Let A = <table><tr><td>1</td><td>0</td><td>0</td></tr><tr><td>2</td><td>1</td><td>0</td></tr><tr><td>3</td><td>2</td><td>1</td></tr></table>. If u<sub>1</sub> and u<sub>2</sub> are column matrices such that Au<sub>1</sub> = <table><tr><td>1</td></tr><tr><td>0</td></tr><tr><td>0</td></tr></table> and Au<sub>2</sub> = <table><tr><td>0</td></tr><tr><td>1</td></tr><tr><td>0</td></tr></table>, then u<sub>1</sub> + u<sub>2</sub> is equal to:
(A) <table><tr><td>1</td></tr><tr><td>-1</td></tr><tr><td>-1</td></tr></table>
(B) <table><tr><td>-1</td></tr><tr><td>1</td></tr><tr><td>0</td></tr></table>
(C) <table><tr><td>-1</td></tr><tr><td>1</td></tr><tr><td>-1</td></tr></table>
(D) <table><tr><td>-1</td></tr><tr><td>-1</td></tr><tr><td>0</td></tr></table>
Step-by-Step Solution
Key Concept: Solve the system of linear equations Au1 = b1 and Au2 = b2 using forward substitution or by finding the inverse of A.
Given A = [[1, 0, 0], [2, 1, 0], [3, 2, 1]]. <br> For Au1 = [1, 0, 0]^T: <br> x1 + 0y1 + 0z1 = 1 => x1 = 1 <br> 2x1 + y1 + 0z1 = 0 => 2(1) + y1 = 0 => y1 = -2 <br> 3x1 + 2y1 + z1 = 0 => 3(1) + 2(-2) + z1 = 0 => 3 - 4 + z1 = 0 => z1 = 1. So u1 = [1, -2, 1]^T. <br> For Au2 = [0, 1, 0]^T: <br> x2 + 0y2 + 0z2 = 0 => x2 = 0 <br> 2x2 + y2 + 0z2 = 1 => 2(0) + y2 = 1 => y2 = 1 <br> 3x2 + 2y2 + z2 = 0 => 3(0) + 2(1) + z2 = 0 => 0 + 2 + z2 = 0 => z2 = -2. So u2 = [0, 1, -2]^T. <br> u1 + u2 = [1+0, -2+1, 1-2]^T = [1, -1, -1]^T.
Correct Answer: 1