Suppose a matrix has\[12\] identical elements, then what are the possible orders it can have?
A. \[3\]
B. \[1\]
C. \[6\]
D. None of the above.

Hint:
We know the multiplication of rows and columns of a matrix is the number of total entries in the matrix. So if a matrix is of the order\[\left( {m \times n} \right)\], the possible identical elements will be \[mn\]. For example if we take $6$ identical elements we can get \[4\] possible ordered matrices such as \[\left( {6 \times 1} \right), \left( {3 \times 2} \right), \left( {2 \times 3} \right), \left( {1 \times 6} \right)\].

Complete step by step solution:
To find all the possible orders of \[12\] identical elements we have to find all the ordered matrices of a natural number whose product is \[12\].
We have to find the two numbers whose product is $12$.
So the matrices that can be formed are of these orders \[\left( {1 \times 12} \right),\left( {2 \times 6} \right),\left( {3 \times 4} \right),\left( {4 \times 3} \right),\left( {6 \times 2} \right),\left( {12 \times 1} \right)\].
All of them will have the same 12 identical elements.
So, the possible orders it can have is \[6\].

Hence, option (c) is correct.

Note:
The common mistake all we do is sometimes we forget to count the repeating matrices such as $(1 \times 12)$ and $(12 \times 1)$, $(3 \times 4)$ and $(4 \times 3)$, $(2 \times 6)$ and $(12 \times 6)$. In this case rows and columns are interchanging so the matrix formation will be completely different. Hence, if we forget to count the repeating matrices the answer will be \[3\] which is completely wrong. We have to count all the possible matrices that can be formed.