$\operatorname{Rank}(ABC) = 1$ true or false

Question

Suppose $A, B, C$ are $3 × 3$ real matrices with $\operatorname{Rank}( A )= 2, \operatorname{Rank} (B) = 1,\operatorname{Rank} (C) = 2$. Then $\operatorname{Rank}(ABC) = 1$.

We have to find out whether the statement is true or false.

Now $\operatorname{Rank}(ABC)\leq \min\bigl\{\operatorname{Rank}(A),\operatorname{Rank}(B), \operatorname{Rank}(C)\bigr\}=1$. So it should be true!

Answer 1

Consider A with all zero entries except $a_{11}$ and $a_{22}$ ($rank(A)=2$). $B$ is all zeros except $b_{33}$ ($rank(B)=1$). $AB= 0$, thus $ABC=0$ and has rank $0$.