Matrix Inverse via the Adjugate Method

April 13, 2026

Problem

Find the inverse of A = [[2,1],[5,3]] using A⁻¹ = (1/det A) · adj(A). Verify A · A⁻¹ = I.

Explanation

What is a matrix inverse?

A square matrix AA is invertible if there exists a matrix A−1A^{-1} with AA−1=A−1A=IA A^{-1} = A^{-1} A = I

Not every square matrix has an inverse. AA is invertible if and only if det⁡A≠0\det A \ne 0. A matrix without an inverse is called singular.

The 2×2 formula

For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} with det⁡A=ad−bc≠0\det A = ad - bc \ne 0: A−1=1det⁡A(d−b−ca)A^{-1} = \dfrac{1}{\det A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

Swap the diagonal, negate the off-diagonal, divide by the determinant.

Step-by-step

A=(2153)A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}

Step 1 — Compute det⁡A\det A: det⁡A=(2)(3)−(1)(5)=6−5=1\det A = (2)(3) - (1)(5) = 6 - 5 = 1

Step 2 — Build the adjugate (for 2×22 \times 2): swap diagonal, negate off-diagonal: adj(A)=(3−1−52)\mathrm{adj}(A) = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Step 3 — Divide by determinant: A−1=11(3−1−52)=(3−1−52)A^{-1} = \dfrac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Verification

AA−1=(2153)(3−1−52)A A^{-1} = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

  • Row 1, col 1: 2(3)+1(−5)=6−5=12(3) + 1(-5) = 6 - 5 = 1 ✓
  • Row 1, col 2: 2(−1)+1(2)=−2+2=02(-1) + 1(2) = -2 + 2 = 0 ✓
  • Row 2, col 1: 5(3)+3(−5)=15−15=05(3) + 3(-5) = 15 - 15 = 0 ✓
  • Row 2, col 2: 5(−1)+3(2)=−5+6=15(-1) + 3(2) = -5 + 6 = 1 ✓

AA−1=(1001)=IA A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I ✓

Generalization: adjugate method for any nn

For a general n×nn \times n invertible AA: A−1=1det⁡A adj(A)A^{-1} = \dfrac{1}{\det A} \, \mathrm{adj}(A)

where adj(A)\mathrm{adj}(A) is the adjugate (or classical adjoint) — the transpose of the cofactor matrix. The (i,j)(i, j) entry of the cofactor matrix is (−1)i+jMij(-1)^{i+j} M_{ij} where MijM_{ij} is the (n−1)×(n−1)(n-1) \times (n-1) determinant obtained by deleting row ii and column jj.

For n≥3n \ge 3, this is often slower than using Gauss–Jordan elimination on [A∣I][A \mid I], but the adjugate formula gives a clean closed-form.

Why the inverse matters

  • Solving Ax=bA \mathbf{x} = \mathbf{b}: if AA is invertible, x=A−1b\mathbf{x} = A^{-1} \mathbf{b}.
  • Change of basis: converting coordinates between different bases uses inverses.
  • Undoing linear transformations: rotation, scaling, shear — each has an inverse that undoes it.

Common mistakes

  • Forgetting to check det⁡A≠0\det A \ne 0. If the determinant is zero, A−1A^{-1} does not exist.
  • Not negating the off-diagonal. It's (−b,−c)(-b, -c), not (b,c)(b, c).
  • Dropping the 1/det⁡A1/\det A factor. Without it you compute adj(A)\mathrm{adj}(A), not A−1A^{-1}.
  • Trying to invert a non-square matrix. Only square matrices can have a (two-sided) inverse; non-square matrices have pseudo-inverses instead.

Try it in the visualization

Modify entries of AA with sliders; the determinant and A−1A^{-1} update live. When det⁡A=0\det A = 0, the inverse panel turns red and shows "singular — no inverse exists." The product AA−1A A^{-1} also updates to confirm II.

Interactive Visualization

Parameters

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1.00
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Your turn

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Matrix Inverse via the Adjugate Method | MathSpin