MATH70 final Math70-FinalReviewANS

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Published on 31 Jan 2019
Department
Course
Professor
Math 46
Solutions for Final Exam Review
1. In the interests of brevity, I’m not going to write out either the question
or the answer to the various parts of problem one. The answers to the parts
(a), (b), and (c) should all be in your notes, and as for part (d), well ... I
don’t think I should answer that just yet. Maybe after the exam.
2. Let AMm×n. Assume that for all xRnthat Ax=0. Prove that A
is the zero matrix.
Solution: Well, we know that Ax=0for every single vector xRn. That
means we can pick any xwe want, and plug it in. Me, I want to pick x=ei.
Because then we see that Ax=Aeiis the ith column of A. But that’s equal
to 0by hypothesis! So every column of Ais zero, so the whole matrix has
to be zero.
3. Solve the following linear system:
2x+y2z= 10
3x+ 2y+ 2z= 1
5x+ 4y+ 3z= 4
(a) by row reduction.
(b) by Cramer’s Rule.
Solution: I’m going to omit the details of this, since you can see multitudi-
nous examples of row reduction and Cramer’s Rule in the textbook and in
your notes. However, the answer is that there is only one solution to this
system, and it’s given by x= 1, y= 2, and z=3.
4. Let Abe an m×nmatrix and let TA:RnRmbe defined by TA(x) = Ax.
Are the following statements true or false? If true, give a proof. If false,
explain why.
(a) dim NulAn.
Solution: True. The null space of Ais a subspace of Rn, so its dimension
has to be at most n.
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(b) rankAm.
Solution: True. The rank of Ais equal to the number of pivots in the
reduced row-echelon form of A. Since each pivot is in a different row, the
number of pivots is no more than the number of rows of A, which is m.
(c) If n > m then the linear transformation TAcannot be one-to-one.
Solution: True. Recall that TAis one-to-one if and only if Ahas a pivot
in every column (after row-reduction). This is only possible if Ahas at least
as many rows as columns ... which means that if TAis one-to-one, then we
can’t have n > m.
(d) If n < m then TAcannot be onto.
Solution: True. Recall that TAis onto if and only if Ahas a pivot in every
row (after row-reduction). This is only possible if Ahas at least as many
columns as rows ... which means that if TAis onto, then we can’t possibly
have n < m.
(e) If TAis one-to-one and m=n, then TAmust be onto.
Solution: True. (Yes, all the answers to question four were “true”.) If
m=n, then Ais a square matrix. If Ais square, then the IMT applies, and
that means that TAis onto if and only if it’s one-to-one. So we’re done.
5. Let Abe an n×nmatrix satisfying A3=In, where Inis the n×nidentity
matrix. Show that det A= 1.
Solution: Since A3=I, we know that det A3= det I= 1. This means that
(det A)3= 1, since det(AB) = det(A) det(B) for any two matrices Aand B.
But if (det A)3= 1, then assuming that the entries in Aare real numbers
(which, in this course, they always will be), we must have det A= 1, as
desired.
6. For the following problems, prove the statement or give a specific coun-
terexample:
(a) W={pP2|(p(3))2+p(3) = 0}is a subspace of P3.
Solution: This is obviously not true, since Wis defined to be a set of
polynomials in P2, so it can’t be a subspace of P3.
If you ignore that typo, and ask if Wis a subspace of P2instead, then
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the answer is still no. To check if Wis a subspace, you need to check three
things: the existence of 0W, closure under addition, and closure under
scalar multiplication. As it turns out, the zero vector is in W, but Wfails
the second and third tests.
For example, let p(t) = 1, the constant polynomial. Then certainly
(p(3))2+p(3) = (1)2+ (1) = 1 1 = 0, so p(t)W. But if we multiply
p(t) by the scalar 3, we see that (3p(3))2+ 3p(3) = 9 + 3(1) = 6 6= 0. Thus,
3p(t) is not in W, so Wis not closed under scalar multiplication.
(b) W={(x, y)R2|x+y= 0}is a subspace of R2.
Solution: This is true. Recall that to check if Wis a subspace of R2, we
need to check three things: the existence of 0W, closure under addition,
and closure under scalar multiplication. Let’s check them, one by one.
First, it is clear that 0= (0,0) is in W. This follows immediately from
the fact that 0 + 0 = 0.
Second, we need to check that Wis closed under addition. To do this,
pick any two vectors (x1, y1) and (x2, y2) in W. We want to show that
(x1, y1) + (x2, y2) = (x1+x2, y1+y2) is also in W. This means showing that
x1+x2+y1+y2= 0, because that’s the definition of W.
But we already know that x1+y1= 0, because (x1, y1)W. And
x2+y2= 0, because (x2, y2)W. So that means that x1+x2+y1+y2= 0,
so (x1+x2, y1+y2)W, as desired.
Lastly, we need to check that Wis cloesd under scalar multiplication. To
do this, pick any vector (x, y)W, and pick any scalar αR. We want to
show that α(x, y) = (αx, αy) is in W.
This amounts to showing that αx+αy = 0. But that follows immediately
from the fact that x+y= 0 (which is the definition of what it means for
(x, y) to be in W), so we’re done.
Since we’ve shown that Wsatisfies all three criteria for being a subspace
of R2, we’re done.
(c) Let Vbe a vector space and let f1:VRand f2:VRbe linear
transformations. Define T:VR2by T(v) = (f1(v), f2(v)). Then Tis
linear.
Solution: This is true.
In order to prove that Tis linear, we need to check two things: that T
respects addition, and that it respects scalar multiplication.
First, we need to show that Trespects addition. This means showing
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Document Summary

Solutions for final exam review: in the interests of brevity, i"m not going to write out either the question or the answer to the various parts of problem one. The answers to the parts (a), (b), and (c) should all be in your notes, and as for part (d), well I don"t think i should answer that just yet. Maybe after the exam: let a mm n. Assume that for all x rn that ax = 0. Solution: well, we know that ax = 0 for every single vector x rn. That means we can pick any x we want, and plug it in. Me, i want to pick x = ei. Because then we see that ax = aei is the ith column of a. So every column of a is zero, so the whole matrix has to be zero: solve the following linear system:

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