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.
1. Suppose A=(a_{ij}) is a 200 by 200 matrix. All of the
entries of A are 0 except Diagonals Each a_{ii}=2 for 1<= i<=200. Weirdos a_{ij}=10 when i=100 and j=150, and a_{ij}=x when i=150 and j=30. What is the determinant of A? Comment A matrix with relatively few nonzero entries is called a sparse matrix.
Answer Sent by C. Graziadio at 2 Apr 2004 20:20:43 
2. Suppose this matrix is in RREF:
(* 7 * * * * * *) (* * 0 1 * * * *) (* * * * 0 0 0 0)Change each * to one of the following: 0 if the entry must be 0; 1 if it must be 1; A if it can be anything (0 or not 0: no restrictions). Explain your answers. What's the rank of this matrix?
Answer [ 1 7 A 0 A A A A ] [ 0 0 0 1 A A A A ] [ 0 0 0 0 0 0 0 0 ]Position (1,1) has to be a one because all leading entries in rows in RREF are one. It also follows that positions (2,1) and (3,1) must be zero because in RREF all entries above and below a leading 1 entry must be zero. Matrices in RREF also have an upper triangular look to it so, all entries in row 2, before position (2,4) must be zero. So in row 3, all entries must be zero. These are the properties of a matrix in RREF, therefore the other entries in rows one and two could be any value A.
The rank of this matrix = 2
Sent by J. Sirak at Sun, 4 Apr 2004 15:00:20 
3. Suppose (1,2,3,0,0) and (0,0,1,1,1) are both solutions of
one specific system of 5 linear equations in 5 unknowns. a) If this system is written as AX=B where A is a matrix of coefficients, X is a column vector of unknowns, and B is a column vector of specific real numbers (you can't deduce much about the entries of A and B from what's given), then what are the dimensions of A and X and B? b) Does the system AX=0 (here 0 is a column vector of 0's) have any solutions? Be as specific about the solutions as you can. c) How many solutions does the original system AX=B have? Be as specific about the solutions as you can. d) What is the determinant of A? What are the possible values of the rank of A? Answer The two vectors given will be called X_{1}=(1,2,3,0,0) and X_{2}=(0,0,1,1,1). a) A is 5 by 5 from the fact that it is 5 eqns. in 5 unknowns X is 5 by 1 and B is also 5 by 1. b) AX=0 has solutions for example A(X_{1}X_{2})=AX_{1}AX_{2}=BB=0 thus the vector X*=(1,2,2,1,1) is a solution to AX=0. c) AX=B has infinitely many solutions of the form eX*+X_{p} where is any real number and Xp is any vector that solves the equation Ax=B. for example since AX*=0 and AX_{2}=B we have that A(eX*+X_{2})=B where again e can be any real number. d) since A is 5 by 5 and the system AX=B has more then one solution then inverse(A)=A^{1} does not exist because X not equal to A^{1}*B. Since A does not have an inverse then detA=0 and therefore the rank of A is less then 5 and greater then zero Sent by Ben Cohen at Mon, 29 Mar 2004 17:44:46 
4. Suppose that A and B are both 2 by 2 matrices which have
inverses, and that A^{1}=( 3 1) (8 1)and B^{1}= (2 1) (1 7)If C=A^{2}B, what is the inverse of C? Comment A small amount of thought can save much irritation. Answer [ 17 4] [32 9]so since (B^{1})= [2 1] [1 7]we see that (C^{1})= [ 66 17] [241 67]Sent by C. Hunt Thu, 1 Apr 2004 19:28:13
Comment from the Management

5. Suppose that A is a 50 by 70 matrix whose rank is 40. What is the
dimension of the subspace S of R^{N} consisting of the
solutions of the homogeneous system AX=0? What is the dimension of
the subspace S of R^{M} consisting of all possible
Y's for which the equation AX=Y can be solved? What are
N and M?
