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to form a basis of R4, the RREF of A must be I4. | to form a basis of R4, the RREF of A must be I4. | ||
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| + | MA 351 Homework 9 (10/28/2010) | ||
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| + | 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. | ||
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| + | 3.3 #69 | ||
| + | Eliminate the non-independent row. then all the linearly independent rows will form the basis of the row space. | ||
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| + | 3.4 #2 | ||
| + | [x]B=[c1 | ||
| + | c2] so that c1v1+c2v2=x | ||
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| + | 3.4 #14 | ||
| + | Same principle as #2 | ||
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| + | 3.4 #16 | ||
| + | Same principle as #2 | ||
| + | |||
| + | 3.4 #20 | ||
| + | a. Use the equation S-1AS. S=[v1 v2] | ||
| + | |||
| + | c. Follow example 4 | ||
Latest revision as of 08:35, 28 October 2010
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
