##### 33A pactice midterm answers
Jackhe
Jia Le He Discussion 1C UID = 204460520 Here are my answers for the practice midterm. I am sure about number 7 since my TA showed it in class. Feel free to point out errors or ask me any questions about my procedure 1) x_{1} = -34 - 7x_{4} x_{2} = -5 - x_{4} x_{3} = 21 + 4x_{4} x_{4} = x_{4} 2) Kernel's dimension = 1 Kernel = span\begin{bmatrix} 7/6\\ -17/3\\ 1/2\\ 1 \end{bmatrix} Dimension of image = 3 from rank nullity theorem 3) \begin{bmatrix} 2/3 & -1/2 & 1/3\\ -8/3 & 2 & -1/3\\ 1 & -1/2 & 0 \end{bmatrix} 4) a) linear b) not linear c) not linear d) linear 5) Image = span\begin{bmatrix} \begin{pmatrix} 1\\ 3\\ 1 \end{pmatrix} & \begin{pmatrix} 2\\ -2\\ -6 \end{pmatrix} \end{bmatrix} kernel = span\begin{pmatrix} -5/4\\ -1/8\\ 1 \end{pmatrix} 6) To prove that it is a basis just prove that it is linearly dependent. For the second part show that they are not linearly independet. 7) \begin{pmatrix} 7/2a +1/2b\\ 2a \end{pmatrix} 8) \begin{pmatrix} 1 & 2 & 1/2\\ 1 & 1 & 1\\ 2 & 2 & 0 \end{pmatrix} 9) \begin{pmatrix} -7/5 & 1/5 & 29/5\\ -1 & 11/5 & 19/5\\ -4/5 & -2/5 & 17/5 \end{pmatrix}
Jackhe: May 5, 2015, 4:26 p.m.
Sorry for bad formatting, still trying to get used to it..
Jackhe: May 5, 2015, 4:35 p.m.
I found a mistake in number 9. I multiplied the matrices in the wrong order and since it's not commutative it's wrong.. The actual matrix should be: \begin{pmatrix} 3 & 33/5 & 21/5\\ 5/2 & 52/5 & 19/5\\ -3/2 & -41/5 & -12/5 \end{pmatrix} I hope we don't get all these fractions in the actual midterm..
Jackhe: May 5, 2015, 4:36 p.m.
oh, and in number 6 I meant linearly independent, not dependent
danielkim: May 5, 2015, 5:25 p.m.
I have a question. If you have matrices of the same nxn dimensions such as problem 9, (S^-1) (B) (S), do you have to multiply the matrices from left to right? ie multiply (S^-1) (B) first
thaiarthur: May 5, 2015, 5:57 p.m.
I think you do (B)(S) first, then (S^-1)(Answer from (B)(S)). The order does matter, so you have to remember what we are doing. Multiplying by B with S gets us the output of the transformation in our basis. Multiplying that output by S^-1 will get us the output of the transformation in our friends basis.
danielkim: May 5, 2015, 6:14 p.m.
@thaiarthur thanks for the explanation, it makes sense
Jackhe: May 5, 2015, 6:29 p.m.
Matrices are associative so if you have ABC you can do (AB)C or A(BC).
Jackhe: May 5, 2015, 6:37 p.m.
Matrices are associative so if you have ABC you can do (AB)C or A(BC).
dannystapleton: May 5, 2015, 7:43 p.m.
@Jackhe For number 9, I keep seeing mixed ways of having (1,0,2) (2,3,2) (2,1,0) as either being the IDentitiy matrix from gamma to beta or the Identity matrix of beta to gamma. With your final answer ^, I believe you set the Id Matrix from gamma to beta, but can you explain to me why that must be the case?
Jackhe: May 5, 2015, 9:49 p.m.
I Believe that the problem says that those are the basis of the friend with respect to you, which means that by multiplying by that vector you can convert from your basis to your friends basis that is beta to gamma. At least that's how I understood it
Jackhe: May 5, 2015, 9:56 p.m.
Wait, I just contraddicted myself lol, sorry, what I mean is that you your friend can find the coordinates in your basis using the matrix composed of those. So it is gamma to beta which is my final answer. But this is a super confusing problem and I'm not sure anymore...
Jackhe: May 5, 2015, 10 p.m.
I Believe that the problem says that those are the basis of the friend with respect to you, which means that by multiplying by that vector you can convert from your basis to your friends basis that is beta to gamma. At least that's how I understood it
syp881122: May 6, 2015, 4:13 p.m.
I still confuse about #2 answer, I think the answer is not ⎡⎣⎢⎢⎢⎢7/6−17/31/21⎤⎦⎥⎥⎥⎥ my answer is -7/6 17/3 -1/2 could anyone can tell me why it comes contrary numbers?
Jackhe: May 8, 2015, 1:35 a.m.
It's simple. Let's say you get a row in a matrix that's let's say (1 0 3 4). So you have x_{1} + 3x_{3} + 4x{4). Now, you are looking for the span of the kernel and by definition the kernel is the set of all elements mapped to zero therefore x_{1} = -3x_{3} - 4x{4) and thus the signs get inverted
Jackhe: May 8, 2015, 1:36 a.m.
Dammit, subscript doesn't work, hope it's understandable
Jackhe: May 8, 2015, 1:38 a.m.
Oh, I think I forgot a step: since it's the kernel they get mapped to 0, so x1 + 3x_{3} + 4x{4} = 0