Matrix Inverse by Gauss-Jordan

Invert a square matrix by Gauss-Jordan elimination, without np.linalg.inv.

Glue an identity matrix onto the right of A to get an (n, 2n) block, then row reduce the left half all the way to the identity. Whatever the right half becomes is A^-1, because every row operation is a left multiplication and together they spell out A^-1 A | A^-1 I.

Use partial pivoting: at each column, swap in the row whose entry has the largest absolute value. If the best pivot is smaller than 1e-12 the matrix is singular, so return None.

Unlike row echelon form you have to eliminate above each pivot as well, so the left block ends up as the identity and not just triangular.

Function signature

def matrix_inverse(A):

Example

Input

matrix_inverse(np.array([[4.0, 7.0], [2.0, 6.0]]))

Output

array([[ 0.6, -0.7],
       [-0.2,  0.4]])

Reasoning

The determinant is 4 * 6 - 7 * 2 = 10. The inverse is [[6, -7], [-2, 4]] / 10, and Gauss-Jordan elimination gives the same result.

Constraints and notes

• Use partial pivoting.
• If the best pivot is below 1e-12, return None.
• Do not use np.linalg.inv.