Question

Find the number of all possible matrices of order 2 $\times$ 2 with each entry 0 or 1.

Answer 1

Hint: In order to find the number of all possible matrices of order 2 × 2, we first try to understand the concept of a 2 × 2 matrix. A 2 × 2 matrix has four spaces to be filled. Given each space is to be filled by 0 or 1. We use the concept of permutations and combinations.

Complete step-by-step answer:
Given Data,
A 2 × 2 matrix, to be filled with either 0 or 1.
A 2 × 2 matrix typically looks like one as follows:
 $ \left[ {\begin{array}{*{20}{c}}
  \_&\_ \\
  \_&\_
\end{array}} \right] $
It has two rows and two columns to be filled.
Given each space takes either 0 or 1.

So there are 4 spaces in total and each space takes either 0 or 1. Therefore we can say there are two ways of filling each of the four spaces.The total number of such matrices that exist will be nothing but the number of ways of all the possibilities of arranging 0 and 1 in four spaces.
There are two digits out of which one is to be selected in each space, i.e. $ {}^2{{\text{C}}_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = \dfrac{2}{1} = 2 $ ways.
Therefore the total number of ways 0 and 1 can be arranged are 2 × 2 × 2 × 2 = 16 ways.
Therefore the number of all possible matrices of order 2 × 2 with each entry 0 or 1 is 16.

Note: In order to solve this type of questions the key is to clearly know the number of possibilities in which the given quantities can be arranged in the prescribed number of spaces.This type of an arrangement where the order of the elements does not matter is called a combination and is given by the formula, $ {}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}} $, where n is the number of total available options and r is the number of objects to be arranged.
n! means the factorial of a number n and it is given by, n! = n × (n-1) × (n-2) × …….. × (n – (n-1)).