Maximum and minimum value of Determinant of $3 \times 3$ Matrix with entries $\pm 1$

Question

Find Maximum value of Determinant of $3 \times 3$ Matrix with entries $\pm 1$

My try:

I considered a matrix as :

$$A=\begin{bmatrix} 1 &-1 &-1 \\ -1 &1 &-1 \\ -1&-1 &1 \end{bmatrix}$$

we have $$Det(A)=-4$$ and maximum is $4$, but how can we show that these are max and min values?

I also tried as follows:

By definition Determinant of a matrix is dot product of elements of any row with corresponding Cofactors

hence

$$Det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$$

By cauchy Scwartz Inequality we have

$$a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13} \le \left(\sqrt{a_{11}^2+a_{12}^2+a_{13}^2}\right)\left(\sqrt{C_{11}^2+C_{12}^2+C_{13}^2}\right)$$

any way to proceed here?

Answer 1

Let's assume the matrix is $$\textbf A=\begin{bmatrix}(-1)^a & (-1)^b & (-1)^c \\ (-1)^d & (-1)^e & (-1)^f \\ (-1)^g & (-1)^h & (-1)^i\end{bmatrix}$$ and thus $$\det(\textbf A)=(-1)^{a+e+i}+(-1)^{b+f+g}+(-1)^{c+d+h}-(-1)^{a+h+f}-(-1)^{b+d+i}-(-1)^{c+e+g}$$ From this we can see that $\det(\textbf A)$ is always even because switching 1 of (+1) with (-1) will change the sum by 2. Now as you have shown that $\det(\textbf A)=4$ is possible, we need to check for the possibility of 6. When the determinant is 6, we want, $$\begin{matrix}\textbf{even} &\textbf{odd} \\ a+e+i & a+h+f\\ b+f+g & b+d+i\\ c+d+h & c+e+g\end{matrix}$$ But now left side says that sum of a,b,c,d,e,f,g,h and i must be even and the right part says that it must be odd. Contradiction! $\longrightarrow \longleftarrow$

Answer 2

Let $\mathcal{E} = \{-1,1\}$. Given any $A \in {\rm Mat}_{3\times 3}(\mathcal{E})$, let $v_1, v_2, v_3 \in \mathcal{E}^3$ be its column vectors.
Notice

$$\det A = \det[ v_1, v_2, v_3 ] = 4 \det\left[ v_1, \frac{v_1+v_2}{2}, \frac{v_1+v_3}{2} \right]$$

and $\frac{v_1+v_2}{2}, \frac{v_1+v_3}{2} \in \mathbb{Z}^3$. We find $\det A = 4k$ for some $k \in \mathbb{Z}$.

When one expand $\det A$ completely, it is a sum of $6$ terms from $\mathcal{E}$. This implies $|\det A| \le 6$. As a result, $\det A$ can only take values $0, \pm 4$.

Since the value $4$ is achieved by matrix $\left[\begin{smallmatrix} +1 & +1 & +1\\ -1 & -1 & +1\\ +1 & -1 & +1 \end{smallmatrix}\right]$, the maximum value of such determinants equals to $4$.