjugate is the transpose of the cofactor matrix; using

C

itself gives a wrong answer whenever

A

is not symmetric
• Losing the checkerboard signs —

C_{i,j}

carries

(-1)^{i+j}

; at

3 \times 3

the four edge-centre cofactors are negated and the corners and centre are not
• Dividing before checking the determinant — if

\det A = 0

the adjugate still exists but the inverse does not; the adjugate identity gives

A \cdot \operatorname{adj} A = 0

• Mixing up the $2 \times 2$ shortcut — swap the diagonal entries, negate the off-diagonal ones, then divide; swapping the off-diagonal entries instead is the classic slip
• Reversing the order in a product —

(AB)^{-1} = B^{-1} A^{-1}

, not

A^{-1} B^{-1}

• Assuming every square matrix is invertible — singular matrices are common, and any matrix with a zero row, two equal rows, or proportional columns is one
• Expecting integer inverses — the entries of

A^{-1}

are cofactors divided by

\det A

, so fractions are the rule unless

\det A = \pm 1