<p>Let <br>\( A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \). If \( u_1 \) and \( u_2 \) are column matrices such that \( Au_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \) and \( Au_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \), then \( u_1 + u_2 \) is equal to</p>
<p>\(\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\)</p>
<p>\(\begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}\)</p>
<p>\(\begin{pmatrix} -1 \\ -1 \\ 0 \end{pmatrix}\)</p>
<p>\(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\)</p>
Step-by-Step Solution
Key Concept: To find u₁ and u₂, solve the linear systems Au₁ = e₁ and Au₂ = e₂ by exploiting that A is lower triangular—use forward substitution starting from the first equation. Then add the solutions.
<p><strong>Step 1:</strong> Recognize A is lower triangular with all diagonal entries = 1.</p><p><strong>Step 2:</strong> Solve Au₁ = [1, 0, 0]ᵀ by forward substitution:</p><p>From row 1: u₁₁ = 1</p><p>From row 2: 2u₁₁ + u₁₂ = 0 ⟹ u₁₂ = -2</p><p>From row 3: 3u₁₁ + 2u₁₂ + u₁₃ = 0 ⟹ 3 - 4 + u₁₃ = 0 ⟹ u₁₃ = 1</p><p>So u₁ = [1, -2, 1]ᵀ</p><p><strong>Step 3:</strong> Solve Au₂ = [0, 1, 0]ᵀ by forward substitution:</p><p>From row 1: u₂₁ = 0</p><p>From row 2: 2(0) + u₂₂ = 1 ⟹ u₂₂ = 1</p><p>From row 3: 3(0) + 2(1) + u₂₃ = 0 ⟹ u₂₃ = -2</p><p>So u₂ = [0, 1, -2]ᵀ</p><p><strong>Step 4:</strong> u₁ + u₂ = [1, -2, 1]ᵀ + [0, 1, -2]ᵀ = [1, -1, -1]ᵀ</p><p>∴ Answer: A</p>
Correct Answer: A