Question

If A and B are two matrices such that rank of A = m and rank of B = n, then
\[\begin{align}
  & A.\text{Rank}\left( AB \right)=\text{inn} \\
 & \text{B}\text{.Rank}\left( AB \right)\ge \text{Rank}\left( A \right) \\
 & C.\text{Rank}\left( AB \right)\ge \text{Rank}\left( B \right) \\
 & D.\text{Rank}\left( AB \right)\le \text{min}\left( \text{RankA,RankB} \right) \\
\end{align}\]

Answer 1

Hint: In this question, we have to find the nature of the rank of a product of two matrices if the rank of individual matrices is given. We will use the property of the rank of the matrix which are:
(i) Rank of a matrix is the dimension of range R(M) of matrix M. So rank (M) = dim (R(M)).
(ii) $\text{dim}\left( \text{V} \right)\le \text{dim}\left( \text{W} \right)$ if V is a subset of vector space W.
(iii) $\text{R}\left( \text{AB} \right)\le \text{R}\left( \text{A} \right)$ i.e. range of matrix AB is less than equal to range of matrix A.

Complete step-by-step solution:
Here we are given two matrices A and B whose ranks are m and n respectively.
We know that the ratio of a matrix M is the dimension of the range of the matrix M. Range of the matrix M is written as P (M). Therefore, we can write the rank of the matrix M as rank (M) = dim R(M).
Now for finding the rank of matrix AB.
$\text{rank}\left( \text{AB} \right)=\text{dim}\left( \text{R}\left( \text{AB} \right) \right)\cdots \cdots \cdots \left( 1 \right)$.
Also for rank of matrix A, $\text{rank}\left( \text{A} \right)=\text{dim}\left( \text{R}\left( \text{A} \right) \right)\cdots \cdots \cdots \left( 2 \right)$.
We know that, range of product of two matrices A and B is less than or equal to the range of the individual matrix i.e. we can say $\text{R}\left( \text{AB} \right)\le \text{R}\left( \text{A} \right),\text{R}\left( \text{AB} \right)\le \text{R}\left( \text{B} \right)\cdots \cdots \cdots \left( 3 \right)$.
Now we know that for V to be subset of vector space W, dimension of V is less than that or equal to dimensions of W. Therefore, $\text{dim}\left( \text{V} \right)\le \text{dim}\left( \text{W} \right)\cdots \cdots \cdots \left( 4 \right)$.
Hence, using (1), (2), (3) and (4) we get:
$\text{rank}\left( \text{AB} \right)=\text{dim}\left( \text{R}\left( \text{AB} \right) \right)\le \text{dim}\left( \text{R}\left( \text{A} \right) \right)=\text{rank}\left( \text{A} \right)$.
So we conclude that, $\text{rank}\left( \text{AB} \right)\le \text{rankA}$.
Similarly, we can say that $\text{rank}\left( \text{B} \right)=\text{dim}\left( \text{R}\left( \text{B} \right) \right)\text{,R}\left( \text{AB} \right)\le \text{R}\left( \text{B} \right)$.
Hence, $\text{rank}\left( \text{AB} \right)=\text{dim}\left( \text{R}\left( \text{AB} \right) \right)\le \text{dim}\left( \text{R}\left( \text{B} \right) \right)=\text{rank}\left( \text{B} \right)$.
So we conclude that $\text{rank}\left( \text{AB} \right)\le \text{rankB}$.
Now, as we can see that $\text{rank}\left( \text{AB} \right)\le \text{rankA and rank}\left( \text{AB} \right)\le \text{rankB}$. So if rank A is less than rank B then rank AB will be less than or equal to rank A.
Hence we can say that rank AB is less than or equal to the minimum of rank A and rank B.
Hence option D is the correct answer.

Note: Students should know all definitions of rank of a matrix, range of a matrix and dimension of matrix to solve this question. They should note that, if matrix B is nonsingular then $\text{rank}\left( \text{AB} \right)=\text{rankA}$.