The inverse of a square matrix A is the matrix A⁻¹ that undoes it: A × A⁻¹ = I, the identity matrix. Only matrices with a non-zero determinant have one. For a 2×2 matrix there's a shortcut (swap the diagonal, negate the other two, divide by the determinant). For 3×3 you need a few more steps, but they're all mechanical.
Here's the matrix we'll invert:
[ 2 −1 0 ]
A = [ 1 3 2 ]
[ 0 1 4 ]
Step 1: the matrix of minors
The minor of an entry is the determinant of the 2×2 matrix left when you cover up that entry's row and column. There are nine of them.
For the top-left entry (row 1, column 1), cover row 1 and column 1, leaving [[3, 2], [1, 4]]. Its determinant is 3 × 4 − 2 × 1 = 10.
| Entry | Remaining 2×2 | Minor |
|---|---|---|
| row 1, col 1 | [[3, 2], [1, 4]] | 3·4 − 2·1 = 10 |
| row 1, col 2 | [[1, 2], [0, 4]] | 1·4 − 2·0 = 4 |
| row 1, col 3 | [[1, 3], [0, 1]] | 1·1 − 3·0 = 1 |
| row 2, col 1 | [[−1, 0], [1, 4]] | −1·4 − 0·1 = −4 |
| row 2, col 2 | [[2, 0], [0, 4]] | 2·4 − 0·0 = 8 |
| row 2, col 3 | [[2, −1], [0, 1]] | 2·1 − (−1)·0 = 2 |
| row 3, col 1 | [[−1, 0], [3, 2]] | −1·2 − 0·3 = −2 |
| row 3, col 2 | [[2, 0], [1, 2]] | 2·2 − 0·1 = 4 |
| row 3, col 3 | [[2, −1], [1, 3]] | 2·3 − (−1)·1 = 7 |
Step 2: apply the sign pattern to get cofactors
Multiply each minor by + or − according to this chequerboard. The top-left is always +.
[ + − + ] [ 10 −4 1 ]
[ − + − ] gives cofactors [ 4 8 −2 ]
[ + − + ] [ −2 −4 7 ]
Only the minors in the "−" positions change sign: 4 became −4, −4 became 4, 2 became −2, 4 became −4.
Step 3: the determinant
Multiply each entry of the first row of A by its cofactor and add:
det(A) = 2 × 10 + (−1) × (−4) + 0 × 1
= 20 + 4 + 0
= 24
It's not zero, so the inverse exists. You can expand along any row or column and get the same answer; pick one with zeros in it to save work.
Step 4: transpose to get the adjugate
Swap rows and columns of the cofactor matrix: the first row becomes the first column, and so on. The result is called the adjugate (or classical adjoint), adj(A).
[ 10 4 −2 ]
adj(A) = [ −4 8 −4 ]
[ 1 −2 7 ]
Forgetting this transpose is the most common mistake in the whole method. For a symmetric matrix it doesn't matter, which is why textbook examples sometimes hide it.
Step 5: divide by the determinant
[ 10 4 −2 ] [ 0.4167 0.1667 −0.0833 ]
A⁻¹ = 1/24 × [ −4 8 −4 ] = [ −0.1667 0.3333 −0.1667 ]
[ 1 −2 7 ] [ 0.0417 −0.0833 0.2917 ]
It's usually neater to leave the 1/24 outside, especially in exams.
The 30-second check
Multiply A by adj(A). If everything's right you'll get det(A) on the diagonal and zeros everywhere else, which is 24 × I here. You don't have to do all nine entries; two or three spot checks catch nearly every error:
Row 1 of A · column 1 of adj(A): 2(10) + (−1)(−4) + 0(1) = 24 ✓
Row 1 of A · column 2 of adj(A): 2(4) + (−1)(8) + 0(−2) = 0 ✓
Row 3 of A · column 3 of adj(A): 0(−2) + 1(−4) + 4(7) = 24 ✓
Enter the matrix in the calculator to see the same cofactors and adjugate.
Open with this matrixUsing the inverse: solving three equations
Matrices earn their keep with simultaneous equations. The system
2x − y = 1
x + 3y + 2z = 10
y + 4z = 13
is A × (x, y, z) = (1, 10, 13). Multiply both sides by A⁻¹:
[x] [ 10 4 −2 ] [ 1 ] [ 10 + 40 − 26 ] [ 24 ] [ 1 ]
[y] = 1/24 [ −4 8 −4 ] [ 10 ] = 1/24 [ −4 + 80 − 52 ] = 1/24 [ 24 ] = [ 1 ]
[z] [ 1 −2 7 ] [ 13 ] [ 1 − 20 + 91 ] [ 72 ] [ 3 ]
So x = 1, y = 1, z = 3. Substitute back to check: 2 − 1 = 1 ✓, 1 + 3 + 6 = 10 ✓, 1 + 12 = 13 ✓.
When the determinant is zero
If det(A) = 0 there's no inverse, and the matrix is called singular. It means the rows aren't independent: one row is a combination of the others. For example, in [[1, 2, 3], [4, 5, 6], [7, 8, 9]], row 3 = 2 × row 2 − row 1. The matching system of equations then has either no solution or infinitely many, and you'll need elimination to find out which.
Is there a faster way?
For larger matrices, yes: Gauss–Jordan elimination. Write [A | I] side by side, use row operations to turn the left half into I, and the right half becomes A⁻¹. It takes far fewer operations for 4×4 and up, and it's how computers do it. For 3×3 by hand, the cofactor method is easier to keep organised and much easier to check, so most courses teach it first.