Let $A$ be a square matrix of order $3$ such that $a_{ij} \in \{1, 0, -1\}$. Find the minimum of $\det A$.

Question

In this problem, i'm thinking of divide this into 2 cases, which are 'the matrix is LI' and 'the matrix is not LI'. If the matrix is not LI, then we have its determinant is zero, so i only consider the second case.

So im thinking that, with a diagonal matrix of order 3 such as below:

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

With only 2 types of elementary row operation are:

• Multiply a row by -1 or 1.
• Add 1 time or -1 time a row to another.

I can generate any matrix of order 3 that satisfy given requirements. With the determinant of the new matrix is either 1 or -1 time the diagonal matrix. With that i conclude that its determinant will have the minimum of -1. But this just feel so wrong.

I really appreciate any help.

Answer 1

After an exhaustive search, it seems that the minimal possible determinant is $-4$. One matrix for which this minimum is attained is $$ A = \left(\begin{array}{ccc} -1 & -1 & -1\\ -1 & -1 & 1\\ 1 & -1 & -1 \end{array}\right). $$ Here is the Matlab code with which I arrived at this answer.

n = 3;

m = 0;
A_m = zeros(n);

for i = 0:3^(n^2)-1
    numstr = dec2base(i,3,n^2);
    A = zeros(n);
    for j = 0:n-1
        for k = 0:n-1
            A(j+1,k+1) = str2double(numstr(1 + 3*j + k));
        end
    end
    A = A-1;
    d = det(A);
    if d < m
        A_m = A;
        m = d;
    end
end

disp('minimal determinant:')
m
disp('minimizing matrix:')
A_m