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hw2_solu

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

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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 , 0) = ( 1 , 1 , 0) and −→ PR = (0 , 0 , 1) (1 , 0 , 0) = ( 1 , 0 , 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 0 0 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright hop on the point P + s 1 1 0 + t 1 0 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 0 1 1 1 R 1 clockwise R 2 −−−−→ 1 1 0 2 0 1 1 1 1 - 2 R 1 + R 2 −−−−−−→ R 1 + R 3 1 1 0 0 2 1 0 2 1 R 2 + R 3 −−−−→ 1 1 0 0 2 1 0 0 0 - 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 0 R 1 clockwise R 3 −−−−→ 2 2 0 3 5 1 3 1 1 1 2 R 1 −−−→ 1 1 0 3 5 1 3 1 1 - 3 R 1 + R 2 −−−−−−→ - 3 R 1 + R 3 1 1 0 0 2 1 0 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. 1

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3. Q28,P.84 The augmented matrix of the system is as follows: 2 3 1 4 0 3 1 0 1 1 3 4 1 1 2 After performing the row operations (I’m skipping the row operation steps here). We get 2 3 1 4 0 0 11 2 3 2 5 1 0 0 2 11 8 11 5 11 There are three piv ots (cor re spond ing to the vari ables w,x and y ) but
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