Suppose $B$ is a non-zero real skew-symmetric matrix of order $3$ and $A$ is a non-singular matrix with inverse $C$. Then rank of $ABC$ is:

(A) $0, 1, 2$

(B) definitely $1$

(C) definitely $2$

(D) definitely $3$

Here we are given $B^{T}=-B$ and $A$ is non-singular i.e. $A^{-1}$ exists and $A^{-1}=C$

Now, $rank(ABC)=rank(ABA^{-1})=rank(B)$ Since $B$ is non-zero, option (A) is incorrect but what about (B), (C) and (D)?

  • $\begingroup$ Can we use this result: The rank of non-zero real skew-symmetric matrix must be even $\endgroup$ – Kns Jun 25 '13 at 5:42

Can a skew-symmetric $3\times 3$ matrix be non-singular? HINT: If $A$ is $n\times n$ and skew-symmetric, then

$$\det=\det A^T=\det(-A)=k\,\det A\;,$$

where $k=\ldots$ what?

That gives you part of what you need; for the rest see this question.

  • $\begingroup$ $k=(-1)^{n}$ it means if $n$ is odd then $det(A)=0$. $\endgroup$ – Kns Jun 25 '13 at 5:58
  • $\begingroup$ @Kns: That’s right: when $n$ is odd, the matrix must be singular, so in this case the rank must be $1$ or $2$, and the link takes care of the rest. $\endgroup$ – Brian M. Scott Jun 25 '13 at 6:03

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