Matrix again

Since the determinant is equal to $\sum_{\pi \in S_3} \mathrm{sgn}(\pi)a_{1,\pi1}a_{2,\pi2}a_{3,\pi3}$ it is equal to a sum of six terms, each of which is equal to $\pm 1$ . Thus $-6 \le |A| \le 6$ . On the other hand, using row operations, we can show that $|A|$ is equal to the determinant of a matrix of the form $\left(\begin{array}{ccc} a & b & c \\ 0 & d & e \\ 0 & f & g \end{array} \right)$ where $a,b,c \in \{-1,1\}$ and $d,e,f,g \in \{-2,0,2\}$ . Thus $|A| \; = \; a(dg - ef)$ is a multiple of $4$ . The largest possible value of $|A|$ is therefore $\boxed{4}$ , and this is possible, for example with $A \; = \; \left(\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & -1 & 1 \end{array}\right)$

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