Hints and Selected Answers to Homework Sets

Hints and Selected Answers to Homework Sets Phil R. Smith, Ph.D. March 8, 2 Test I: Systems of Linear Equations; Matrices Lesson. Here s a linear equation in terms of a, b, and c: a 5b + 7c 2 ( and here s another in terms of z, z 2, z, and z 4 : π z + 2 2 π ) z 2 sin z 7z 4 4 π. e Generate at least three more examples of linear equations on your own. 2.(a) Remember that the variables in a linear equation can only be raised to the power and that a term of a linear expression only consists of a single constant multiplied by single variable. (b) i. linear ii. linear iii. not linear: which term causes the problem? iv. not linear: which term causes the problem? v. not linear: which term causes the problem? vi. linear (c) i. always linear, no matter the value of k. ii. there s a problem with one of the terms; which one and why? iii. always linear, no matter the value of k..(a) i. * x 7 + 5 s 4 t ii. x 2 s x t iii. * (c) * 4. * Lesson 2 5x 4 5 x 2 5x.. Here s one system of linear equations: 2.x πx 2 + 2 π 2 + x x + x 4.72 Give three more. Use different unknowns and and different num-

hints and selected answers to homework sets 2 bers of equations. 5 2.(a) u (5,, 2) 2 (c) * a b (d) r (a, b, c, d, e) c d e 5 2. a b c d e. (e) * 2 2.(a) Augmented matrix: 7 6 2x + 2x System of linear equations: x + 7x 6x + x 2 x 2 2 x Matrix equation: 7 x 2 6 x 2 2 Column vector equation: x + x 2 + x 7 6 (c) * (d) * 4. * 5. Review your lecture notes to help you construct the three 2-D graphs to represent a linear system with no solutions, exactly one solution, or infinitely many. 6.(a) Notice that the second equation of the system can be transformed with an elementary row operation of multiplying by 2, that is, 2 R 2 R 2, giving us: x y 5 x y k 2

hints and selected answers to homework sets When k, the two equations are identical, which means that they overlap everywhere. Thus, every solution of the first equation is also a solution of the second equation, an infinite number of solutions. When k, the two equations are parallel but never intersect. Thus, there are no solutions when k. Because these are the only two possibilities for k, either k or k, the two lines can only be coincident or parallel. They can never intersect in exactly one point, so there is no k in which there is exactly one solution. (c) Look closely at the coefficients for x and y. Perhaps doing so will suggest a way for you to look at the coefficients of z and the constant column. See if you can figure out a way to equate some of the coefficients. Lesson. * 2. *. * 4. There are a couple of ways to approach this using intermediate algebra methods. One strategy is elimination solve for one variable and substitute to reduce the number of equations from to 2. For example, perhaps solve for x in the first equation and substitute the equivalent expression for x wherever x occurs in the next two equations. This will create a system of linear equations in terms of y an z only, which can be solved using beginning algebra techniques. Another approach might be to add the first two equations together to get rid of z and to add the second and third equations together to get rid of z. The two new equations will be in terms of x and y only and can be solved with beginning algebra techniques. 5. If you get stuck, here s list of row operations that will lead to the matrix being in RREF. They are, however, NOT in the correct order: R 2 + R R 2 R R 2R + R 2 R 2 R + R R

hints and selected answers to homework sets 4 R + R 2 R 2 2 R 2 R 2 R 2 + R R R + R R 6. To compute RREF on a graphing/symbolic calculator, review the calculator s manual or find a tutorial for the TI-89 (or equivalent) calculator online. For the rest of the term, it will be assumed that you know how to input a matrix into a calculator and find its RREF. 7. Here s a row vector with a leading : 2 7. Here s one without a leading : 2. On your own, give three additional examples of row vectors with leading s and three additional rows without. 8. * 9. *. Of the matrices below, there are 4 RREF matrices, 5 REF only matrices, and 4 matrices that are neither REF only nor RREF. x 8 + 7t x 2 2 t. (a) * (b) (c) Inconsistent because x + y + z x 5 t x 4 t has no solution. 2. * (d) * (e) (x, y, z) (9,, ) (Note that this is the same as writing x 9, y, and z.) Did the extra row of zeros throw you off? This is an example of an overdetermined system. Lesson 4.(a) * 2 8 (c) i. Start with the augmented matrix 2 and 7 4 perform the following row operations:

hints and selected answers to homework sets 5 ii. * R + R 2 R 2 R + R R R 2 R 2 R 2 + R R 52 R R 5R + R 2 R 2 2R + R R R 2 + R R iii. * 5 2 6 iv. Start with the augmented matrix and perform the following row 2 operations: 5 R R 2R + R 2 R 2 5R 2 R 2 2 5 R 2 + R R 2.(a) (x, y, z, w) (β, 2α, α, β) (b) (x, x 2, x ) ( 4, 2, 7) (c) inconsistent ).(a) Observe that terms like 2 y and ( z can be rewritten as 2 ( ) y and, respectively. Then consider making a series of z substitutions to turn the system of equations into a system of linear equations. (b) Plug the specific points into the general cubic equation to get a system of linear equations. Then solve. (c) * Lesson 5. * 2. *. Here s a trivial solution with 5 unknowns: x x 2 x x 4 x 5. Now give a trivial solution with 8 unknowns. 4. Here s a nontrivial solution with unknowns: a, b, c. Explain why this solution is considered nontrivial. Now give a nontrivial solution with 7 unknowns.

