Every equivalent condition for invertibility

Both directions lean on one algebraic fact borrowed from determinant theory: for any two n\times n matrices A and B, \det(AB) = (\det A)(\det B).

If A is invertible, then \det A \ne 0. By definition AA^{-1} = I_n. Taking the determinant of both sides and applying the product rule above, \det(A)\det\left(A^{-1}\right) = \det I_n = 1. A product of two real numbers can only equal 1 if neither factor is 0 (anything times 0 is 0, never 1), so \det A \ne 0 — and this same equation already isolates \det\left(A^{-1}\right) = \frac{1}{\det A}, which is the stated corollary.

If \det A \ne 0, then A is invertible. Row-reduce A to an echelon form U. Each elementary row operation used along the way changes \det A by a known nonzero factor — a row swap multiplies the determinant by -1, scaling a row by a nonzero constant c multiplies the determinant by c, and adding a multiple of one row to another leaves the determinant unchanged — so \det U is \det A times a nonzero constant. Since \det A \ne 0 by hypothesis, this forces \det U \ne 0 as well.

U is upper triangular, so \det U is exactly the product of its diagonal entries. For that product to be nonzero, every diagonal entry of U must be nonzero — none of them collapsed to a zero pivot during elimination. An n\times n triangular matrix with n nonzero diagonal entries has a pivot in every row and every column, so U row-reduces further, by scaling each row to make its pivot 1 and clearing every entry above each pivot, all the way down to I_n. Since A was row equivalent to U and U is row equivalent to I_n, A is row equivalent to I_n.

Every elementary row operation is itself invertible — undoing a swap is another swap, undoing a scaling by c\ne0 is scaling by 1/c, undoing a row addition is subtracting the same multiple back off — and each can be realized as left-multiplication by an elementary matrix E_k. Chaining the operations that carried A to I_n gives E_k \cdots E_1 A = I_n, so the product B = E_k\cdots E_1 satisfies BA = I_n. Since B is itself a product of invertible elementary matrices, B is invertible with inverse B^{-1} = E_1^{-1}\cdots E_k^{-1}; multiplying BA = I_n on the left by B^{-1} gives A = B^{-1}, so A is invertible with A^{-1} = B.

Sanity check: for the 1\times 1 case A = [a], \det A = a, and the theorem reduces to the familiar fact that a single number is invertible (has a reciprocal) exactly when it is nonzero — the determinant test is the n\times n generalization of that one observation.