This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: M340L First Midterm Exam Solutions, February 18, 2010 1.Let A = 1 2 5 2 1 7 6 3 3 12 6 . a) Compute A rref . A rref = 1 3 4 1 1 2 . b) Find all solutions to A x = 0. The equations become z 1 = 3 x 3 4 x 4 , x 2 = x 3 + 2 x 4 , with x 3 = x 3 and x 4 = x 4 . In other words x = x 3  3 1 1 + x 4  4 2 1 , where x 3 and x 4 are arbitrary. c) Find all solutions to A x = 1 1 . This is similar, only with an extra term on the righthand side. Row reducing [ A  b ] gives 1 3 4   1 1 1 2  1  . This means z 1 = 3 x 3 4 x 4 1, x 2 = x 3 + 2 x 4 + 1, with x 3 = x 3 and x 4 = x 4 . In other words x = x 3  3 1 1 + x 4  4 2 1 +  1 1 , where x 3 and x 4 are arbitrary. d) Find all solutions to A x = 1 1 1 . Rowreducing [ A  b ] gives 1 3 4   1 1 1 2  1  1 , so there are no so lutions....
View Full
Document
 Spring '08
 PAVLOVIC
 Equations, Matrices

Click to edit the document details