<p>Let \(A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}\). 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: Since A is lower triangular with 1's on diagonal, A is invertible (det = 1). Find u₁ and u₂ by solving Au₁ = e₁ and Au₂ = e₂ using back-substitution, which is efficient for triangular systems.
<p><strong>Step 1:</strong> Solve Au₁ = (1, 0, 0)ᵀ using back-substitution on the lower triangular system:</p><p>Row 1: u₁₁ = 1</p><p>Row 2: 2(1) + u₂₁ = 0 ⟹ u₂₁ = -2</p><p>Row 3: 3(1) + 2(-2) + u₃₁ = 0 ⟹ u₃₁ = 1</p><p>∴ u₁ = (1, -2, 1)ᵀ</p><p><strong>Step 2:</strong> Solve Au₂ = (0, 1, 0)ᵀ using back-substitution:</p><p>Row 1: u₁₂ = 0</p><p>Row 2: 2(0) + u₂₂ = 1 ⟹ u₂₂ = 1</p><p>Row 3: 3(0) + 2(1) + u₃₂ = 0 ⟹ u₃₂ = -2</p><p>∴ u₂ = (0, 1, -2)ᵀ</p><p><strong>Step 3:</strong> u₁ + u₂ = (1, -2, 1)ᵀ + (0, 1, -2)ᵀ = <strong>(1, -1, -1)ᵀ</strong></p><p>∴ Answer: A</p>
Correct Answer: A