How do you find the inverse of a 3x3 matrix?

Method: A-1 = adj(A) / det(A) Steps: 1. Calculate det(A) using cofactor expansion. If det = 0, stop — no inverse. 2. Calculate all 9 cofactors: Cij = (-1)^(i+j) * det(Mij) where Mij is the 2x2 matrix with row i and column j removed. 3. Form the cofactor matrix C (3x3 grid of all Cij values). 4. Transpose C to get adj(A): adj(A) = C-transpose 5. Divide every entry by det(A): A-1 = adj(A) / det(A) Verify by computing A x A-1 = I (identity matrix). Example: A = [[1,2,3],[0,1,4],[5,6,0]], det = 1 A-1 = [[-24,18,5],[20,-15,-4],[-5,4,1]]

What is the cofactor matrix and how is it formed?

The cofactor matrix C has the same dimensions as A (3x3). Each entry Cij = (-1)^(i+j) * Mij, where: - (i,j) is the row and column position (0-indexed: i,j in {0,1,2}) - (-1)^(i+j) is the sign: + if (i+j) is even, - if odd - Mij is the determinant of the 2x2 minor (delete row i and col j) Sign pattern: | + - + | | - + - | | + - + | The cofactor captures both the magnitude (from the 2x2 det) and the sign from the position. Cofactors are the building blocks of both the determinant and the inverse.

What is the adjugate (classical adjoint) of a matrix?

The adjugate adj(A) is the TRANSPOSE of the cofactor matrix: adj(A)[i,j] = C[j,i] (swap row and column indices) In other words: take the cofactor matrix and flip it along the main diagonal. Key property: A * adj(A) = det(A) * I This means adj(A) / det(A) = A-1 when det != 0. For 2x2: adj([[a,b],[c,d]]) = [[d,-b],[-c,a]] (a special case of the general formula). For 3x3, the adjugate has 9 entries, each a signed 2x2 determinant. Note: 'adjugate' and 'classical adjoint' mean the same thing. Do not confuse with the 'Hermitian adjoint' (conjugate transpose) used in quantum mechanics.

When does a 3x3 matrix have no inverse?

A 3x3 matrix has no inverse when det(A) = 0 (singular matrix). Geometric meaning: the matrix collapses 3D space to a 2D plane, 1D line, or point. Information is lost and the transformation cannot be reversed. Algebraic indicators of a singular 3x3 matrix: - One row is a linear combination of the other two - Two rows are identical - One row is all zeros - The columns are linearly dependent Example: [[1,2,3],[4,5,6],[7,8,9]] Row 3 = Row 1 + 2*(Row 2 - Row 1), so det = 0. For systems Ax = b with singular A: - Infinitely many solutions if b is in the column space of A - No solution otherwise

What is the relationship between the inverse and the determinant?

Several key relationships: 1. A-1 = adj(A) / det(A) The inverse is proportional to the adjugate; det scales it. 2. det(A-1) = 1 / det(A) If A scales volume by k, A-1 scales it by 1/k. 3. det(A * A-1) = det(I) = 1 Consistent with det(AB) = det(A) * det(B): det(A) * det(A-1) = 1. 4. (kA)-1 = (1/k) * A-1 for scalar k 5. (AB)-1 = B-1 * A-1 (reverse order) 6. (AT)-1 = (A-1)T Practical implication: a matrix with very small |det| is nearly singular. Inverting such matrices amplifies numerical errors — this is the condition number problem in numerical linear algebra.

3x3 Matrix Inverse Calculator — Step-by-Step with All 9 Cofactors

The 3x3 Matrix Inverse Calculator finds A-1 = adj(A)/det(A) with complete step-by-step working. Enter any 3x3 matrix and see all nine cofactors computed individually, the adjugate matrix formed by transposing them, and every entry of the inverse. The verification panel confirms A x A-1 = I by showin...

ENTER 3×3 MATRIX

Formula

A⁻¹ = adj(A) / det(A)

adj(A) = transpose of cofactor matrix C

[

]

QUICK EXAMPLES

DETERMINANT

A⁻¹ exists · multiply adj(A) by 1/1

STEPS

9 COFACTORS Cᵢⱼ = (−1)^(i+j) × det(Mᵢⱼ)

C[1,1] +|M11|

-24

C[1,2] -|M12|

C[1,3] +|M13|

-5

C[2,1] -|M21|

C[2,2] +|M22|

-15

C[2,3] -|M23|

C[3,1] +|M31|

C[3,2] -|M32|

-4

C[3,3] +|M33|

ADJUGATE adj(A) = Cᵀ

[

-24

18

5

20

-15

-4

-5

4

1

]

÷ 1

A⁻¹ (Inverse Matrix)

[

-24

18

5

20

-15

-4

-5

4

1

]

VERIFICATION A × A⁻¹ = I

[

]

[

-24

18

5

20

-15

-4

-5

4

1

]

[

]

✓

Created with❤️byeaglecalculator.com

HOW TO USE

1
Enter the nine values of your 3x3 matrix row by row. The inverse is computed automatically using the cofactor-adjugate method: A-1 = adj(A) / det(A).
2
The right panel shows all nine cofactors Cij = (-1)^(i+j) * Mij, where Mij is the determinant of the 2x2 matrix formed by deleting row i and column j. Green = positive, red = negative.
3
The adjugate adj(A) is the transpose of the cofactor matrix — rows and columns of the cofactors are swapped. The adjugate divided by det(A) gives the inverse.
4
The inverse matrix A-1 is displayed on a dark background. Entries may be shown as fractions (e.g. 1/2) for cleaner reading.
5
The verification panel confirms A x A-1 = I by computing the product explicitly. All nine entries of the product are shown — they should be 1 on the diagonal and 0 off it. If det = 0, a clear message explains why no inverse exists.

WORKED EXAMPLE

A = [[1,2,3],[0,1,4],[5,6,0]], det = 1. Cofactors: C[0,0] = +det([[1,4],[6,0]]) = +(0-24) = -24 C[0,1] = -det([[0,4],[5,0]]) = -(0-20) = +20 C[0,2] = +det([[0,1],[5,6]]) = +(0-5) = -5 C[1,0] = -det([[2,3],[6,0]]) = -(0-18) = +18 C[1,1] = +det([[1,3],[5,0]]) = +(0-15) = -15 C[1,2] = -det([[1,2],[5,6]]) = -(6-10) = +4 C[2,0] = +det([[2,3],[1,4]]) = +(8-3) = +5 C[2,1] = -det([[1,3],[0,4]]) = -(4-0) = -4 C[2,2] = +det([[1,2],[0,1]]) = +(1-0) = +1 adj(A) = transpose of cofactor matrix = [[-24,18,5],[20,-15,-4],[-5,4,1]]. A-1 = adj(A)/det = [[-24,18,5],[20,-15,-4],[-5,4,1]] / 1 = same matrix. Verify: A x A-1 = I ✓.

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Last updated: April 29, 2026 · Formula verified by EagleCalculator team · Eagle-eyed accuracy for every calculation.