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MA 351 Homework 8

3.2 #24

When a vector [V] is in the span of Ker(A), it means that the linear transformation of [V]([A])=the zero vector.

So... The vector that makes the vector A zero is in the span of Ker(A)

3.2 #28

Use theorem 3.2.4.

But first determine whether each column is linearly independent.

3.2 #45

Use summary 3.1.8 on Pg. 109

Note that ker(A)=zero vector, that means all columns in A are linearly independent.

3.3 #28

to form a basis of R4, the RREF of A must be I4.


MA 351 Homework 9 (10/28/2010)

3.3 #30 reverse solve x1, x2, x3 and x4 in terms of r,s,t. Then take out the r,s and t as coefficients, the matrix becomes your basis.

3.3 #69 Eliminate the non-independent row. then all the linearly independent rows will form the basis of the row space.

3.4 #2 [x]B=[c1

     c2]         so that c1v1+c2v2=x 

3.4 #14 Same principle as #2

3.4 #16 Same principle as #2

3.4 #20 a. Use the equation S-1AS. S=[v1 v2]

c. Follow example 4

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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