Let's break down the problem step by step: Given: - \( n = 1, 2, \ldots, 20 \) - \( x \in \mathbb{R}^{20} \) is a vector where the \( n \)th entry is 20 and all other entries are 1. - \( A \) is a \( 20 \times 20 \) matrix where all diagonal entries are 1 and all off-diagonal entries are 2. - The reduced echelon form (RREF) of \( A \) is the \( 20 \times 20 \) identity matrix. **Question:** For which of the following vectors \( b \in \mathbb{R}^{20} \) is the matrix equation \( Ax = b \) consistent? Let's analyze the matrix \( A \): ### Step 1: Structure of \( A \) - Diagonal entries: 1 - Off-diagonal entries: 2 So, \( A \) looks like: \[ A = \begin{bmatrix} 1 & 2 & 2 & \cdots & 2 \\ 2 & 1 & 2 & \cdots & 2 \\ 2 & 2 & 1 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \cdots & 1 \end{bmatrix} \] ### Step 2: RREF of \( A \) is the Identity Matrix Since the RREF is the identity, \( A \) is invertible. **This means for every \( b \in \mathbb{R}^{20} \), there exists a unique solution \( x \) to \( Ax = b \).** ### Step 3: Options (a) The vector \( b \) whose entries are all equal to 20 (b) The vector \( b \) whose entries are all equal to 21 (c) The vector \( b \) whose entries alternate between 1 and 20 (d) All of the above (e) None of the above ### Step 4: Conclusion Since \( A \) is invertible, **for any vector \( b \), \( Ax = b \) is consistent**. **Correct Answer: (d) all of the above**