Why does inv() work on a rank deficient matrix?

I have an nxn square matrix A with rank n-1. When I call inv(A), MATLAB computes the inverse without complaining. How is this possible? Shouldn't a matrix that is rank deficient be impossible to invert?

Answer 1

[Walter Roberson]

If I recall correctly, inv() does not use rank() to detect rank deficiency.

"inv performs an LU decomposition of the input matrix (or an LDL decomposition if the input matrix is Hermitian). It then uses the results to form a linear system whose solution is the matrix inverse inv(X). For sparse inputs, inv(X) creates a sparse identity matrix and uses backslash, X\speye(size(X))."

These algorithms have their own internal settings as to whether they complain about a marginal matrix or not. Depending on the numeric noise in the calculations, sometimes they do not detect a matrix as singular when it is singular.

... You probably should not be using inv() anyhow.

[N/A]

Thanks, yeah, this answers my question. I switched to using the backslash operator. Of course, now I am wondering if there are cases where the backslash operator will fail and whether I can prevent this somehow by ensuring the matrix is well-conditioned.

[Walter Roberson]

Yes, and the tolerance of the backslash operator has changed in the not-so-distant past. Changed both ways: mostly to reject more problematic matrices, but it also started accepting some matrices it should probably have rejected.