Question

Find the number of matrices $A$ of order $3\times 2$ whose elements are from the set $\{\pm2,\pm1,0\}$, if $\mathrm{Tr}(A^TA)=5$.

• A. 310
• B. 312
• C. 320
• D. 325

Hint:

$\mathrm{Tr}(A^TA)$ equals the sum of squares of all entries of $A$. Always reduce such problems to counting valid square-sum combinations.

Answer 1

The Correct Option is B

The question requires us to find the number of matrices \( A \) of order \( 3 \times 2 \) whose elements are from the set \(\{ \pm 2, \pm 1, 0 \}\), and satisfy the condition \(\mathrm{Tr}(A^TA) = 5\).

To solve this problem, let's understand the requirement and the process step-by-step:

1. A^T is the transpose of matrix A. For a \(3 \times 2\) matrix \(A\), its transpose \(A^T\) will be a \(2 \times 3\) matrix. Therefore, the product \(A^TA\) results in a \(2 \times 2\) matrix.
2. The trace of a matrix, \(\mathrm{Tr}(A^TA)\), is the sum of elements on the main diagonal. For a \(2 \times 2\) matrix, \(\text{Tr}(A^TA) = (a_{11}^2 + a_{21}^2) + (a_{12}^2 + a_{22}^2) + (a_{13}^2 + a_{23}^2)\).
3. Using the given elements \(\{\pm 2, \pm 1, 0\}\), each \(a_{ij}^2\) will be one of \(\{4, 1, 0\}\).
4. The condition states \(\mathrm{Tr}(A^TA) = 5\), meaning:
   \[ a_{11}^2 + a_{21}^2 + a_{12}^2 + a_{22}^2 + a_{13}^2 + a_{23}^2 = 5 \]
5. Now, we need to count the possible combinations of the squares of elements in the matrix that sum to 5, using elements \(\{4, 1, 0\}\).
6. Possible configurations of these squares can be categorized:
   • One element being 4 and the rest being 1 or 0.
   • Several configurations where these values add up to 5.
7. The enumeration of such combinations effectively represents a combinatorial counting problem.
8. An orderly counting of valid configurations yields a total of \(312\) matrices.

The correct answer, therefore, is 312.