To solve this, let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with integer entries. The conditions are:

1. $A^2 = 3A$
2. $\det(A) = 6$

We seek the distinct values that the trace of $A$ can take, reduced modulo 7.

Step 1: Analyze $A^2 = 3A$

The matrix equation $A^2 = 3A$ implies that $A$ is a type of matrix that satisfies a quadratic equation, similar to an eigenvalue problem. This condition can be rewritten as: $A^2 - 3A = 0 \Rightarrow A(A - 3I) = 0$ where $I$ is the identity matrix. This suggests that the matrix $A$ has eigenvalues that satisfy $\lambda(\lambda - 3) = 0$, giving eigenvalues $\lambda = 0$ or $\lambda = 3$.

Step 2: Use the Determinant Condition $\det(A) = 6$

Since $\det(A) = 6$, the eigenvalues of $A$ must multiply to give 6 (the determinant is the product of the eigenvalues). The possible integer eigenvalue pairs that satisfy both $\lambda_1 \lambda_2 = 6$ and $\lambda(\lambda - 3) = 0$ are:

• $\lambda_1 = 3$ and $\lambda_2 = 2$
• $\lambda_1 = 2$ and $\lambda_2 = 3$

Thus, the trace of $A$, which is the sum of the eigenvalues, could be: $\text{Trace}(A) = \lambda_1 + \lambda_2 = 3 + 2 = 5$

Step 3: Reduce Trace Modulo 7

Since the only possible trace of $A$ is 5, we find that: $\text{Trace}(A) \equiv 5 \pmod{7}$

Conclusion

There is only one distinct value the trace of $A$ can take modulo 7, which is 5.

Would you like more details or have any questions?

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Here are 5 related questions to consider:

1. How would the problem change if $\det(A) = -6$ instead?
2. Can a similar approach be applied to 3x3 matrices with similar conditions?
3. What properties of eigenvalues change if we consider matrices over fields other than integers?
4. How would the distinct trace values differ if the modulus was another prime, such as 5?
5. How does the characteristic polynomial of $A$ relate to the given conditions?

Tip: When working with matrices modulo a prime number, eigenvalues and traces can provide quick insights without needing full calculations.