Question

The number of \( 3 \times 3 \) matrices with entries \( -1 \) or \( +1 \) is:

• A. \( 2^4 \)
• B. \( 2^5 \)
• C. \( 2^6 \)
• D. \( 2^7 \)
• E. \( 2^9 \)

Hint:

This logic applies to any combinatorial matrix problem. For an \( m \times n \) matrix with \( k \) possible choices for each entry, the total number of distinct matrices is \( k^{m \times n} \).

Answer 1

Answer: (E)

Concept: A \( 3 \times 3 \) matrix has 9 independent positions (3 rows and 3 columns). According to the Fundamental Principle of Counting, if each position can be filled in \( n \) ways, the total number of ways to fill the entire matrix is \( n^{\text{total positions}} \).

Step 1: Identify the total number of elements in the matrix.
A \( 3 \times 3 \) matrix \( A \) is represented as: \[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \] Total number of entries = \( 3 \times 3 = 9 \).

Step 2: Calculate the total combinations.
Each of the 9 entries has 2 possible choices: \(\{ -1, +1 \}\). Total number of matrices = \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \) \[ = 2^9 \] Note: \( 2^9 = 512 \).