Question

If \(3\times3\) matrices are formed by using \(0,\pm1,\pm2\) as their elements, then the number of matrices whose trace is \(0\) is

• A. \(19\cdot5^6\)
• B. \(16\cdot5^9\)
• C. \(15\cdot5^6\)
• D. \(14\cdot5^8\)

Hint:

For matrix counting problems, \[ \boxed{\text{Total count}= (\text{Valid diagonal choices}) \times (\text{Choices for remaining entries})} \] The off-diagonal entries are independent of the trace.

Answer 1

The Correct Option is A

Step 1: Find the possible diagonal entries. Each diagonal entry can be chosen from \[ \{-2,-1,0,1,2\}. \] Let the diagonal entries be a,b,c. Since the trace is zero, \[ a+b+c=0. \] The ordered triples satisfying this condition are \[ (0,0,0), \] \[ (2,-2,0) \text{ and all its permutations}, \] \[ (1,-1,0) \text{ and all its permutations}, \] \[ (2,-1,-1) \text{ and all its permutations}, \] \[ (-2,1,1) \text{ and all its permutations}. \] Hence, \[ 1+6+6+3+3=19 \] ordered triples are possible. Thus, \[ \boxed{19} \] choices exist for the diagonal entries.

Step 2: Choose the remaining entries. A \(3\times3\) matrix has \[ 9-3=6 \] off-diagonal entries. Each can be chosen independently in 5 ways. Hence, \[ 5^6 \] possible choices.

Step 3: Find the total number of matrices. Therefore, \[ 19\times5^6 \] matrices have trace zero. Hence, \[ \boxed{19\cdot5^6} \] is the required number. Thus, \[ \boxed{(A)} \] is the correct answer.