This preview shows pages 1–3. 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: Math 167 homework 1 solutions October 6, 2010 1.3.14a: Construct a 3 by 3 system that needs two row exchanges to reach a triangular form and a solution. There are many examples, but this is one: 3 y 2 z = 12 z = 3 4 x +3 y = 3 1.3.14b: Construct a 3 by 3 system that needs a row exchange to keep going, but breaks down later. 3 y 2 z = 12 4 x +3 y = 3 4 x +3 y = 1 1.3.18a: It is impossible for a system of linear equations to have exactly two solutions. If ( x, y, z ) and ( X, Y, Z ) are two solutions, what is another? Again, there are many solutions, i.e. any point on the line generated by the two given points. For example ( x + X 2 , y + Y 2 , z + Z 2 ) 1.3.18b: If 25 planes meet at two points, where else do they meet? These plane most also intersect at the line containing these two points. 1.3.20: Find the pivots and the solution for these four equations: 2 x + y = 0 x +2 y + z = 0 y +2 z + t = 0 z +2 t = 5 Using Gaussian Elimination we get 1 2 x + y = 0 3 y +2 z = 0 4 z +3 t = 0 5 t = 20 So the pivots are 2 , 3 , 4 , and 5. Using backsubstitution we find that ( 1 , 2 , 3 , 4) is the solution to the system. 1.3.26: Solve by elimination the system of two equations 2 x y = 0 3 x +6 y = 18 Draw a graph representing each equation as a straight line in the x y plane; the lines intersect at the solution. Also, add one more linethe graph of thethe lines intersect at the solution....
View Full
Document
 Fall '11
 Wiley
 Math, Differential Equations, Equations

Click to edit the document details