Analyzing the Setup
We begin with the quadratic equation x2+x−1=0. Its roots, α and β, serve as the building blocks of the set T={1,α,β}.
From Vieta's formulas, we know that
α+β=−1. This leads us to the master identity:
1+α+β=0
This identity is the heartbeat of the problem, confirming that the sum of the elements in set T is zero. This property will simplify all subsequent matrix operations.
The Matrix Puzzle
Part P
We construct a 3×3 matrix M where every row sum Ri and every column sum Cj equals zero. Because the sum of elements in T is zero, each row and column must contain exactly one instance of 1, α, and β.
The first row can be any permutation of these three elements, providing 3!=6 possibilities. For the second row, we must ensure no element is repeated in any column, which requires the second row to be a derangement of the first.
For three elements, there are exactly two such derangements. Once the first two rows are fixed, the third row is uniquely determined to satisfy the zero-sum condition. Thus, the total number of such matrices is:
6×2×1=12
Symmetry and Constraints
Part Q
Next, we consider symmetric matrices where Cj=0. A symmetric matrix satisfies M=MT, which implies Ri=Ci.
If the column sums are zero, the row sums are automatically zero. We seek the subset of our 12 matrices that are symmetric. For a matrix to be symmetric, the elements across the main diagonal must be equal.
If we fix the first row, there is only one way to arrange the remaining elements to maintain symmetry while satisfying the zero-sum condition. Since there are 6 choices for the first row, we find exactly 6 symmetric matrices.
The Skew-Symmetric Mystery
Part R
We now examine skew-symmetric matrices, where the diagonal elements are zero and aij=−aji. We consider the system of equations Mx=b.
Because M is a 3×3 skew-symmetric matrix, its determinant is zero. This implies that the system is linearly dependent.
When we expand the matrix multiplication, we obtain three linear equations. These equations are consistent, and because the determinant is zero, the system does not have a unique solution. Instead, it possesses infinite solutions.
The Determinant's Vanishing Act
Part S
Finally, we calculate the determinant of a matrix where Ri=0. We apply the column operation C1→C1+C2+C3.
Because the sum of elements in each row is zero, the new first column becomes a column of zeros. A matrix with a column of zeros has a determinant of zero.
Thus, the absolute value of the determinant is 0. This concludes our journey through the properties of roots, permutations, and matrix algebra.