Number of Possible Matrices Explained

The question asks for the total number of possible matrices of order $3 \times 3$ where each entry (or element) of the matrix can only be either the number 1 or the number 2.

A matrix of order $3 \times 3$ has 3 rows and 3 columns. The total number of entries in such a matrix is calculated by multiplying the number of rows by the number of columns:

Total entries = Number of rows $\times$ Number of columns = $3 \times 3 = 9$ entries.

Each of these 9 entries in the matrix can be filled with one of two possible values: either 1 or 2.

Since the choice for each entry is independent of the choices for the other entries, we can find the total number of possible matrices by multiplying the number of options for each entry together.

For the first entry, there are 2 options (1 or 2).

For the second entry, there are 2 options (1 or 2).

...and so on, for all 9 entries.

So, the total number of possible matrices is the product of the number of options for each of the 9 entries:

Total number of matrices = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ (9 times)

This can be written in a more compact form using exponents as $2^9$.

Now, we need to calculate the value of $2^9$:

• $2^1 = 2$
• $2^2 = 4$
• $2^3 = 8$
• $2^4 = 16$
• $2^5 = 32$
• $2^6 = 64$
• $2^7 = 128$
• $2^8 = 256$
• $2^9 = 512$

Therefore, there are 512 possible matrices of order $3 \times 3$ with each entry being either 1 or 2.