Linear Algebra Solutions

Linear Algebra Solutions - 1 1.1 SOLUTIONS Notes The key...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 1.1 SOLUTIONS Notes : The key exercises are 7 (or 11 or 12), 1922, and 25. For brevity, the symbols R1, R2,, stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section. 1 . 1 2 1 2 5 7 2 7 5 x x x x + = = 1 5 7 2 7 5 Replace R2 by R2 + (2)R1 and obtain: 1 2 2 5 7 3 9 x x x + = = 1 5 7 3 9 Scale R2 by 1/3: 1 2 2 5 7 3 x x x + = = 1 5 7 1 3 Replace R1 by R1 + (5)R2: 1 2 8 3 x x = = 1 8 1 3 The solution is ( x 1 , x 2 ) = (8, 3), or simply (8, 3). 2 . 1 2 1 2 2 4 4 5 7 1 1 x x x x + = + = 2 4 4 5 7 11 Scale R1 by 1/2 and obtain: 1 2 1 2 2 2 5 7 1 1 x x x x + = + = 1 2 2 5 7 11 Replace R2 by R2 + (5)R1: 1 2 2 2 2 3 2 1 x x x + = = 1 2 2 3 2 1 Scale R2 by 1/3: 1 2 2 2 2 7 x x x + = = 1 2 2 1 7 Replace R1 by R1 + (2)R2: 1 2 12 7 x x = = 1 12 1 7 The solution is ( x 1 , x 2 ) = (12, 7), or simply (12, 7). 2 CHAPTER 1 Linear Equations in Linear Algebra 3 . The point of intersection satisfies the system of two linear equations: 1 2 1 2 5 7 2 2 x x x x + = = 1 5 7 1 2 2 Replace R2 by R2 + (1)R1 and obtain: 1 2 2 5 7 7 9 x x x + = = 1 5 7 7 9 Scale R2 by 1/7: 1 2 2 5 7 9/7 x x x + = = 1 5 7 1 9/7 Replace R1 by R1 + (5)R2: 1 2 4/7 9/7 x x = = 1 4/7 1 9/7 The point of intersection is ( x 1 , x 2 ) = (4/7, 9/7). 4 . The point of intersection satisfies the system of two linear equations: 1 2 1 2 5 1 3 7 5 x x x x = = 1 5 1 3 7 5 Replace R2 by R2 + (3)R1 and obtain: 1 2 2 5 1 8 2 x x x = = 1 5 1 8 2 Scale R2 by 1/8: 1 2 2 5 1 1/4 x x x = = 1 5 1 1 1 / 4 Replace R1 by R1 + (5)R2: 1 2 9/4 1/4 x x = = 1 9/4 1 1/4 The point of intersection is ( x 1 , x 2 ) = (9/4, 1/4). 5 . The system is already in triangular form. The fourth equation is x 4 = 5, and the other equations do not contain the variable x 4 . The next two steps should be to use the variable x 3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with 5 times R3. 6 . One more step will put the system in triangular form. Replace R4 by its sum with 3 times R3, which produces 1 6 4 1 2 7 4 1 2 3 5 1 5 . After that, the next step is to scale the fourth row by 1/5....
View Full Document

This note was uploaded on 06/15/2011 for the course M 340L taught by Professor Pavlovic during the Fall '08 term at University of Texas.

Page1 / 423

Linear Algebra Solutions - 1 1.1 SOLUTIONS Notes The key...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online