Find the number of 3×3 matrices A such that the entries of A belong to the set Z of all integers, and such that the trace of $A^t$*$A$ is 6 . ($A^t$ denotes the transpose of the matrix A).
For this problem,I can't understand how to calculate the number of the 3*3 matrices satisfying the given condition. Any help would be appreciable. A little wordy solution is appreciable.
Thank you.
Let $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$. Then by a matrix multiplication, one obtains that $$\operatorname{tr}(A^t A)=\sum_{i,j} a_{ij}^2=6.$$ Hence all entries must be between $-2$ and $2$. First, consider the case where one of the entries has absolute value $2$. Then there must be $2$ other entries with absolute value $1$, and all other entries are zero. The number of such matrices is $$\binom{9}{1}\cdot 2\cdot \binom{8}{2}\cdot 2^2,$$ where we have multiplied by $2^3$ to take into account that these entries may also be negative. Now consider the case where there does not exist an entry with absolute value $2$. Then there are $6$ entries with absolute value $1$ and $3$ are zero. The number of such matrices is $$\binom{9}{6}\cdot 2^6.$$ The sum of these is the result $$\binom{9}{1}\binom{8}{2}\cdot 2^3+\binom{9}{6}\cdot 2^6=7392.$$