Question

The number of $ 3\times 3 $ matrices with entries -1 or +1 is
A. $ {{2}^{-4}} $
B. $ {{2}^{5}} $
C. $ {{2}^{6}} $
D. $ {{2}^{9}} $

Answer 1

Hint: The given question involves the concept of combination. A $ 3\times 3 $ matrix contains nine elements and we can choose either -1 or +1 for each of the nine slots. Find the number of ways we can do this by using the above information.

Complete step-by-step answer:
A matrix is an arrangement of some data in the form of rows and columns. A $ 3\times 3 $ matrix is a matrix where the data (say some numbers) is arranged in three rows and three columns. The numbers or variables inside a matrix are called elements of the matrix.
It is asked to find the number of $ 3\times 3 $ matrices with entries -1 or +1. This means that the matrix can have either -1 or +1 as the matrix’s elements.
Since there are three rows and three columns, the total number of elements (entries) of the matrix are $ 3\times 3=9 $ .
Now, it is said that each of the 9 elements can be either -1 or +1. Therefore, there are two choices for each of the elements of the matrix.
This problem
The number of ways in which we can choose one number from two different numbers is 2.
Therefore, the total numbers of the combinations for the nine slots inside the slot will be $ \Rightarrow 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{9}} $ .
Each of these combinations will represent a matrix. Therefore, the number of $ 3\times 3 $ matrices with entries -1 or +1 is $ {{2}^{9}} $ .
Hence, the correct option is D.
So, the correct answer is “Option D”.

Note: Note that the given problem involves the concept of permutations and combinations. Here, we found the total number of combinations found with entries -1 or +1.
The number of ways in which we can choose ‘m’ objects from ‘n’ different objects is given as $ {}^{n}{{C}_{m}}=\dfrac{n!}{(n-m)!(m!)} $ .