Solutions to 33A Practice Midterm #1
Sam_Lai
I expect to have made some mistake somewhere, so these are probably not the definitive answers. I am especially unsure about 7,8,9. I'm just putting this out there. \\ 1) Row Reduce \\ x = -7w - 34 \\ y = -w - 5 \\ z = 4w + 21 \\ 2) Row Reduce \\ dim(im(A)) = 3 \\ dim(ker(A)) = 1 by rank nullity theorem \\ ker(A) = span $\bigl(\begin{smallmatrix} 7/6 \\ -17/3\\1/2\\1 \end{smallmatrix} \bigr)$ \\ 3) \\ \begin{matrix} 2/3 & -1/2 & 1/3 \\ -8/3 & 2 & -1/3 \\ 1 & -1/2 & 0 \end{matrix} \\ 4) Linear, not linear, not linear, linear \\ 5) Im(A) = span {(1,3,1), (2,-2,6)} \\ ker(A) = span {(-5/4,1/8,0)} \\ 6) check for linear independence \\ 7) \\ \begin{matrix} (9/4)a + (1/4)b \\ 2a \end{matrix} \\ 8) \begin{matrix} 1/2 & 1/2 & 0 \\ 1 & -2 & -1 \\ -3/2 & 3/2 & -1/2 \end{matrix} \\ 9) \begin{matrix} 1/5 & -2/5 & 2/5 \\ -1/5 & 2/5 & 1/10 \\ 3/5 & -1/5 & -3/10 \end{matrix}
connorkenny: May 4, 2015, 2:12 p.m.
I think that number 5 should read "Im(A) = span {(1,3,1), (2,-2,-6)}" The original question has a negative sign in front of the 6 in the second vector. I'm fairly sure this matters, but please let me know if it doesn't.
Brianlui: May 4, 2015, 2:23 p.m.
In number 5, I believe that the ker(A) should be the span{ ( -5/4 , 1/8 , 1 ) }.
carRobinson: May 4, 2015, 2:27 p.m.
For 8 I got 1 -2 1/2 1 1 1 -2 4 0
carRobinson: May 4, 2015, 2:27 p.m.
whoa that did not come out right, give me a second to fix that
carRobinson: May 4, 2015, 2:33 p.m.
I can't figure how to make it into a nice looking matrix so whatever: the first row is (1,-2, 1/2) then the 3rd row below that is (1,1,1) and then the final row below that is (-2,4,0).
nabihahshuaib: May 4, 2015, 4:53 p.m.
I got the same answer as carRobinson for #8
Sam_Lai: May 4, 2015, 5:01 p.m.
I agree with connorkenny and Brianlui. Unfortunately, I can't edit the post. \\ I also have some corrections to make to questions 7 and 9. \\ I think the answer to 7 should be $\bigl(\begin{smallmatrix} (7/2)a+(1/2)b \\ 2a \end{smallmatrix} \bigr)$ \\ Also I think the answer to 9 should be $\bigl(\begin{smallmatrix} 3&33/5&21/5 \\ 5/2 & 52/5 & 19/5 \\ -3/2 & -41/5 & -12/5 \end{smallmatrix} \bigr)$ \\ Of course, I could still be wrong. \\ carRobinson, can you explain how you got your result? I did number 8 by finding what the transformation does to $\beta$ and then finding the identity matrix from $\beta$ to $\gamma$ and multiplying the two matrices together.
connorkenny: May 4, 2015, 5:15 p.m.
How do you set up number 7? I have tried so many times and can't seem to get it right.
Sam_Lai: May 4, 2015, 5:18 p.m.
In light of the comments about question #8, I redid the problem and arrived at the same answer. Thanks everyone.
elaine_lei: May 4, 2015, 5:29 p.m.
I got the same answer for number 3.
joseph_koh: May 4, 2015, 8:51 p.m.
I think that you forgot to include the vector that maps to zero when multiplied by A in problem #2.
joseph_koh: May 4, 2015, 9:03 p.m.
I also got the same answer for #3.
joseph_koh: May 4, 2015, 9:05 p.m.
Oops I just realized that you did solve for it in #2. My bad.
TLindberg: May 4, 2015, 10:53 p.m.
For #8 I got a different answer for my matrix. I got: \begin{pmatrix} 1 & $\frac{-1}{2}$ & $\frac{1}{2}$ \\ 1 & -2 & 1 \\ -2 & 4 & 0 \end{pmatrix} I did this by taking the linear transformation of { x+1, x-2, $x^{2}$+x } and then transforming each of those answers into the basis set of the vectors {2, $x^{2}$+1, 1-x }. If I simply did the calculations wrong I wouldn't be surprised but I'm just wondering if this is the same way you guys approached it.
TLindberg: May 4, 2015, 10:54 p.m.
Oops, looks like the fraction notation got messed up. Just ignore the dollar signs above.
kylecitko: May 5, 2015, 2:37 a.m.
For question #4, I understand the answers, but how do you justify them?
Cullen_Im: May 5, 2015, 3:44 a.m.
For 6, to show that {x + 1, x - 2, $x^{2} + 2$} is a basis, I believe you also have to show that the bases span the set of polynomials of degree less than or equal to 2. The bases span a$x^2$ + bx +c when gamma = a, beta = (b-a-c)/3, and alpha = (2b-2a+c)/3. I did this by setting a$x^2$ + bx +c = to alpha(x - 1) + beta(x - 2) + gamma($x^2 + x$) and solving for gamma, beta, and alpha.
kylecitko: May 5, 2015, 3:48 a.m.
Never mind, I figured out how to justify the answers for #4.
Cullen_Im: May 5, 2015, 3:53 a.m.
@kylecitko: to justify the answers for #4, to show a function is linear, you have to show that T(X) + T(Y) = T(X + Y) and that kT(X) = T(kX). If at least one of these statements is not true, then the function is not linear. For example, for part d., T(a + bx + c$x^2$ + d + ex + f$x^2$) = T(a + bx + c$x^2$) + T(d + ex + f$x^2$) and T(k(a + bx + c$x^2$)) = kT(a + bx + c$x^2$)
Cullen_Im: May 5, 2015, 4:27 a.m.
@connorkenny: for #7, you start by setting (a,b) = x(1,1) + y(1,-1) and solving for "x" and "y" in terms of "a" and "b" to find out what direction (a,b) is in terms of the beta bases. You should get (x,y) = ((a+b)/2,(a-b)/2). Then multiply the matrix by (x,y) to get the product (a, (5a+b)/2). Finally, see what the product is with respect to your gamma bases by setting a(1,2) + ((5a+b)/2)(1,0) to get your answer of ((7/2)a+(1/2)b2a)
Cullen_Im: May 5, 2015, 4:28 a.m.
final answer is for #7 is ((7a +b)/2, 2a)
Cullen_Im: May 5, 2015, 4:37 a.m.
@TLindberg: I used the same methodology for #8 but got the same answer as @carRobinson so I think there are just some calculation mistakes
TLindberg: May 6, 2015, 12:21 a.m.
@Cullen_Im: Yeah, there was. I believe I messed up the second row if I remember correctly. My fault