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 Subject: Linear Algebra Question Category: Science > Math Asked by: bildy-ga List Price: \$12.00 Posted: 29 Oct 2002 12:50 PST Expires: 28 Nov 2002 12:50 PST Question ID: 92331
 ```I am in need of help with the following three linear algebra questions. I need to study from these questions for a test, so I need to see the work to solve these questions, not just answers. I am more interested in how to solve these problems, then what the answers are, but the answers must be correct as well. Thanks for all of your help. 1. Let v1=(1,0,2), v2=(2,1,3), and v3=(3,L,1). Answer the following: (a) For what value(s) of L are the vectors v1, v2, v3 linearly dependent? (b) Find at least two vectors in span {v1,v2,v3} other than v1, v2, v3. 2. For what values of a is the vector (a^2, -3a, -2) in span {(1,2,3),(0,1,1),(1,3,4)}? 3. Let v be the set of all 2x2 matrices [a,b;c,d], where d=bc-a. Answer the following: (a) Is V closed under the operation (matrix addition)? (b) Is V closed under the operation (scalar multiplication)?```
 ```Thank you for your question. I have done a lot of linear algebra over the years - my dissertation had oodles of it, and it is certainly a subject that takes a lot of practice. One of the most tricky (but interesting) things about linear algebra is that the same entity is often thought of in many different ways. For example, sometimes a matrix is just a table of numbers; sometimes a matrix is a linear transformation. With practice, things become clearer. By the way, if you can you might want to look at a software package called Matlab. It is extremely powerful in linear algebra - basically almost everything it does involves matrices. Once you start in with more complicated things like eigenvectors and so on, you might find Matlab fun to play with. Now for the questions: -------------------------- QUESTION 1 1. Let v1=(1,0,2), v2=(2,1,3), and v3=(3,L,1). Answer the following: QUESTION 1(a) For what value(s) of L are the vectors v1, v2, v3 linearly dependent? -------------------------- ANSWER TO 1(a) The simplest way to tell if these three vectors are linearly dependent is to take the determinant of the matrix formed from the vectors, since a matrix has nonzero determinant if and only if its rows are linearly independent. In this case, the determinant would be: det(M) where M= 1 0 2 2 1 3 3 L 1 Now, the determinant of M is (I assume you can take the determinant of a 3x3 matrix) 1-3L+2(2L-3) = 1-3L+4L-6 = L-5 So, the determinant is 0 if and only if L=5. ------------------------------------------------ QUESTION 1(b) Find at least two vectors in the span of {v1,v2,v3} other than v1, v2, v3. [not that I changed "span" to "the span of".] ---------------------------------------------- ANSWER TO 1(b) The SPAN of a set of vectors is just the set of the linear combinations. That is, just the set of all vectors of the form a*v1+b*v2+c*v3, where a, b, and c are real numbers. Obviously there are many solutions; the simplest is to take a=4, a'=5, the other coefficients 0. So the first vector is: V=4*v1+0*v2+0*v3=(4,0,8) The second vector would be W=5*v1+0*v2+0*v3=(5,0,10) . --------------------QUESTION 2------------------------------ 2. For what values of a is the vector (a^2, -3a, -2) in span {(1,2,3),(0,1,1),(1,3,4)}? ------------------------------------------------------------- ANSWER TO QUESTION 2 ------------------------------------------------------------ Let v=(1,2,3), w=(0,1,1), z=(1,3,4), r=(a^2,-3a,-2) . Since v+w=z, r is in the span of {v,w,z} if and only if r is in the span of {v,w} . Now, r is in the span of {v,w} if and only if {v,w,r} is linearly dependent. We already know how to check if a set of vectors is linearly dependent from question 1: we take the determinant of the matrix whose rows are those vectors and we set it to 0. Thus, v,w,r are linearly dependent if and only if the determinant of M is zero, where M= 1 2 3 0 1 1 a^2 -3a -2 The determinant of M is: -2+3a +a^2(2-3) =-2+3a-a^2 =-(a^2-3a+2) =-(a-1)(a-2) This is zero if and only if a=1 or a=2, which is the desired answer. ------------------------------------------------------------------ QUESTION 3 3. Let V be the set of all 2x2 matrices a b c d where d=bc-a. [Note: I have changed v to V in this question, and redid the formatting] ------------------------------------------------------------------ QUESTION 3(a) (a) Is V closed under the operation (matrix addition)? ----------------------------------------------------------------------- ANSWER TO 3(a) A set is CLOSED under an operation if an only if, whenever two elements x and x' are in the set, the result of performing the operation on x and x' remains in the set. So, V is closed under the operation if and only if, for any two matrices x and x', their sum, x+x', is also in V. Suppose that x and x' are in V. By definition of V, we can write, for some real numbers a,b,c,d and a',b',c',d' such that d=bc-a and d'=b'c'-a' x= a b c d y = a' b' c' d' So x+y = a+a' b+b' c+c' d+d' Is x+y in V? Only if (d+d')=(b+b')(c+c')-(a+a') If we multiply this out, this is true if and only if: d+d'=bc + b'c + b'c' + bc' -a - a' . = (bc-a) + (b'c'-a') +b'c + bc' = d + d' + b'c + bc' The last equality holds because x and y are in V. This equation is certainly not true in general; indeed, it is false if b'=c=b=c'=1 . So V is not closed under the operation. ---------------------------------------------------------------------------- QUESTION 3(b) Is V closed under the operation (scalar multiplication)? ---------------------------------------------------------------------------- V is closed under scalar multiplication if, whenever x is in V, and t is some real number, then tx is in V. This turns out to be false. I will give an explicit example here: Let x= 0 1 1 1 Then x is in V since 1=1*1-0 . But 2x is not in V, since 2x = 0 2 2 2 and 2 is not equal to 2*2-0. So V is not closed under scalar multiplication. I hope this is helpful, and feel free to ask for clarification if necessary.``` Request for Answer Clarification by bildy-ga on 29 Oct 2002 16:34 PST ```1. I have come up with a two different solutions to the second question to verify that the solution is correct, but I seem to be having a problem with the cardinality of my solutions. My solutions are as follows: x(1,2,3) + y(0,1,1) +z(1,3,4) =(a^2,-3a,-2) where x,y and z are integers. We get the following three equations by equating three coordinates x+z = a^2 2x+y+3z=-3a 3x+y+4z=-2 Subtracting first equation from third we get 2x+y+3z=-2-a^2 the left hand of this equation and second equation is same so we can substitute -2-a^2 for 2x+y+3z in the second equation -2-a^2=-3a a^2+3a+2=0 Now by solving quadratic equation we get a^2+a+2a+2=0 (a+2)(a+1)=0 a=-1 or a=-2 So for the above values of a the vector (a^2,-3a,-2) is in the span of given vectors. My second solution is as follows: (a^2, -3a, -2) is in the span of (1,2,3),(0,1,1) and (1,3,4) when: (a^2,-3a,-2) = b(1,2,3)+c(0,1,1)+d(1,3,4) where b,c and d are some constants. This gives us that a^2=b+d, -3a=2b+c+3d and -2=3b+c+4d Doing the same as above and putting the corresponding matrix in to echelon form will give a you the follwoing polynomial a^2 - 3a + 2 = (a-1)(a+2) so the solutions of a are 1 and -2. So my clarification asks, what is the correct soltion: a. Solutions of a are 1 and 2 (As you first stated) b. Solutions of a are -1 and -2 c. Solutions of a are 1 and -2 Thank you very much for the clarification.``` Clarification of Answer by rbnn-ga on 29 Oct 2002 16:49 PST ```Your analysis is right! (You are basically writing out, almost "re-proving", showing why some of the tools I was using worked.) You just made one careless sign error, which is why your two solutions were different. With the correct sign you will find that your solution is the same as mine. Some quick comments on your clarification request. 1. You write: are as follows: ...where x,y and z are integers In a linear combination, the coefficients can be any real number. They do not have to be integers. 2. The two equations below are given as equivalent in your derivation, but they are not. -2-a^2=-3a a^2+3a+2=0 The second equation should read: a^2-3a+2=0``` Request for Answer Clarification by bildy-ga on 29 Oct 2002 18:29 PST ```If you could possibly answer a very simple question I would be most graciously appreciative (I have many other large questions that need to be answered in the field of linear algebra as well as calculus III, so if you are interested within the next week or so please let me know). The following question should take 5 minutes of your time tops, and I would just like to verify that I am doing this correctly before I go in to take the test. Thank you very much in advance. The question: Consider the set W of all vectors in R^4 of the form (a,b,c,d), where a=b+c and d=a+1. Is W a subspace of r^4? Complete parts (a) and (b) of Theorem 6.2. Theorem 6.2 is as follows: Let V be a vector space with operations (matrix addition) and (scalar multiplication) and let W be a nonempty subset of V. Then W is a subspace of V if and only if the following conditions hold: (a)If u and v are any vectors in W, then u+v is in W ( + = matrix addition) (b)If c is any real number and u is any vector in W, then c * u is in W ( * is scalar multiplication)``` Clarification of Answer by rbnn-ga on 29 Oct 2002 18:56 PST ```Regarding your request for clarification. W is not a subspace because if z=(0,0,0,1) then z is in W but z+z is not in W. Regarding the second part of the question, the theorem is stating properties of the subspace W that are often taken as defining a subspace. It would not be possible to answer that question without knowing exactly what the text is using as a definition for "subspace".```
 bildy-ga rated this answer: ```Awesome answer. This question was answered almost immediately, and the answer was right on target. I could not have asked for anything more. Thank you so much rbnn-ga.```