# Checking the singularity of a matrix

Is there any simple way of telling if a matrix $A$ is singular or not?

I saw somewhere the following:

If $rank(A)<n \Rightarrow$ $A$ is singular.

If $rank(A)\geqslant n \Rightarrow$ A is non-singular.

But I don't know how to verify if is this is always true. Also, how does the determinant relate to the singularity of a matrix?

• $\text{det}(A)=0$ if and only if $A$ is square and singular. – Adrian Keister Feb 27 '18 at 19:08
• For small size of $A$ the determinant is easy to compute, see for example the rule of Sarrus. Then you know quickly whether $A$ is singular or not. – Dietrich Burde Feb 27 '18 at 19:10
• @AdrianKeister Isn't that only if $A$ does not have an inverse? – Amateur Mathematician Feb 27 '18 at 19:11
• There are a bunch of equivalent conditions: $\det(A)=0$ is equivalent to $A$ singular, equivalent to $A$ does not have an inverse, equivalent to $\text{rank}(A)<n$ (assuming $A$ is $n\times n$), equivalent to non-trivial kernel, and the list goes on and on. – Adrian Keister Feb 27 '18 at 19:12
• In exact arithmetic, the determinant or the rank are useful. In inexact arithmetic, things are slightly murkier, and one has to be contented with determining the condition number. (See e.g. this.) – J. M. is a poor mathematician Feb 28 '18 at 3:22