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= chpater3&4 MA351Spring2011= | = chpater3&4 MA351Spring2011= | ||
| + | This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help. | ||
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| + | 3.1 11 Ax = 0 to find the span,very basic but very important | ||
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| + | 3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already. | ||
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| + | 3.2 4 Fact 3.2.2. | ||
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| + | 3.2 12 Linearly dependent | ||
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| + | 3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first. | ||
| + | 3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3 | ||
| − | + | 4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't. | |
| + | |||
| + | 4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1. | ||
Latest revision as of 14:35, 30 April 2011
chpater3&4 MA351Spring2011
This is the review for Chapter 3 and 4. The problems posted here are the one that I consider hard. Hope it can help.
3.1 11 Ax = 0 to find the span,very basic but very important
3.115 By Fact 3.1.3, the image of A is the span of the columns of A, any two of these vectors span all of R2 already.
3.2 4 Fact 3.2.2.
3.2 12 Linearly dependent
3.3 5 The first two vectors are non-redundant, but the third is a multiple of the first.
3.4 17 By inspection, we see that in order for x to be in V , x = 1v1 + 1v2 + 1~v3
4.1 11 Not a subspace: I3 is in rref, but the scalar multiple 2 I3 isn't.
4.2 53 Thus the kernel consists of all constant polynomials f(t) = a(when b = c = 0), and the nullity is 1.
