The sum of the elements in each row of \(A \in \mathbb{R}^{n \times n}\) is 1. If \(B = A^{3} - 2A^{2} + A\), which one of the following statements is correct (for \(x \in \mathbb{R}^{n}\))?
Answer
(C)
Explanation
Let \(e = (1, 1, \dots, 1)^{T}\). Each row of \(A\) sums to 1, so \(Ae = e\); then \(A^{2}e = e\) and \(A^{3}e = e\) as well. Hence
\[Be = A^{3}e - 2A^{2}e + Ae = e - 2e + e = 0.\]
(Equivalently, \(B = A(A - I)^{2}\) and \((A - I)e = 0\).) So \(Bx = 0\) has the non-zero solution \(e\), and every multiple \(ce\) solves it too: infinitely many solutions.
- A fails: \(x = 0\) always solves \(Bx = 0\).
- B is impossible: the solutions form a subspace, so there is exactly one solution or infinitely many.
- D fails: \(e \ne 0\) is a second solution.
Answer: C