The number of different possible order of matrices having 18 identical elements is
(a) 3
(b) 1
(c) 6
(d) 4
Answer
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Hint: First, we have to understand the meaning of different possible order of matrices having the same element means by multiplying values of rows and columns and we get the same number of identical elements. For example taking 4 identical elements, we can get 4 elements in $2\times 2$ , $1\times 4$ , $4\times 1$ . On multiplying this we get 4 elements so; here different possible order of matrices is 3. Similarly, we can find 18 identical elements.
Complete step-by-step answer:
Here, we have to find how many different types of matrices can be formed having the same elements in it, i.e. interchanging rows and columns of the matrix.
Now taking example of $2\times 2$ matrix which can be written as
$\left( \begin{matrix}
1 & 2 \\
3 & 4 \\
\end{matrix} \right)$
Now, this means that there are 4 elements in the $2\times 2$ matrix. So, other possible way to write these 4 elements can be $1\times 4$ given as
$\left( \begin{matrix}
1 & 2 & 3 & 4 \\
\end{matrix} \right)$
Also, again order can be changed and written as $4\times 1$i.e.
$\left( \begin{matrix}
1 \\
2 \\
3 \\
4 \\
\end{matrix} \right)$
Thus, there are 3 different possible orders of matrices having 4 identical elements.
Similarly, we have total 18 elements which means $m\times n$ i.e. rows and column can be $1\times 18$ , $2\times 9$ , $3\times 6$ , $6\times 3$ , $9\times 2$ , $18\times 1$ . All this size matrix will have the same 18 identical elements.
So, the number of different possible orders of matrices having 18 identical elements is 6.
Hence, option (c) is correct.
Note: Be careful while counting all the different possible orders. Sometimes, students forget to count the repeating matrix i.e. $1\times 18$ , $2\times 9$ , $3\times 6$ , $6\times 3$ , $9\times 2$ , $18\times 1$ out of these, they do not count $6\times 3$ , $9\times 2$ , $18\times 1$ as they are just interchanged values of rows and column. But this whole matrix representation is changed, so we have to count these matrices in different orders. So, be clear about this type of problem while solving. Otherwise the answer will be 3 which is wrong.
Complete step-by-step answer:
Here, we have to find how many different types of matrices can be formed having the same elements in it, i.e. interchanging rows and columns of the matrix.
Now taking example of $2\times 2$ matrix which can be written as
$\left( \begin{matrix}
1 & 2 \\
3 & 4 \\
\end{matrix} \right)$
Now, this means that there are 4 elements in the $2\times 2$ matrix. So, other possible way to write these 4 elements can be $1\times 4$ given as
$\left( \begin{matrix}
1 & 2 & 3 & 4 \\
\end{matrix} \right)$
Also, again order can be changed and written as $4\times 1$i.e.
$\left( \begin{matrix}
1 \\
2 \\
3 \\
4 \\
\end{matrix} \right)$
Thus, there are 3 different possible orders of matrices having 4 identical elements.
Similarly, we have total 18 elements which means $m\times n$ i.e. rows and column can be $1\times 18$ , $2\times 9$ , $3\times 6$ , $6\times 3$ , $9\times 2$ , $18\times 1$ . All this size matrix will have the same 18 identical elements.
So, the number of different possible orders of matrices having 18 identical elements is 6.
Hence, option (c) is correct.
Note: Be careful while counting all the different possible orders. Sometimes, students forget to count the repeating matrix i.e. $1\times 18$ , $2\times 9$ , $3\times 6$ , $6\times 3$ , $9\times 2$ , $18\times 1$ out of these, they do not count $6\times 3$ , $9\times 2$ , $18\times 1$ as they are just interchanged values of rows and column. But this whole matrix representation is changed, so we have to count these matrices in different orders. So, be clear about this type of problem while solving. Otherwise the answer will be 3 which is wrong.
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