inverse from Cayley-Hamilton
Here is a slick consequence of Cayley-Hamilton: when an operator A is invertible, its inverse is a POLYNOMIAL in A. You never have to leave the algebra generated by A to find A^-1 — it was hiding there all along, expressible with the very same matrix you are inverting.
Write the characteristic polynomial as chi(x) = x^n + c_(n-1) x^(n-1) + ... + c_1 x + c_0. Cayley-Hamilton gives A^n + c_(n-1) A^(n-1) + ... + c_1 A + c_0 I = 0. The constant term c_0 equals (-1)^n det(A), which is nonzero exactly when A is invertible. Factor an A out of all but the constant term and rearrange: A times ( -(1/c_0)(A^(n-1) + ... + c_1 I) ) = I, so A^-1 = -(1/c_0)(A^(n-1) + c_(n-1) A^(n-2) + ... + c_1 I).
Read that formula slowly: A^-1 is a specific linear combination of I, A, A^2, ..., A^(n-1). The invertibility condition c_0 != 0 is the same as det(A) != 0, so the construction succeeds precisely when an inverse should exist. You can even use the minimal polynomial instead of chi for a lower-degree, cheaper formula, as long as its constant term is nonzero (which again means 0 is not an eigenvalue).
Beyond being a cute trick, this matters theoretically: it proves A^-1 commutes with everything A commutes with (since it is a polynomial in A), it shows the inverse lives in the commutative algebra F[A], and it underpins symbolic and exact-arithmetic inversion. Numerically it is poor — Gaussian elimination is far more stable — but as structure it is gold.
Rearranging Cayley-Hamilton expresses A^-1 as the polynomial (3I - A)/2 in A, no separate inversion algorithm needed.
A is invertible if and only if its constant term c_0 = (-1)^n det(A) is nonzero, equivalently 0 is not an eigenvalue. If 0 IS an eigenvalue the formula breaks exactly because there is no inverse to find.