hw2_solu - Solutions for Selected problems of Homework 2 A....

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Unformatted text preview: Solutions for Selected problems of Homework 2 A. Agarwal 1016-331 Linear Algebra-1 Spring 2009-10 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....
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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.

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hw2_solu - Solutions for Selected problems of Homework 2 A....

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