Calculating the Number of Matrices Satisfying a Given Trace Condition

To determine the number of matrices A in M_3mathbb{Z} that satisfy the condition tr A^t A 6, we can explore the structure and properties of these matrices. Given that the entries of A are integers, this problem involves a combination of linear algebra and number theory.

Defining the Problem

Consider the 3x3 matrix A defined as

A begin{bmatrix} a_{11} a_{12} a_{13} a_{21} a_{22} a_{23} a_{31} a_{32} a_{33} end{bmatrix}

The condition tr A^t A 6 implies that the sum of the squares of the elements in the main diagonal of the matrix product A^t A equals 6.

Matrix Multiplication and Trace

The product A^t A can be computed as follows:

A^t A begin{bmatrix} a_{11} a_{12} a_{13} a_{21} a_{22} a_{23} a_{31} a_{32} a_{33} end{bmatrix} begin{bmatrix} a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} end{bmatrix}

The trace of this matrix is the sum of the elements on its main diagonal, which can be expressed as:

tr A^t A a_{11}^2 a_{22}^2 a_{33}^2 6

Case Analysis

We need to find the number of integer solutions to the equation a_{11}^2 a_{22}^2 a_{33}^2 6.

Case i: One Element Equals ±2

If one of the diagonal entries of A equals ±2, then the other two diagonal entries must be ±1, and all other entries must be 0. The number of such cases is calculated as follows:

- There are 3 choices for which entry can equal ±2.

- There are 2 choices for the sign of the ±2 entry.

- There are ( binom{9}{1} ) ways to choose the position of the ±2 entry.

- There are 2 ways for each of the two remaining diagonal entries to be ±1.

- The number of ways to choose positions for the two ±1 entries is ( binom{8}{2} ).

Thus, the total number of matrices for this case is:

2 times binom{9}{1} times 2^2 times binom{8}{2} 2 times 9 times 4 times 28 2016

Case ii: Six Elements Equal ±1

If six of the diagonal entries of A equal ±1, and all other entries must be 0, then the number of ways to choose the positions for these elements is calculated as follows:

- There are ( binom{9}{6} ) ways to choose the positions of the ±1 entries.

- There are 2^6 ways to assign the signs to these 6 elements, as each can be either 1 or -1.

Thus, the total number of matrices for this case is:

2^6 times binom{9}{6} 64 times 84 5376

Conclusion

The total number of matrices A in M_3mathbb{Z} that satisfy tr A^t A 6 is the sum of the matrices from both cases:

2016 5376 7392