### Students' answers to review problems for the second exam in Math 421, spring 2004

 Very few of these problems should require extensive computation. The actual exam will ask students for several of the definitions in problem 14. The idea for the list of New Jersey Native Trees and its use here was copied from Professor B. Latka, whose linear algebra exams at Lafayette College featured the same device.

means no answer is available yet. The answers are all here! Very good job, people!
Various corrections have been made to the answers (6:48 PM, 4/8/2004). I currently believe the answers are right.

 8. TREE Are (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) linearly independent? Answer For the vectors (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) to be linearly independent, we can look at the equation C1(-2,-4,8,-7) + C2(1,2,1,1) + C3(1,2,-1,2) = 0. If the vectors are linearly independent, each component of the vector (-2C1+C2+C3, -4C1+2C2+2C3, 8C1+C2-C3, -7C1+C2+2C3) must be zero:```-2C1 + C2 + C3 = 0 -4C1 +2C2 +2C3 = 0 8C1 + C2 - C3 = 0 -7C1 + C2 +2C3 = 0``` To solve this, we can write it was a matrix, and look at its reduced form (see SYCAMORE):```(-2 1 1) (1 0 -1/5) (-4 2 2)~(0 1 3/5) ( 8 1 -1) (0 0 0) (-7 1 2) (0 0 0)``` This gives us the new equations```C1-(1/5)C3 = 0 C2+(3/5)C3 = 0``` Since there are other equations with all zeros, lets plug in 1 for C1:1 = (1/5)C3 and C2 = -(3/5)(5) We find that C1 = 1, C2 = -3 and C3 = 5, and if we plug these solutions into the original equations, they work. So since there are non-trivial solutions to this equation, the vectors (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) are not linearly independent. We can also prove this by trying to find one of these vectors as a linear combination of the others: To do this we should look at the reduced form (see HICKORY)```(-2 -4 8 -7) (1 2 0 3/2) ( 1 2 1 1)~(0 0 1 -1/2) ( 1 2 -1 2) (0 0 0 0)``` The first 2 vectors are reduced to (1,2,0,3/2) and (0,0,1,-1/2). If we subtract the second one from the first, we get the vector (1,2,-1,2) which is the 3rd. Therefore, the vector (1,2,-1,2) is a linear combination of the first 2 vectors, and these vectors are not linearly independent. Sent by L. Kohut at Fri, 2 Apr 2004 15:02:52

 10. TREE What is the rank of A=```(0 2 5 2 0) (2 -1 1 0 1) (1 0 2 1 1) (1 -1 -1 -1 0)```? Find a basis for the subspace of all solutions of AX=0 in RP. What is P? Answer By OAK,```[0 2 5 2 0 ] [1 0 0 -3 -3] [2 -1 1 0 1 ]~[0 1 0 -4 -5] [1 0 2 1 1 ] [0 0 1 2 2] [1 -1 -1 -1 1 ] [0 0 0 0 0]``` The rank of this matrix = 3. Sent by J. Sirak at Sun, 4 Apr 2004 16:13:29 Using OAK, the basis for all solutions to Ax = 0 is:``` [x1] [ 3x4+3x5] [ 3] [ 3] [x2] [ 4x4+5x5] [ 4] [ 5] x =[x3]=[-2x4-2x5]=x4[-2]+x5[-2] [x4] [ x4 ] [ 1] [ 0] [x5] [ x5 ] [ 0] [ 1]``` So the basis for Ax = 0 is (3,4,-2,1,0) and (3,5,-2,0,1) in RP, where P = 5. Sent by M. Woodrow at Tue Apr 6 15:22:56 2004

 12. TREE Verify that (1,2,0,2,1) is in the linear span of (1,1,0,0,2) and (2,1,2,-2,1) and (4,2,5,-4,0). Answer To see that (1,2,0,2,1) is in the linear span of (1,1,0,0,2) (2,1,2,-2,1) and (4,2,5, -4,0), look at the tree corresponding to BIRCH. In RREF the matrix has a row of 0's on the bottom where (1,2,0,2,1) was positioned, which means that the vector is linearly dependent on the other 3 vectors. To verify this you can find the 3 constants multiplied against the 3 spanning vectors of R^5 to form the vector under question. ```1 = A + 2B + 4C 2 = A + B + 2C 0 = 2B + 5C 2 = -2B - 4C 1 = 2A + B ``` Then ```(1-B)/2 = A (-2/5*B) = C``` Substitute A and C into the first equation to get a value for B ```1=(1-B)/2 + 2B +4(-2/5*B) B=-5 A=3 C=2 ``` Which proves (1,2,0,2,1) is a linear combination and in the span of (1,1,0,0,2) (2,1,2, -2,1) and (4,2,5, -4,0). Sent by M. McNair at Tue, 30 Mar 2004 17:41:38 Comment from the Management The problem could also be done using PINE. This RREF shows that the rank of the system x1(1,1,0,0,2)+x2(2,1,2, -2,1)+x3(4,2,5, -4,0)+x4(1,2,0,2,1)=0 is 3. It also shows that this system is row equivalent to ```x1+3x4=0 x2-5x4=0 x3+2x4=0``` so that if x4=1, we will get the vector as a linear combination of the other vectors, with coefficients -3 and 5 and -2. These are x4, so writing the vector as a linear combination means adjusting all the signs. I think Mr. McNair's solution is fine. I just wanted to show an alternate method.

 15. TREE Consider the system``` 3x1+5x2-1x3=a 2x1+6x2+1x3=b 4x1+4x2-3x3=c``` a) Find a specific vector (a,b,c) in R3 so that the system has no solution. Explain. b) Find a specific vector (a,b,c) in R3 so that the system has at least one solution. Find all solutions for that vector. Explain. Answer I) rewrite the equations into an augmented matrix ```(3 5 -1 : a) (2 6 1 : b) (4 4 -3 : c)``` II)RREF of the augmented matrix (see WILLOW)=``` (1 0 -11/8 :3a/4-5b/8) (0 1 5/8 :-a/4+3b/8 ) (0 0 0 :-2a+b+c)``` III)```X(1)-11/8X(3)= 3a/4-5b/8 X(2)+5/8X(3)= -a/4+3b/8 0= -2a+b+c``` a) a specific vector (a,b,c) in R^3 so that the system has no solution is when -2a+b+c does not equal to zero (the system is inconsistent): (a,b,c)=(2,1,1). b) a specific vector (a,b,c) in R^3 for the system to have at least one solution occurs when 0= -2a+b+c, the system is consistent: (a,b,c)=(1,1,1). Sent by A. Lu at Tue, 30 Mar 2004 19:29:07

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 18. TREE a) Explain why the vectors (2,-3,5), (-7,1,-2), and (3,-5,7) are a basis of R3. b) Write a linear combination of the vectors in a) whose sum is equal to the vector (2,1,3). Answer a) Looking at ASH, we see that the rref of the matrix consisting of the three vectors as its columns is I3. Therefore, the vectors are linearly independent. Clearly, the vectors also span R3 since any vector in R3 can be written as a combination of i,j,k (unit vectors). Since the vectors span R3 and are linearly independent, they form a basis of R3. b) In order for (2,1,3) to be a linear combination of the three vectors from part a, we must find the coefficients such that x1(2,- 3,5)+x2(-7,1,-2)+x3(3,-5,7)=(2,1,3). We can solve this as the augmented matrix ASH with (a,b,c) replaced by (2,1,3). Then we get x1=133/25 (note that the coefficient of the b term should be positive), x2=-6/25, x3=-86/25. Sent by N. Wilson at Mon, 5 Apr 2004 15:06:41