Answer 40 30 [  ] 4 [ I40  what ] 0 [  ever ] [________________________] 1 [ all 0's ] 0 [ all 0's ]So, for a homogeneous system Ax = 0, the dimension of the subspace is equal to the rank of A in R^{n} where n = 70, because A is 50 by 70 so x is 70 by 1. Now, the dimension of the subspace S in R^{m} for all Y's that satisfy Ax = Y is 30, because it is the number of free variables in A, and m = 50 simply because A is 50 by 70, x is 70 by 1, so Y is 50 by 1 so Y's subspace will sit in R^{50}. Sent by M. Woodrow at Tue Apr 6 15:22:56 2004 
6. Find all values of x so that the matrix(x 0 1 0) (2 0 1 1) (x 0 0 2) (0 1 1 1)has an inverse. Answer 0 + 0 + 0 + (1)*(1)^(6)*det (x 1 0 = 2x + (4  x) = 3x  4. 2 1 1 x 0 2)Therefore, x must equal 4/3, in order for the original matrix to NOT have an inverse. Sent by X. Sosa at Tue, 06 Apr 2004 16:56:08 
7.The matrix (2 u v w) (0 1 0 s) (0 0 3 t) (0 0 0 4)is invertible for all u, v, w, s, and t. What is its inverse? Comment This matrix is in uppertriangular form with an uppertriangular inverse. First use row operations to make the diagonal elements =1, then pivot and eliminate. Answer (2 u v w  1 0 0 0) (0 1 0 s  0 1 0 0) (0 0 3 t  0 0 1 0) (0 0 0 4  0 0 0 1)Divide First Row by 2 Multiply Second Row by 1 (1 u/2 v/2 w/2  1/2 0 0 0) (0 1 0 s  0 1 0 0) (0 0 3 t  0 0 1 0) (0 0 0 4  0 0 0 1)Divide Third Row by 3 Subtract u/2 * Second Row from First Row (1 0 v/2 w/2+su/2  1/2 u/2 0 0) (0 1 0 s  0 1 0 0) (0 0 1 t/3  0 0 1/3 0) (0 0 0 4  0 0 0 1)Divide Fourth Row by 4 Subtract v/2 * Third Row from First Row (1 0 0 w/2+su/2vt/6  1/2 u/2 v/6 0 ) (0 1 0 s  0 1 0 0 ) (0 0 1 t/3  0 0 1/3 0 ) (0 0 0 1  0 0 0 1/4)Subtract w/2+su/2vt/6 * Fourth Row from First Row Add s * Fourth Row to Second Row Subtract t/3 * Fourth Row from Third Row (1 0 0 0  1/2 u/2 v/6 w/8su/8+vt/24) (0 1 0 0  0 1 0 s/4 ) (0 0 1 0  0 0 1/3 t/12 ) (0 0 0 1  0 0 0 1/4 )Solution: (1/2 u/2 v/6 w/8su/8+vt/24) ( 0 1 0 s/4 ) ( 0 0 1/3 t/12 ) ( 0 0 0 1/4 )may check answer by multiplying the original matrix by the solution and you will get the identity matrix of a 4 x 4 matrix Sent by S. Brynildsen at Wed Mar 31 21:07:20 2004 
8. TREE Are (2,4,8,7), (1,2,1,1), and (1,2,1,2) linearly
independent?
Answer If the vectors are linearly independent, each component of the vector (2C_{1}+C_{2}+C_{3}, 4C_{1}+2C_{2}+2C_{3}, 8C_{1}+C_{2}C_{3}, 7C_{1}+C_{2}+2C_{3}) must be zero: 2C_{1} + C_{2} + C_{3} = 0 4C_{1} +2C_{2} +2C_{3} = 0 8C_{1} + C_{2}  C_{3} = 0 7C_{1} + C_{2} +2C_{3} = 0To 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 C_{1}(1/5)C_{3} = 0 C_{2}+(3/5)C_{3} = 0Since there are other equations with all zeros, lets plug in 1 for C_{1}:1 = (1/5)C_{3} and C_{2} = (3/5)(5) We find that C_{1} = 1, C_{2} = 3 and C_{3} = 5, and if we plug these solutions into the original equations, they work. So since there are nontrivial 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 
9. Is the sum of invertible matrices always invertible? Is the sum of
singular matrices always singular?