hints and selected answers to homework sets 6 5. Here is an example of homogeneous system of 2 equations in unknowns:. Now give a specific homogeneous 2x y 5z x + y + z system of equations in 4 unknowns. 6.(a) i. infinitely many solutions; guaranteed to have more solutions than just the trivial solution because there are four unknowns and three equations. ii. The theorem doesn t guarantee any solutions beyond the trivial one. iii. infinitely many solutions; guaranteed to have more solutions than just the trivial solution because there are three unknowns and two equations. iv. Explain why this is a trick question. Is there a way to rewrite the equations so that the Number of Solutions to a Homogeneous System of Equations Theorem would apply? (b) i. * 7. * ii. * iii. * iv. * Lesson 6. Just remember that the number of rows is always stated before the number of columns. The phrase rose columns can serve as 5 2 mnemonic device to help you remember. So, 2 is a 5 2 9 matrix because it has 5 rows and 2 columns, and 5 8 9 is a 6 matrix because it has only row and 6 columns. Now, generate specific example matrices of size 2, 4 2, and on your own. 2 4 2. is a square matrix because it has 2 rows and 2 columns; 5 7 5 8 9 is row matrix as it has only one row; and 5 4 is a column matrix because it has only one column. Now, give at least one example each of a square, column, and row matrix

hints and selected answers to homework sets 7 on your own. (BTW, note that column matrices are sometimes called column vectors, and row matrices are sometimes called row vectors.).(a) tr(a) 6 + ( 4) + 7 9 (b) Trick question! The trace function is not defined for nonsquare matrices. tr(c) 2λ 2 + + 7λ (c) (2λ + )(λ + ) λ 2 or 4.(a) a 5, b, c 4, d λ 5.(a) * (b) k 2 k B 2 k 2 k 2. 5 8 4 (c) C + D +. 2 7 5 8 (d) Before beginning this problem, check the sizes of E and F. (e) * 2 6. A T, B T 5 2 7.(a) i. 2 ii. undefined iii. undefined iv. 4 4 2 7 8.5, and C T 5 b. a 9 c (b) Start by giving arbitrary sizes to matrices A and B. For example, let A be a p q matrix and let B be an r s matrix. Then determine the sizes of the products AB and BA. 8.(a) * 4x + mq xp + m (b) 8 + q 2p + (c) Not possible to multiply. Can you explain why?. ac b (d) 2b b + c 9.(a) *

hints and selected answers to homework sets 8 (c) It s not possible to multiply a 2 4 matrix and a 4 matrix. What specifically prevents them from being multiplied?.(a) * (b) In order to multiply B by itself, s t. In order to add A + B 2, the number of rows of A and the B 2 have to be same and so do the number of columns. So, A + B 2 is defined provided that p q s t. (c) * (d) If BA is defined, then it will be an s q matrix. Then (BA) T must be a q s matrix. (e) * a a 2 a a 4 () 2 + () 2 + 2 () 2 + () 2 + 4 a.(a) i. A 2 a 22 a 2 a 24 a a 2 a a 4 (2) 2 + (2) 2 + 2 (2) 2 + (2) 2 + 4 () 2 + () 2 + 2 () 2 + () 2 + 4 a 4 a 42 a 4 a 44 (4) 2 + (4) 2 + 2 (4) 2 + (4) 2 + 4 2 4 5 5 6 7 8 2 7 8 9 2 ii. * iii. * (b) In the definition of a ij for the matrix I, consider breaking up the definition into two cases. Lesson 7. There are several ways to explain noncommutativity of matrix mutliplication, but consider looking at the sizes of specific matrices A and B in order to generate a counterexample. 2. *. 2 2 ;. Now give specific examples of zero matrices at the following additional sizes: 4 and 2. 4. I. Now, write the 5 5 identity matrix. 5. * 6.(a) 5 2 5 2

