Find maximum determinant of $3 \times 3$ matrix whose entries are $1$ or $2$.
This problem was from an old algebra exam at my university. After some coding, I found the answer to be $5$. But I am not sure how to prove it on paper. In case the entries are $0$ or $1$, I know how to prove.
I tried some case work but it's too long and there are too many cases to consider. But maybe I didn't know how to split the cases optimally.
Any help would be very appreciated. Thanks in advance.
Call your matrix $A$. Its maximum possible determinant is positive (because $\det I_3$ is already positive). Denote by $m_{ij}$ the minor obtained by deleting the $i$-th row and the $j$-th column of $A$. By Laplace expansion along row $i$, we have $$ \det A=\sum_ja_{ij}\left[(-1)^{i+j}m_{ij}\right]. $$ Given an optimal $A$, we must have $a_{ij}=2$ if $(-1)^{i+j}m_{ij}>0$, and $a_{ij}=1$ if $(-1)^{i+j}m_{ij}<0$. If $m_{ij}=0$, we can modify the value of $a_{ij}$ to $2$ without changing the determinant, but the number of $2$s in $A$ is increased. Since $A$ has only nine elements, if we keep doing such modification, we will reach, in finite steps, a modified optimal $A$ such that for all $(i,j)$, $$ (-1)^{i+j}m_{ij}\text{ is } \begin{cases} \ge0&\text{iff }a_{ij}=2,\\ <0&\text{iff }a_{ij}=1.\\ \end{cases} $$ Now look at this modified optimal $A$. Since all elements of $A$ are either $1$ or $2$, we have $|m_{ij}|\le3$. Sort any row of $A$ in ascending order. If some sorted row of $A$ is $(1,1,1)$ or $(1,1,2)$, then $\det A\le 1(-1)+1(-1)+2(3)=4$. The similar holds for any sorted column of $A$.
If, instead, all sorted rows/columns of $A$ are either $(1,2,2)$ or $(2,2,2)$, since $A$ has no repeated rows/columns (because the optimal determinant is positive), $A$ must be equal to either $$ P\ \underbrace{\pmatrix{1&2&2\\ 2&1&2\\ 2&2&1}}_{\det=5}\ Q \quad\text{or}\quad P\ \underbrace{\pmatrix{1&2&2\\ 2&1&2\\ 2&2&2}}_{\det=2}\ Q $$ for some permutation matrices $P$ and $Q$. Hence the maximum possible value of $\det A$ is $5$ and every modified optimal $A$ is in the form of $P(2ee^T-I_3)Q$ for some permutation matrices $P,Q$ such that $\det(PQ)=1$. Since all two-rowed minors of $2ee^T-I_3$ are nonzero, the unmodified optimal $A$s are also in the same form. That is, the optimal solution is unique up to appropriate permutations of rows and columns of $2ee^T-I_3$.
P.S. There is no way I could cook up this proof during an examination.
I asked my professor and I got this answer. We separate the proof into $3$ cases. Since we can switch the rows (columns) and only make the determinant change signs. We can combine a lot of symmetries into these cases.
Case $1$: There is a row (column) that is a multiple of $\begin{pmatrix} 1 & 1 & 1\end{pmatrix}$, then we have:
$|A| = \lambda\begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ d & e & f\end{vmatrix} = \lambda\begin{vmatrix}1 & 0 & 0 \\ a & b - a & c - a \\ d & e - d & f- d\end{vmatrix} \le \lambda \cdot 2 = 4$
Case $2$: There exists a row (column) that is a permutation of $\begin{pmatrix} 1 & 1 & 2\end{pmatrix}$, then we have:
$|A| = \begin{vmatrix}1 & 1 & 2 \\ a & b & c \\ d & e & f\end{vmatrix} = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ d & e & f \end{vmatrix} + \begin{vmatrix}0 & 0 & 1 \\ a & b & c \\ d & e & f \end{vmatrix} \le 2 + 3 = 5$
Case 3: The rows are $\begin{pmatrix} 1 & 2 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 2 & 1 \end{pmatrix}$. The matrix will now have the form:
$|A| = \begin{vmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{vmatrix} = 5$
The rows (columns) shouldn't repeat since the determinant will be zero. All other matrix will have $5$ or $-5$ as its determinant since we can switch the rows and columns around.
After all cases considered, the maximum determinant is $5$.
Let $A$ be an arbitrary $3 \times 3$ matrix with entries $0,1$, so $B = A+\mathbf{1}\mathbf{1}^\top$ is an arbitrary $3 \times 3$ matrix with entries $1,2$. Here $\mathbf{1}$ refers to the length $3$ column vector with all $1$'s and $\mathbf{1} \mathbf{1}^\top$ is an outer product, namely the $3 \times 3$ matrix with all $1$'s.
By the matrix determinant lemma, we have $$\det(B) = \det(A) + \mathbf{1}^\top \mathrm{adj}(A) \mathbf{1}.$$
By this post, $\det(A)$ has a maximum of $2$, which can be obtained at $$A = \begin{pmatrix}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix} \qquad\Rightarrow\qquad B = \begin{pmatrix}2 & 1 & 2 \\ 2 & 2 & 1 \\ 1 & 2 & 2\end{pmatrix}.$$
This $B$ has determinant $5$ and is a solution to the original problem.