Answer b) A square matrix is said to be nonsingular if it has an inverse. If it has no inverse, the matrix is called singular. The answer to this question is no. This can be proved by summing two singular matrices and obtaining an invertible matrix. For example, (0 0)+(1 0)=(1 0) (0 1) (0 0) (0 1)Comment from the Management Of course the invertibility or noninvertibility of such small matrices is easy to check by using the determinant. Sent by J. Malek at Tue, 6 Apr 2004 19:02:14 
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 R^{P}. What is P? Answer [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: [x_{1}] [ 3x_{4}+3x_{5}] [ 3] [ 3] [x_{2}] [ 4x_{4}+5x_{5}] [ 4] [ 5] x =[x_{3}]=[2x_{4}2x_{5}]=x_{4}[2]+x_{5}[2] [x_{4}] [ x_{4} ] [ 1] [ 0] [x_{5}] [ x_{5} ] [ 0] [ 1]So the basis for Ax = 0 is (3,4,2,1,0) and (3,5,2,0,1) in R^{P}, where P = 5. Sent by M. Woodrow at Tue Apr 6 15:22:56 2004 
11. Briefly explain why the vectors (3,4,1), (2,2,1), (1,2,0),
and (0,1,4) in R^{3} must be linearly dependent. Do this
in two ways:
Answer [3 2 1 0] A = [4 2 2 1] [4 1 0 4]We see that we get a 3 x 4 matrix. In order for the vectors to be linearly independent, rank = # columns = dim(A). In this case, we can easily see that the maximum possible rank of the matrix will be 3. However, the number of columns exceeds this rank. A free variable is present, indicating that that the vectors are in fact linearly dependent ( in this case, in the x4 variable ). In general, when the # columns > # rows , the vectors will be dependent. (ii) By placing the matrix A into RREF we get the following : [1 0 0 (7/6)] RREF of A = [0 1 0 (2/3) ] [0 0 1 (13/6)]Then solving the homongeneous system. We get the following 3 equations : x_{1} = (7/6) x_{4} x_{2} = (2/3) x_{4} x_{3} = (13/6) x_{4} x_{4} = x_{4}So a general solution can be the following : [ 7/6] [ 2/3] [13/6] x_{4} [ 1 ]where x_{4} can be any value and hence there are nontrivial solutions. Sent by J. Sirak at Wed, 31 Mar 2004 13:56:10
Comment from the Management 
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 1 = A + 2B + 4C 2 = A + B + 2C 0 = 2B + 5C 2 = 2B  4C 1 = 2A + BThen (1B)/2 = A (2/5*B) = CSubstitute A and C into the first equation to get a value for B 1=(1B)/2 + 2B +4(2/5*B) B=5 A=3 C=2Which 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 x_{1}+3x_{4}=0 x_{2}5x_{4}=0 x_{3}+2x_{4}=0so that if x_{4}=1, we will get the vector as a linear combination of the other vectors, with coefficients 3 and 5 and 2. These are x_{4}, 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. 
13. Suppose A is a 3 by 3 matrix, and det(A)=2. Compute
det(A^{5})
and det(5A)
and det(A^{1})
and det(AA^{t}).
Answer
Since det(A^5) is equivalent to det(A*A*A*A*A) which equals det(A)*det
(A)*det(A)*det(A)*det(A), det(A^5)=2^5=32. (5 0 0) (0 5 0) (0 0 5)and the determinant of this matrix is 125, so det(5A)=125*2=250 det(A^{1})=1/2 because A*A^{1}=I and taking det of both sides gives detA*det(A^{1})=detI, where detI=1. Lastly, since detA=det(A^t), det(AA^t)=2*2=4 Sent by Michele Mudrak at Mon, 29 Mar 2004 00:13:48 
14. Be prepared to discuss the meaning of homogeneous ... inhomogeneous ... basis ... linear independence ... linear combination ... subspace ... dimension ... spanning ... rank ... eigenvalue ... eigenvector ... transpose ... matrix addition and multiplication ... symmetric ... diagonalization ... life.
Answers
Comment from the Management 
15. TREE
Consider the system3x_{1}+5x_{2}1x_{3}=a 2x_{1}+6x_{2}+1x_{3}=b 4x_{1}+4x_{2}3x_{3}=ca) Find a specific vector (a,b,c) in R^{3} so that the system has no solution. Explain. b) Find a specific vector (a,b,c) in R^{3} so that the system has at least one solution. Find all solutions for that vector. Explain.
