1
more than 2
0
2
Let A be a 3x3 symmetric matrix with entries 0 or 1. Five entries are 1 and four entries are 0.
The system of linear equations AX = (1, 0, 0)^T is inconsistent if and only if the rank of the augmented matrix [A | (1, 0, 0)^T] is greater than the rank of A. Since A is a 3x3 matrix, this means that rank(A) < 3 and rank([A | (1, 0, 0)^T]) = 3.
Let A = [[a, b, c], [b, d, e], [c, e, f]]. Since A is symmetric, there are 6 entries. Five are 1 and four are 0. This is a contradiction. There must be a mistake in the problem statement. Let's assume that the question intended to state that there are 6 entries, 5 are 1 and 1 is 0.
If det(A) = 0, then the system is inconsistent if (1,0,0) is not in the column space of A. Let's consider the case where the first column of A is (0, b, c). Then for the system to be inconsistent, we need the first column of A to be linearly dependent on the second and third columns. This implies that the first column is all zeros. The only way for a 3x3 symmetric matrix with 5 ones and 4 zeros to have a column of zeros is if the entire matrix is diagonal.
If A is a diagonal matrix, then AX = (1,0,0) has a solution if and only if the first diagonal entry is 1. Thus, the system is inconsistent if the first entry is 0. The other entries must be 1 to satisfy the condition that five entries are 1 and one is 0.
If A is a diagonal matrix with the first entry 0, then A = [[0, 0, 0], [0, 1, 0], [0, 0, 1]]. In this case, AX = (1, 0, 0)^T is inconsistent.
If A has a column of all zeros, then the system is inconsistent if and only if the first entry of the right hand side vector is 1. In a symmetric matrix, this means that the first row and first column are all zeros. Since we have five 1s and one 0, this is only possible if the first row and column are all zeros, the rest of the entries are 1. This corresponds to one matrix.
Therefore, there is only 1 such matrix A for which the system is inconsistent.