# Simple proof that a $3\times 3$-matrix with entries $s$ or $s+1$ cannot have determinant $\pm 1$, if $s>1$.

Let $s>1$ and $A$ be a $3\times 3$ matrix with entries $s$ or $s+1$. Then $\det(A)\ne \pm 1$. The determinant has the form $as+b$ with integers $a$,$b$ and it has to be proven that $a>0$ if and only if $b>0$ and $a<0$ if and only if $b<0$. (The case $b=0$ is clear because the determinant is a multiple of $s$). I checked this with brute force, but I would like to have a simple proof.

• How do I get a plus sign over a minus-sign ? – Peter Jul 26 '14 at 19:26
• You do that with \pm – naslundx Jul 26 '14 at 19:27
• For square matrices with a size greater than $3\ x\ 3$, the above claim is not true. – Peter Jul 26 '14 at 19:57
• Your statement that ${\rm det}A$ has the form $as+b$ for integers $a$ and $b$ seems to assume that $s$ is an integer, which you have not stated in the problem. – Geoff Robinson Jul 26 '14 at 20:51
• s need not be an integer, the determinant has always the form as+b with a,b integers. But for non-integral s, the determinant might be non-integral. – Peter Jul 26 '14 at 20:55

$A$ is in the form $A=sU+V$ where the $u_{i,j}$ are $1$ and the $v_{i,j}\in\{0,1\}$. Then $\det(A)=as+b$ where $a=trace(Uadj(V))$ and $b=\det(V)$. If $\det(V)=0$, then we are done ; thus we assume that $V$ is invertible. Note that $ab$ has the same signum as $trace(UV^{-1})$ and it remains to show that $trace(UV^{-1})>0$ (the sum of the entries of $V^{-1}$). That can be done by brute-force with a PC in 0.2 second.