Using geometry to find all possible values of the determinant of a $3\times 3$ matrix whose entries are $0$ or $1$

Question

The matrix $A_{3\times 3}$ has all its entries $0$ or $1$. All possible values of $\det A$ are ...

From trial method, I know the answer to be $\pm2$, $\pm 1$, $0$. For a more complicated matrix the trial would be tedious, so I was looking for a geometric method.

I know the columns are vectors and the determinant is the volume of the parallelepiped formed. Can anyone help me with this thinking process? I can tell $0$, $\pm1$ as values easily, but how to think for greater values for other such determinant (limit to $3\times 3$)?

Answer 1

Up to a sign, the determinant is the volume of a parallelepiped with sides $\in \{ 0, 1, \sqrt{2}, \sqrt{3} \}$.

Since the volume is an integer, if none of the sides is $\sqrt{3}$, the volume is at most $\lfloor \sqrt{2}^3 \rfloor = 2$. In fact, the volume $2$ is realizable. e.g. following matrix has determinant $2$ ${}^{\color{blue}{[1]}}$. $$\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0\end{bmatrix}$$

If the parallelepiped has a side $\sqrt{3}$. One of the column is all $1$. If you project the parallelepiped to a plane perpendicular to the $(1,1,1)$ direction. You obtain a parallelogram. If the parallelogram non-degenerate, it has sides $\sqrt{\frac23}$. In this case, the volume of the parallelepiped is at most $\left\lfloor \sqrt{3} \sqrt{\frac23}^2\right\rfloor = 1$

Combine these, we find the possible values of the determinant is $0$, $\pm 1$ and $\pm 2$.

Notes

• $\color{blue}{[1]}$ - In order for the determinant equals to $\pm 2$, up to permutation of rows/columns, this is the only choice. When the volume is $2$, all sides of the parallelepiped need to be $\sqrt{2}$. Configuration of sides like $( \sqrt{2},\sqrt{2}, 1 )$ won't work because the two longer sides are not perpendicular to each other.