How do I find an inverse matrix?

How it works

The inverse of a square matrix A is the matrix A⁻¹ with A × A⁻¹ = A⁻¹ × A = I, the identity matrix. The calculator finds it by Gauss-Jordan elimination:

1. Write the n × 2n matrix [A | I]: A with the identity matrix to its right.
2. For each column from the left: take the first row (from the top of the rows not yet used) with a non-zero entry in that column as the pivot row, and swap it up if needed.
3. Divide the pivot row by its pivot, so the pivot is 1.
4. Subtract multiples of the pivot row from every other row, so the rest of the column is 0.
5. When the left half is I, the right half is A⁻¹. If a column of the left half has no non-zero entry to pivot on, the determinant is 0 and A has no inverse.

The determinant comes from the same steps. Each swap multiplies it by −1, each scaling by 1 ÷ pivot multiplies it by 1 ÷ pivot, and adding a multiple of a row leaves it unchanged. The steps end at I, whose determinant is 1, so det(A) is the product of the pivots, times −1 for each swap.

Assumptions

• Every entry is read exactly as typed, as a fraction: 0.1 is 1/10. All the arithmetic is exact, so the inverse and the determinant have no rounding.
• The matrix is square, from 1 × 1 to 6 × 6.
• The decimal version rounds each exact entry to 10 significant figures, as JavaScript's toPrecision(10) does: to the nearest, and an exact tie in the entry's double value rounds away from zero. It is then written as the shortest JavaScript number, so trailing zeros go, and an entry below 10⁻⁶ in size (or 10²¹ and above) is written in e notation, such as 5.77962679e-9 or 1e-9.
• The working lists at most the first 30 row operations, separated by "; ", then "and N more" when there are more. Rows count from 1, and fractions are in lowest terms with a plain hyphen: a swap is "R1 ↔ R2", scaling is "R1 → 1/3 × R1" (or "R3 → -1 × R3"), and adding a multiple is "R3 → R3 − 2 × R1" or "R3 → R3 + 2/5 × R1" (the true minus sign between terms, and no "1 ×" for a multiple of 1: "R3 → R3 − R1"). A matrix already equal to I gives an empty working.
• Matrices are written row by row in square brackets, rows separated by "; " and entries by ", ": [3/5, -7/10; -1/5, 2/5]. Fractions are in lowest terms with a plain hyphen for negatives, and the determinant is written the same way (-1, 1/8).

Worked examples by hand

A = [1, 2, 3; 2, 5, 3; 1, 0, 8]. Start from [A | I].

• Column 1, pivot 1: R2 → R2 − 2 × R1 gives [0, 1, −3 | −2, 1, 0]; R3 → R3 − R1 gives [0, −2, 5 | −1, 0, 1].
• Column 2, pivot 1: R1 → R1 − 2 × R2 gives [1, 0, 9 | 5, −2, 0]; R3 → R3 + 2 × R2 gives [0, 0, −1 | −5, 2, 1].
• Column 3, pivot −1: R3 → −1 × R3 gives [0, 0, 1 | 5, −2, −1]; R1 → R1 − 9 × R3 gives [1, 0, 0 | −40, 16, 9]; R2 → R2 + 3 × R3 gives [0, 1, 0 | 13, −5, −3].

So A⁻¹ = [−40, 16, 9; 13, −5, −3; 5, −2, −1]. There were no swaps, and the pivots were 1, 1, and −1, so det(A) = −1.

A = [4, 7; 2, 6]. ad − bc = 4 × 6 − 7 × 2 = 10, so A⁻¹ = 1/10 × [6, −7; −2, 4] = [3/5, −7/10; −1/5, 2/5], or [0.6, −0.7; −0.2, 0.4].

A = [0, 1; 1, 0]. Column 1 has 0 at the top, so swap R1 and R2. That already gives I, so the matrix is its own inverse, and the one swap makes det(A) = −1.

A = [4]. The inverse of a 1 × 1 matrix [a] is [1/a], so A⁻¹ = [1/4] and det(A) = 4.

A = [0.5, 0; 0.1, 0.25]. Read exactly, this is [1/2, 0; 1/10, 1/4]. It is lower triangular, so det(A) = 1/2 × 1/4 = 1/8. The 2 × 2 formula gives A⁻¹ = 8 × [1/4, 0; −1/10, 1/2] = [2, 0; −4/5, 4].