The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is

The correct option is 512.

To find the total number of all possible 3 × 3 matrices where each entry can be either 0 or 1, we can analyze the structure of the matrix step-by-step.

First, let us determine the total number of entry positions (or elements) in a matrix of order 3 × 3.

A matrix of order 3 × 3 has 3 rows and 3 columns. Therefore, the total number of elements in this matrix is:
$$3\times 3=9$$
So, there are 9 distinct positions to fill in the matrix.

Each of these 9 positions can be filled with either the digit 0 or the digit 1. This means there are exactly 2 choices (0 or 1) for each of the 9 entries.

By the basic principle of counting (multiplication principle), if each of the 9 independent positions can be filled in 2 ways, the total number of all possible matrices is given by multiplying the number of choices for each position:
$$2^9$$
Let us calculate this value:
$$2^9=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=512$$
Thus, the total number of all such possible matrices is 512.