How this instrument works
The Moore-Penrose pseudoinverse is a generalization of the matrix inverse, built to exist for matrices that an ordinary inverse can't handle — ones that aren't square, or square ones with a determinant of zero. For any matrix that IS square and invertible, though, the generalization collapses back to something familiar: the pseudoinverse equals the plain inverse, entry for entry, with no separate formula needed. This sheet works with 2×2 matrices [[a,b],[c,d]] that satisfy exactly that condition, so what it reports is the everyday matrix inverse computed by way of a name usually reserved for harder cases.
The arithmetic itself runs off one number, the determinant: det = ad − bc. From there, every output entry is det dividing into a rearranged copy of the original matrix — invA = d ⁄ det, invB = −b ⁄ det, invC = −c ⁄ det, and invD = a ⁄ det. Notice the diagonal entries a and d swap places while the off-diagonal entries b and c both pick up a negative sign; that rearrangement, divided through by det, is the entire recipe.
It would be misleading to oversell this page as something dramatically apart from a plain 2×2 inverse calculator, so here's the honest limitation: on square, invertible input, the two ideas produce numerically identical output every time, and this instrument cannot demonstrate otherwise. The real payoff of the Moore-Penrose concept shows up only outside that comfortable case — for a 3×4 matrix, say, or a square one whose determinant is exactly zero — situations a 2×2, invertible-only calculator like this one is not built to reach. Think of this page as the on-ramp: the formula you'd already trust from a matrix-inverse sheet, relabeled to match the broader idea it happens to agree with here.
Two boundary cases are worth holding in mind while entering numbers. The identity matrix, [[1,0],[0,1]], maps to itself under this operation, since its determinant is 1 and its off-diagonal entries are already zero. And because det appears in every denominator, a matrix with ad exactly equal to bc has no output at all — division by zero the field can't paper over, singular or not.
- Enter the top row of the matrix into a (row 1, col 1) and b (row 1, col 2).
- Enter the bottom row into c (row 2, col 1) and d (row 2, col 2), keeping d away from a value that makes the determinant zero.
- Read Inverse: row 1, col 1 and Inverse: row 1, col 2 for the top row of the result.
- Read Inverse: row 2, col 1 and Inverse: row 2, col 2 for the bottom row of the result.
- Multiply your original four entries by the four reported back, matrix-style, to confirm the identity matrix falls out.
Worked example — inverting [[2, 1], [3, 4]]
Take a = 2, b = 1, c = 3, d = 4. First the determinant: det = (2 × 4) − (1 × 3) = 8 − 3 = 5, safely away from zero. From there each entry is a straight division: invA = d ⁄ det = 4 ⁄ 5 = 0.8, invB = −b ⁄ det = −1 ⁄ 5 = −0.2, invC = −c ⁄ det = −3 ⁄ 5 = −0.6, and invD = a ⁄ det = 2 ⁄ 5 = 0.4. Multiplying the original matrix by this result reconstructs the identity matrix exactly, confirming the four figures are correct.
The identity matrix [[1, 0], [0, 1]] makes a useful second check: here a = 1, b = 0, c = 0, d = 1, so det = (1 × 1) − (0 × 0) = 1, and every output entry equals its matching input entry unchanged — invA = 1.0, invB = 0.0, invC = 0.0, invD = 1.0. A matrix that already behaves like a multiplicative 1 has to be its own inverse, and this sheet reports exactly that with no rounding involved.
Questions
What is the Moore-Penrose pseudoinverse of a matrix?
A generalized inverse defined for any matrix, including ones that aren't square or have no ordinary inverse. For a square, invertible matrix — the only kind this sheet accepts — the pseudoinverse and the plain inverse turn out to be exactly the same four numbers.
Is the pseudoinverse the same as the regular inverse here?
Yes, numerically identical, because this sheet only accepts square, invertible input. The two concepts genuinely diverge only for matrices that aren't square or that have a zero determinant — cases outside what a 2×2, invertible-only calculator can show.
What is the pseudoinverse of [[2, 1], [3, 4]]?
[[0.8, −0.2], [−0.6, 0.4]]. The determinant is (2×4) − (1×3) = 5, and each output entry divides a rearranged copy of the original matrix by that 5: invA = 4⁄5 = 0.8, invB = −1⁄5 = −0.2, invC = −3⁄5 = −0.6, invD = 2⁄5 = 0.4.
Why is the identity matrix its own pseudoinverse?
Because its determinant is 1 and its off-diagonal entries are already zero, so dividing by 1 and swapping zeros with zeros changes nothing. Multiplying the identity matrix by itself, or by any other matrix, leaves that other matrix untouched — the defining behavior of a multiplicative 1.
When does the pseudoinverse actually differ from a plain inverse?
Only outside square, invertible territory — for a rectangular matrix with more rows than columns or vice versa, or a square matrix whose determinant is exactly zero. This site's separate matrix-inverse page covers the same invertible-square case as this one; a full demonstration of the pseudoinverse's extra reach needs input neither page is built to accept.
What happens if the matrix entered here is singular?
Every output entry divides by det = ad − bc, so a matrix with ad equal to bc produces a division by zero rather than a number. A singular 2×2 matrix does still have a Moore-Penrose pseudoinverse in the fuller theory, just not one this determinant-based formula can compute.