hints and selected answers to homework sets 9 5 2 2 (b) 4 4 5 (c) 2 2 5 7 Make a guess about the inverse of, then verify 2 with a calculator. (d) Trick question! Explain why this matrix has no inverse. a b 7. One case occurs when the bottom row is a zero. So, is an example as is, in which the top row is zero. (Having a row a b of zeros means that each elementary product will equal zero and the determinant will equal. If the determinate is equal to, the matrix is not invertible. What other cases are there?) 8.(a) * (5A) 5A (b) 5A A (7A T ) (7A T ) 7A T (c) A T A A (d) * 2 7 2 7 7 2 75 2 5 5 5 5 2 5 2 2 5 7 2 7 7 7 2 7 7 7 7 2 7 5 7 5 7 T 5 7

hints and selected answers to homework sets 9. In a matrix context, the 2 in the function should be treated as a matrix: 2I..(a) * (b) Use the Goobbledygook Principle to treat parts of the equation below as globs of goobledygook surrounding X. Then use inverses applied to the left and right to isolate X. Once isolated, use theorems about inverses to expand the globs of goobledygook: Lesson 8 {}}{{}}{ (BCA T ) X (CB T A) AC T Yep, I made this principle up. The word goobledygook is a corruption of the word gobbledygook, which means unclear, wordy, needlessly complicated jargon. The extra o in the first syllable is intended to distinguish the new word, and it also makes it just a little bit more fun to say. When talking about the Goobledygook Principle, I mean that we take expressions that appear complicated and treat them as simpler, independent units.. * 2. *.(a) First, rewrite the statement of the theorem as a conditional (ifthen) statement. Use the definition of an even number to write an expression for an arbitrary even number. Then use algebra to show that the square will also be even. (b) It helps to use the notation A a ij and B b ij. The tr (A + B) is the sum of the diagonals of A + B a ij + b ij : tr (A + B) (a + b ) + (a + b ) + + (a nn + b nn ). Finish the proof by showing that (a + b ) + (a + b ) + + (a nn + b nn ) tr (A) + tr (B). 4.(a) Only two of the matrices below are symmetric. (b) a, b 9, and c (c) A general formula for all symmetric 2 2 matrices is S 2 2 a b. Note that the diagonal entries can be distinct, but the b c upper right corner and the lower left corner must be the same in the 2 2 case. Find general formulas for S and S 4 4 on your own. (d) Two of the four mattrices A a ij, B b ij, C c ij, and D d ij are symmetric and two are not. 5.(a) * (c) * (d) * (e) *

hints and selected answers to homework sets 6.(a) * (c) * (d) * 7. * Lesson 9.(a) * (c) * 2.(a) E (b) E (c) E ( ) (d) E ().(a) One way is to build the matrix from the identity using row operations first and then convert them into left-multiplications of elementary matrices. It requires that you look carefully at each step along the way and it is somewhat of a trial-and-error process to get the right row operations. R R 4R 2 +R R 4 R 2 R 2 4 R +R 2 R 2 4 So, 4 A E 2 () E 2 () E 2 ( 4) E () I to start with A, reduce it to the matrix I using row operations. Write I as a product of elementary matrices and the matrix A. Then, by taking the inverses of the elementary matrices, we can write A as a product of elementary matrices. In symbols, A more systematic way uses Gauss-Jordan row reduction and the fact that the inverse of an elementary matrix will be an elementary matrix. Here s the procedure: 4

hints and selected answers to homework sets 2 i. Use Gauss-Jordan to reduce A to I. Note the series of row operations used. ii. Convert each row operation above into an elementary matrix and write I as a product of elementary matrices and the matrix A. iii. Take the inverse of each elementary matrix to write A as a product of elementary matrices. 4 R R 4 R +R 2 R 4 R 2 R 2 4 So, I Then, ( ) ( ) ( ) 4 2 E E 2 E 2 ( ) E A E 2 ( ) ( ) 4 E 2 E ( ) ( E E 2 ( ) E 2 I 4 4 R 2+R R 4 ( ) ( ) ( ) 4 2 E E 2 E 2 ( ) E A ( ) 2 ( ) E I A ) ( ) 4 2E I A E () E 2 () E 2 () 2 E ( 4 ) I A 4 4 4 Thus, 4. 4. First write the augmented matrix A I : Then apply the following row operations: (a) R + R 2 R 2 (b) R 2 + R R (c) 2 R R (d) 2R + R 2 R 2 (e) R + R R

hints and selected answers to homework sets Until the following augmented matrix I A is reached: 2 2 2 2 2 2 2 2 Thus, the inverse of A is A. 5.(a) * 25 26 (c) C 4 4 5 4 2 2 2 2 Lesson 4 2 5.(a) i. x x 2 x x 2 x x 2 x x 2 Therefore, x, x 2. 9 4 2 5 9 5 4 4 7 2 7 9 ii. * 5x + x 2 + 2x 4 iii. x + x 2 + 2x 2 x 2 + x 5 5 2 x 4 2 x 2 2 x 5 x 5 2 4 x 2 2 2 x 5 x 2 2 4 x 2 5 2 2 2 2 x 2 5 2 5 x x 2 x 6 Therefore, x, x 2, x 6.