Of course, it's not clear that a maximum for $\det(A)$ will automatically maximize $\det(B)$. The correction term $\mathbf{1}^\top \mathrm{adj}(A) \mathbf{1}$ is the sum of all the terms of the adjugate matrix of $A$. It's easy to see this adjugate must be a $-1,0,1$-matrix, so the sum of its entries are at most $9$, giving $\det(B) \leq 2+9=11$.
The collection $\mathrm{adj}(A)$ here is quite constrained relative to all $3 \times 3$ matrices of $-1,0,1$'s. Brute force says it has a maximum of $3$, obtained only at cyclic reorderings of $A$ above or the identity. That would give $\det B \leq 5$, and the example above would be provably the maximum up to cyclic reordering.
Edit: Here's a way to show that correction term is at most $3$, hence showing $B$ above is maximal. It involves a little calculation, though in principle is doable by hand.
Let $r_i$ be the $i$th row of $A$ and let $c_k$ be the $k$th column of $\mathrm{adj}(A)$. Let $s_k$ be the sum of the coordinates of $c_k$. We want to show $s_1 + s_2 + s_3 \leq 3$.
First we claim for each $i \neq k$ that $s_k$ is at most the number of zero's in $r_i$. To see this, recall the fundamental property of adjugates, $A \mathrm{adj}(A) = \det(A) I_3$, so $r_i c_k = \delta_{ik} \det(A) = 0$. Let $s_k'$ (respectively, $s_k''$) be the sum of the coordinates of $c_k$ for which the corresponding coordinate of $r_i$ is $0$ (respectively, $1$). Then $s_k = s_k' + s_k''$, but $s_k'' = r_i c_k = 0$, so $s_k = s_k'$. Since $c_k$ consists of $-1,0,1$'s, the claim follows.
Now suppose $A$ has a row of all $0$'s. This forces six complementary entries of $\mathrm{adj}(A)$ to be zero, so $s_1+s_2+s_3 \leq 3$.
Next suppose $A$ has two rows each with two $0$'s. The possible $A$'s are, up to reordering, $$A = \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & c\end{pmatrix} \qquad \Rightarrow \qquad \mathrm{adj}(A) = \begin{pmatrix}-b & b & 0 \\ a & -a & 0 \\ 0 & 0 & 0\end{pmatrix}$$ or $$A = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ a & b & c\end{pmatrix} \qquad \Rightarrow \qquad \mathrm{adj}(A) = \begin{pmatrix}-b & -c & 1\\a & 0 & 0 \\ 0 & a & 0\end{pmatrix}.$$ In the first case, $s_1+s_2+s_3=0$. Reordering the rows or columns of $A$ does not change this. In the second case, $s_1+s_2+s_3=1+2a-b-c \leq 3$. Reordering preserves this inequality.
Now we may assume $A$ has at most one row with two $0$'s, with the others having at most one $0$. From the claim, we have $s_k \leq 1$ for each $k$, so $s_1+s_2+s_3 \leq 3$, completing the argument.
This answer is based on Sarrus's calculation of determinant for $3 \times 3$ matrices.
https://en.wikipedia.org/wiki/Rule_of_Sarrus
Rule of Sarrus is diagonally oriented calculation process so in considerations below diagonals play the most important role.
The first phase - presentation of the form of matrix to consider.
In desired matrix we know that there are $1$s, more than one 1 element.
We should check at least two cases:
at least two $1$ values in different row and column,
at least three $1$ values in different rows and columns,
so let us place them on the main diagonal, (if these $1$s are located on other positions always we can group them via permutations which generally don't affect absolute value of determinant - we know that permutation matrix has determinant $1$ or $-1$ what is sufficient for this problem),
other elements can be assumed as unknowns.
We can start from three $1$s.
Notice that we want only 3 potentially diagonal (i.e. which can be shifted to the diagonal via permutations diagonal) elements with value 1.
$$A = \begin{pmatrix}1 & a & b \\ d & 1 & c \\ e & f & 1\end{pmatrix} $$
Presentation of such form of matrix ends first phase of considerations.
The second phase concentrates on algebraic formula for calculations of determinant.
For considered matrix determinant is equal
$\det(A)=1+dfb+eac-ad-be-cf $.
From this form it is visible that we have to maximize $dfb$ and $eac$ so their factors have to be equal $2$.
This gives value of determinant equal $5$.
For clarification, for this case
$\det(A)=1+2 \times 2^3-3 \times 2^2$
$1$ plus $2$ cubes minus $3$ squares. The fact that we have here cubes vs. squares is crucial so the problem was reduced to evaluation of algebraic expression with values from the set $\{1,2\}$.
The second case with two $1$s on the main diagonal can be checked in similar way and it gives determinant below $5$.
$$A_2 = \begin{pmatrix}2 & a & b \\ d & 1 & c \\ e & f & 1\end{pmatrix} $$
$\det(A_2)=2+dfb+eac-ad-be-2cf=(1+dfb+eac-ad-be-cf)+(1-cf) $.
The case with only one potentially diagonal $1$ can be neglected as it leads (what is easy to check) to the determinant equal $0$.
Concluding:
the method does not only solves the problem but also it can generalize problem for matrices $3 \times 3$ with other two values:
for example: ($1$ and $10$)
then maximal $\det(A)= 1+2\times 10^3 - 3\times 10^2=1701$
or ($3$ and $7$) or even ($e$ and $2\pi$).