Answer (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/45b/8) (0 1 5/8 :a/4+3b/8 ) (0 0 0 :2a+b+c)III) X(1)11/8X(3)= 3a/45b/8 X(2)+5/8X(3)= a/4+3b/8 0= 2a+b+ca) 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 
16. Suppose A is a 4 by 4 matrix whose inverse is(2 3 1 1) (1 1 1 0) (2 0 2 3) (4 1 2 2)a) Find all solutions of AX=B if B=(2,1,1,0). b) Find all solutions of the homogeneous system AX=0.
Answer (2 3 1 1) (2) ( 8) X = (1 1 1 0) (1) = ( 4) (2 0 2 3) (1) ( 6) (4 1 2 2) (0) (11)For number 16B, A^{1}*0 = 0
Sent by T. Obbayi at Tue, 06 Apr 2004 17:20:37
(1/7 1/7 3/7 4/7) ( 6/7 8/7 4/7 3/7) (5/7 16/7 1/7 1/7) ( 4/7 10/7 5/7 2/7)and you could then write the system explicitly as AX=B. Of course, to get a solution you'd then turn around and compute A^{1}! So why bother finding A when you have what you need already! 
17. Give an example of a specific matrix which has
exactly 20 nonzero entries and whose characteristic
polynomial is (L5)^{3}(L+2)^{5}(L^{2}+4).
Answer The characteristic polynomial tells us: 1) our matrix is 10 by 10 because the polynomial is degree 10 2) the diagonal entries; three terms are 5, five terms are 2, and the other two are +/2i so the matrix, with 20 nonzero entries, will look like: [5 1 1 1 1 1 1 1 1 1] [0 5 0 0 0 0 0 0 0 0] [0 0 5 0 0 0 1 0 0 0] [0 0 0 2 0 0 0 0 0 0] [0 0 0 0 2 0 0 0 0 0] [0 0 0 0 0 2 0 0 0 0] [0 0 0 0 0 0 2 0 0 0] [0 0 0 0 0 0 0 2 0 0] [0 0 0 0 0 0 0 0 2i 0] [0 0 0 0 0 0 0 0 0 2i]
Sent by M. Woodrow at Tue Apr 6 15:22:56 2004
( 0 1) (1 0)which had +/i eigenvalues. Using this, we could modify the previous answer to the following: [5 1 1 1 1 1 1 1 1 1] [0 5 0 0 0 0 0 0 0 0] [0 0 5 0 0 0 1 0 0 0] [0 0 0 2 0 0 0 0 0 0] [0 0 0 0 2 0 0 0 0 0] [0 0 0 0 0 2 0 0 0 0] [0 0 0 0 0 0 2 0 0 0] [0 0 0 0 0 0 0 2 0 0] [0 0 0 0 0 0 0 0 0 2] [0 0 0 0 0 0 0 0 2 0] 
>
18. TREE
a) Explain why the vectors (2,3,5), (7,1,2), and
(3,5,7) are a basis of R^{3}. 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 I_{3}. Therefore, the vectors are linearly independent. Clearly, the vectors also span R^{3} since any vector in R^{3} can be written as a combination of i,j,k (unit vectors). Since the vectors span R^{3} and are linearly independent, they form a basis of R^{3}. 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 x_{1}(2, 3,5)+x_{2}(7,1,2)+x_{3}(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 x_{1}=133/25 (note that the coefficient of the b term should be positive), x_{2}=6/25, x_{3}=86/25. Sent by N. Wilson at Mon, 5 Apr 2004 15:06:41 
19. a) Diagonalize A=(2 2) (1 1)That is, find the eigenvalues and eigenvectors of A, and find a matrix T so that TAT^{1} is diagonal. b) 3^{5}=243. Use this fact and the diagonalization of A found in a) to compute A^{5}.
