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Original question asked:

If matrix $A$ commutes with $B$, and $B$ commutes with $C$, then does matrix $A$ commute with $C$?

This can easily be disproven by taking $B = I$ and looking at some matrices $A$ and $C$ that doesn't satisfy the condition.

However, I thought: if $B$ is given to be non-invertible, is the case true?

If so, why? If not, can you provide an example?

Also, if not, are there any stronger conditions that would make the case true (for example, $A$ and $C$ has to be both non-invertible)?

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    What if $B$ is the $0$ matrix? – 2017-01-27
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    Awesome. Thanks. That answers my first question. Now my second question: what if $B$ is given to be non-invertible and nonzero? – 2017-01-27

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The answer is no. Consider the following counterexample over any ring with singular matrices first, where $AB=BA=BC=CB=AC=0\ne CA$: $$ A=\pmatrix{0&0&1\\ 0&0&0\\ 0&0&0},\ B=\pmatrix{0&0&0\\ 0&0&1\\ 0&0&0},\ C=\pmatrix{0&0&0\\ 1&0&0\\ 0&0&0}. $$ Now you may add the identity matrix $I$ to $A,B$ or $C$ to make them nonsingular.