Question

Write the number of all possible matrices of order $2\times 2$ with each entry $1,2$ or $3?$

Answer 1

We know that a matrix is an ordered rectangular array of numbers or functions. We also know that a $2\times 2$ matrix is a square matrix containing $4$ elements. We will find the number of ways we can accommodate any of the given entries in a $2\times 2$ matrix when repetition is allowed.

Complete step by step solution:
We are asked to find the number of all possible matrices of order $2\times 2$ with each entry $1,2$ or $3.$
We know that a square matrix of order $2\times 2$ contains $2$ columns and $2$ rows. And thus, it contains $4$ elements.
It is given that the entries are $1,2$ or $3$ with repetition allowed.
We can construct $2\times 2$ matrices with all $4$ elements are the same, $3$ of the elements are the same, $2$ of the elements are the same and all are distinct.
Since, repetition is allowed, we can accommodate any one of the three entries in each of the four positions in the matrix.
Let us consider the first position, that is ${{a}_{11}}.$ As we know, this place can be occupied by any of the three entries. So, we have $3$ choices.
Similarly, in the position ${{a}_{12}},$ we can accommodate any of the three entries. So, we have $3$ choices.
In the third position ${{a}_{21}}$ also, we can accommodate any $3$ of the given entries. This implies that we have $3$ choices again.
So, in the last position ${{a}_{22}},$ we still have $3$ choices because repetition of the entries is allowed.
Thus, we can multiply the number of choices to get the number of $2\times 2$ matrices that can be made using the given entries.

Hence, we can make $3\times 3\times 3\times 3={{3}^{4}}=81$ matrices of order $2\times 2$ with each entry $1,2$ or $3.$

Note: Let us note that the question is about the arrangement of the entries when repetition is allowed. We should remember that we have all the choices for each of the positions we have when repetition is allowed.