Answer ( (2) 2 ) ( 1 (1) )to zero. This characteristic polynomial therefore is (2)(1)2=0, or eigenvalues ^{2} 3=0. The eigenvalues are =3 and =0. Now find eigenvectors that correspond to each eigenvalue by plugging values into C and solving C*X=0 for nonzero X, where X is a vector in R^{2}. For =0: 0=( 2 2 ) (x_{1}) ( 1 1 ) (x_{2})Standard procedure is to reduce matrix C, so reduced row echelon form of C is (1 1) (0 0)from which X can be determined: (x_{1})=x_{2}(1) (x_{2}) ( 1)Therefore (1, 1) is one of the eigenvectors of A that corresponds to the eigenvalue 0. Similarly, (2,1) is one of the eigenvectors of A that corresponds to eigenvalue 3. Defining D= (0 0) (0 3)to be the diagonal matrix, one must define T = (1 2) ( 1 1)Since the first column of D contains the eigenvalue 0, the eigenvector corresponding to eigenvalue 0 must be the first column of T and because the eigenvalue 3 is in the second column of the diagonal matrix D, the eigenvector corrsponding to eigenvalue 3 must be in the second column of T. To find the inverse of T, reduce (T  I_{2}): (1 2  1 0)~(1 0  1/3 2/3 ) (1 1  0 1) (0 1  1/3 1/3 )The inverse of T is T^{1}= (1/3 2/3) ( 1/3 1/3)One can verify that indeed, D=T^{1}*A*T b) Since D=T^{1}*A*T, A=T*D*T^{1} so that A^{n} = (T*D*T^{1})^{n} = T*(D^{n})*T^{1}. Therefore A^{5} = T*(D^{5})*T^{1}, and D^{5} = (0 0) (0 3^{5})then (D^{5})*T^{1}=B= ( 0 0 ) ( 81 81 )and T*B= ([2/3]*3^{5} [2/3]*3^{5})=(162 162) ([1/3]*3^{5} [1/3]*3^{5}) ( 81 81)Which is the same result maple gives when told to evaluate A^{5}. Sent by D. Ivanov at Thu, 1 Apr 2004 00:32:20 
20. If A=(2 1 1 0) (1 1 0 0) (1 0 1 0) (0 0 0 2)then A has eigenvalues 2, 0, 1, and 3 with associated eigenvectors (0,0,0,1), (1,1,1,0), (0,1,1,0), and (2,1,1,0), respectively. Find an orthogonal matrix W so that WAW^{1} is diagonal. Comment from the Management The Management made an error. I have generally been writing W^{1}AC=the diagonal matrix. So what I asked for here is the inverse of the matrix constructed before. I am sorry. This was not intentional.
Answer [0 1 0 2] [0 1 1 1] [0 1 1 1] [1 0 0 0]which arose from the eigenvectors, then normalize it by dividing each column by it's length (ex. B(i,1)/sqrt(B(1,1)^2 + B(2,1,)^2 + B(3,1)^2 + B(4,1)^2)) ) this in turn would give us a diagonalized answer for inv(B)*A*B...BUT we want W*A*inv(W). Yeah, at this point, I just went to your office hours, and the part that asks for W*A*inv(W) was a typo; it should be inv(W)*A*W. So the answer is W = [0 1/sqrt(3) 0 2/sqrt(6)] [0 1/sqrt(3) 1/sqrt(2) 1/sqrt(6)] [0 1/sqrt(3) 1/sqrt(2) 1/sqrt(6)] [1 0 0 0 ] Sent by R. Wadhera (The h is silent...but deadly) at Sun, 4 Apr 2004 17:02:28 
21. Explain why A=( 0 0) (231 0)can't be diagonalized: that is, there is no invertible matrix T so that TAT^{1} is diagonal.
Answer [ 0] det(AI)=det[231]=^{2}So the only eigenvalue of A is =0 Now find the associated eigenvectors by solving Ax=x which yields: [ 0 0][x_{1}] [0] [231 0][x_{2}]=[0]from this we find x_{1} = 0 and x_{2} is free so: (x_{1},x_{2})=(0,x_{2})=x_{2}(0,1). Our associated eigenvector when =0 is (0,1), BUT in order for A, a 2 by 2 matrix, to be diagonalizable it must have two linearly independent eigenvectors so A cannot be diagonalized. Sent by M. Woodrow at Tue Apr 6 15:22:56 2004 