If $A$ and $B$ are two $n - $ rowed square matrices such that $AB = 0$ and $B$ is non-singular, then which of the following options is correct?
A. $A = I$
B. $A = 0$
C. $A \ne 0$
D. $A = nI$

Hint: In the given question, we are provided a product of two square matrices, $A$ and $B$ such that $AB = 0$ . Now use the theorem “The determinant of the product of matrices is equivalent to the product of their respective determinants” to get the required answer.

Complete step by step Solution:
We are given two $n - $ rowed square matrices, $A$ and $B$ such that:
$AB = 0$
Taking the determinants of both sides,
$\left| {AB} \right| = 0$
Now, we have a theorem according to which “The determinant of the product of matrices is equivalent to the product of their respective determinants”, therefore,
$\left| {AB} \right| = \left| A \right|\left| B \right| = 0$
It is also given that $B$ is a non-singular matrix, thus, $\left| B \right| \ne 0$ .
Hence, as $\left| B \right| \ne 0$ and as $\left| A \right|\left| B \right| = 0$ ,
This implies that $\left| A \right| = 0$ .
Now, from the given options, the only option which satisfies $\left| A \right| = 0$ is $A = 0$ .
Hence, $A = 0$ .

Therefore, the correct option is (B).

Note: A singular matrix is a matrix whose determinant is $0$ while a non-singular matrix is a matrix whose determinant is not equal to $0$.. Hence, if $A$ is a singular matrix and $B$ is a non-singular matrix, then $\left| A \right| = 0$ and $\left| B \right| \ne 0$ .