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: Solutions for Selected problems of Homework 2 A. Agarwal 1016331 Linear Algebra1 Spring 200910 1. Q14,P.41 Since points P,Q and R are all in the plane, this means the vectors PQ and PR will lie in the plane. PQ = (0 , 1 , 0) (1 , , 0) = ( 1 , 1 , 0) and PR = (0 , , 1) (1 , , 0) = ( 1 , , 1) So now we have a point P and two direction vectors PQ and PR , so we can use the definition of the vector form of the plane given on page 36 to get x y z = 1 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright hop on the point P + s 1 1 + t 1 1 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright cruise along the directions PQ and PR in the plane 2. Q18,P.83 To show A is row equivalent to B , we need to show that (see theorem 2.1 ) they both can be reduced to the same row ech e lon form. The following shows the row reduction for A : 2 0 1 1 1 1 1 1 R 1 clockwise R 2 1 1 2 0 1 1 1 1  2 R 1 + R 2 R 1 + R 3 1 1 2 1 2 1 R 2 + R 3 1 1 2 1  1( R 2 ) 1 1 0 0 2 1 0 0 0 Likewise we will row reduce B as follows: 3 1 1 3 5 1 2 2 R 1 clockwise R 3 2 2 3 5 1 3 1 1 1 2 R 1 1 1 3 5 1 3 1 1  3 R 1 + R 2  3 R 1 + R 3 1 1 2 1 2 1 R 2 + R 3 1 1 0 0 2 1 0 0 0 The row echelon form of the two matrices are same, hence A and B are row equivalent....
View
Full
Document
This note was uploaded on 05/11/2010 for the course MTH 1016.331 taught by Professor Dr.anuragagarwal during the Spring '10 term at RIT.
 Spring '10
 Dr.AnuragAgarwal
 Linear Algebra, Algebra, Vectors

Click to edit the document details