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Matrix inverse of a 3×3: the adjugate-and-determinant method

Step-by-step inverse of a 3×3 matrix using cofactors — what the formula does, why each step is necessary, and how to spot when a matrix has no inverse.

211132100Adet A = −1⟶adjugate / det001−2133−1−5A⁻¹

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What this shows

For an n×n square matrix A, the inverse A⁻¹ — when it exists — is the matrix that satisfies A · A⁻¹ = A⁻¹ · A = I, the identity matrix. For a 3×3 matrix the closed-form recipe is:

    A⁻¹ = (1 / det A) · adj(A)

where det A is the determinant (a single scalar) and adj(A), the adjugate, is the transposed matrix of cofactors. The cofactor at row i, column j is (−1)^(i+j) times the determinant of the 2×2 submatrix you get by deleting row i and column j from A.

For a worked example, take

    A = | 2  1  1 |
        | 1  3  2 |
        | 1  0  0 |

Step 1 — determinant. Expand along the third row (it has two zeros, which makes the arithmetic easy):

    det A = 1·det| 1 1 | − 0·… + 0·…
                 | 3 2 |
          = 1·(1·2 − 1·3) = 1·(−1) = −1

Step 2 — cofactor matrix. Compute the nine 2×2 sub-determinants with the right ± sign pattern:

    | + −  + |       | 0  2   −3 |
    | − +  − |  →    | 0 −1    1 |
    | + −  + |       | −1 −3   5 |

Step 3 — transpose to get the adjugate:

    adj(A) = |  0   0  −1 |
             |  2  −1  −3 |
             | −3   1   5 |

Step 4 — divide by the determinant:

    A⁻¹ = (1/−1) · adj(A) = | 0   0   1 |
                            | −2  1   3 |
                            | 3  −1  −5 |

Multiply A·A⁻¹ to confirm you get the identity (you do).

The matrix has an inverse only when det A ≠ 0. When det A = 0 the matrix is called singular and the equation Ax = b either has infinitely many solutions or none — never exactly one.

Where it shows up

Matrix inverses are how systems of linear equations are solved when written compactly: Ax = b has the closed-form solution x = A⁻¹b, provided A is invertible. Computer graphics uses inverses to project from world coordinates to camera coordinates and back. Statistics uses them to invert covariance matrices in regression.

In practice, libraries almost never compute A⁻¹ explicitly — they solve Ax = b directly with LU decomposition or Gaussian elimination, which is faster and numerically more stable. The adjugate formula is the right thing to teach, because it shows where the answer comes from, but for large matrices (n ≥ 4) it's wildly inefficient and only used in theory.

Frequently asked questions

What if the determinant is zero?

Then no inverse exists. Geometrically, the matrix collapses some direction in the input space down to a single point, and there's no way to undo that collapse. Algebraically, A·A⁻¹ = I would imply det A · det A⁻¹ = 1, which is impossible if det A = 0.

Why the (−1)^(i+j) signs?

They come from the cofactor expansion of the determinant. The full det A formula is Σⱼ aᵢⱼ · (−1)^(i+j) · Mᵢⱼ where Mᵢⱼ is the minor (the 2×2 sub-determinant after deleting row i, column j). The same signs reappear in the cofactor matrix because the adjugate is built from those very cofactors.

Is there a faster way?

For 3×3, Gauss-Jordan elimination on the augmented matrix [A | I] is usually faster by hand because it skips the 2×2 sub-determinant arithmetic. For computer use, LU decomposition is the standard tool — O(n³) instead of the adjugate's effective O(n!).

Why is the inverse unique when it exists?

If both B and C were inverses of A, then B = B·I = B·(A·C) = (B·A)·C = I·C = C. So any two candidates must be equal — there's at most one inverse, and the adjugate formula either produces it (when det A ≠ 0) or fails to (when det A = 0).

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