# A matrix consists of 30 elements then, what are the possible orders it can have? And how many different orders it can have?

Last updated date: 22nd Mar 2023

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Answer

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Hint: Number of elements in a matrix is the product of the number of rows and number of columns in a matrix. Here no. of elements are given so we write these elements in multiplication of rows and columns.

Complete step-by-step answer:

As we know that the matrix with 30 elements can be a square matrix or any other matrix because it is not specified in the question.

And for any matrix, such that it has m rows and n columns. Its order will be m*n and number of elements in the matrix A will be m*n.

And m and n should be a natural number.

So, we are given that the number of elements in the required matrix is 30.

So, m*n = 30. (1)

So, we had to find different values of m and n, such that it satisfies equation 1.

First, we will find factors of 30. And then after that we will find how such pairs exist which gives 30 after multiplying with each other.

So, factors of 30 are {1, 2, 3, 5, 6, 10, 15, 30}

As we know that m*n and n*m are the two different orders of matrix because,

Matrix with order m*n has m rows and n columns.

But matrix with order n*m has n rows and m columns.

So, now as we can clearly see that {1, 30}, {30, 1}, {5, 6}, {6, 5}, {2, 15}, {15, 2} and {3,10}, {10,3} are the pairs which gives us product equal to 30.

Hence possible orders of the matrix with 30 elements are 1*30, 30*1, 5*6, 6*5, 3*10, 10*3, 2*15 and 15*2.

A matrix with 30 elements can have 8 different orders.

Note: Whenever we come up with this type of problem then first, we have to find the different factors of the total number of elements in the matrix. After that we have to make pairs of two factors such that their product gives us a number that is equal to the total elements in the matrix.

Complete step-by-step answer:

As we know that the matrix with 30 elements can be a square matrix or any other matrix because it is not specified in the question.

And for any matrix, such that it has m rows and n columns. Its order will be m*n and number of elements in the matrix A will be m*n.

And m and n should be a natural number.

So, we are given that the number of elements in the required matrix is 30.

So, m*n = 30. (1)

So, we had to find different values of m and n, such that it satisfies equation 1.

First, we will find factors of 30. And then after that we will find how such pairs exist which gives 30 after multiplying with each other.

So, factors of 30 are {1, 2, 3, 5, 6, 10, 15, 30}

As we know that m*n and n*m are the two different orders of matrix because,

Matrix with order m*n has m rows and n columns.

But matrix with order n*m has n rows and m columns.

So, now as we can clearly see that {1, 30}, {30, 1}, {5, 6}, {6, 5}, {2, 15}, {15, 2} and {3,10}, {10,3} are the pairs which gives us product equal to 30.

Hence possible orders of the matrix with 30 elements are 1*30, 30*1, 5*6, 6*5, 3*10, 10*3, 2*15 and 15*2.

A matrix with 30 elements can have 8 different orders.

Note: Whenever we come up with this type of problem then first, we have to find the different factors of the total number of elements in the matrix. After that we have to make pairs of two factors such that their product gives us a number that is equal to the total elements in the matrix.

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