hints and selected answers to homework sets 4 i. General solution: Specific solution: ii. Specific solution: iii. * iv. * x x 2 x x 2 x x 2 5 b 2 5 2 b 2 2 5 4 7 5 2 5 2 9 5 v. * 2. Take the inverse of 2 and left-multiply it to both sides 2 of the matrix equation. It s easiest to use a graphing/symbolic calculator to perform the inverse and to multiply the large and 5 matrices..(a) i. 7 ii. Not invertible. Why? 8 iii. 5 (b) The matrix products below include at least one diagonal matrix as a factor. Compute the product by inspection. i. Left-multiplying by a diagonal matrix is equivalent to multiplying each diagonal entry by the respective row of the other matrix: 5 2 5(2) 5() 4 ( 4) () 2 5 (2) (5) b b 2 5 4 6 5 ii. Right-multiplying by a diagonal matrix is equivalent to multiplying each column of the other matrix by the diagonal matrix s respective entry along the diagonal: 2 2 2 4 4 9 4 4 6 2 6 iii. * (c) i. A 2, A 2, 9 9 () 2( 2) (4) 2() ( 2) 4(4)

hints and selected answers to homework sets 5 ii. * A k k ( ) k k ( ) k ( ) k

hints and selected answers to homework sets 6 Test II: Determinants Lesson. (i) Two inversions: >, > 2; even permutation. (ii) * ( ) (iii) 5 2 4 Five inversions: 5 >, 5 > 2, 5 > 4, 5 >, 4 > ; odd permutation (iv) * (v) * (vi) * (vii) Trick question: not a permutation because the 5 is repeated. 2.(a) There are! 6 elementary products: b b 22 b, b b 2 b 2, b 2 b 2 b, b 2 b 2 b, b b 2 b 2, and b b 22 b. (b) Challenge problem! There are 4! 24 elementary products for C.. For the matrix B, here are the signed elementary products: ( ) b b 22 b has j-index permutation: 2. This permutation has inversions so it is even. Thus, the signed elementary product is: +b b 22 b. ( ) b b 2 b 2 has permutation: 2 ; one inversion; odd. signed elementary product: b b 2 b 2. ( ) b 2 b 2 b has permutation: 2 ; one inversion; odd. signed elementary product: b 2 b 2 b. ( ) b 2 b 2 b has permutation: 2 ; two inversions; even. signed elementary product: +b 2 b 2 b. ( ) b b 2 b 2 has permutation: 2 ; two inversions; even. signed elementary product: +b b 2 b 2. ( ) b b 22 b has permutation: 2 ; three inversions; odd. signed elementary product: b b 22 b. Note that determinant of B is simply the sum of all of its signed elementary products. So, det (B) +b b 22 b + b 2 b 2 b + b b 2 b 2 b b 2 b 2 b 2 b 2 b b b 22 b. Now, find the signed elementary products of C on your own. 4.(a) * (c) 5 6 2 7 +( 5)( 7) (6)( 2) 5 + 2 47

hints and selected answers to homework sets 7 (d) (e) (f) * (g) * 5. * a 5 a 2 2 7 6 5 2 8 4 +(a )(a 2) (5)( ) a2 5a + 2 +( 2)()(4) + (7)( 2)() + (6)(5)(8) (6)()() ( 2)( 2)(8) (7)(5)(4) 8 42 + 24 8 2 4 6.(a) * (b) det (B) ψ 5 det ψ 4 ψ 4 ψ 9ψ 2 + 8ψ + 6 (ψ + 2)(ψ 5)(ψ 6) ψ 2, ψ 5, or ψ 6 (c) 7.(a) x det (A ) det (A) x 2 det (A 2) det (A) x 2 + x x 2 + x 6 x 2 2x 2 + x 6 (2x )(x + 2) x or 2 2 5 7 2 7 5 7 2 2 x 6 x x 2 26 2

hints and selected answers to homework sets 8 (c) * (d) x det (A ) det (A) x 2 det (A 2) det (A) x det (A ) det (A) 4 2 2 4 4 2 2 4 2 4 4 2 2 4 2 4 8 8 4 4 (e) * Lesson 2. Calculate determinants by inspection. (a) Use basic properties of determinants to calculate the determinants below. Don t use a calculator; look instead for patterns related to the basic properties of determinants. 5 7 i. 4 2 ()(4)( ) 6 because the matrix is (upper) triangular and the determinant of a triangular matrix is the product of its diagonal entries. 7 5 ii. 5 2 The matrix has a row of zeros, every signed elementary product will include a zero factor. If zero is a factor, the resulting product must be zero. Since the determinant is the sum of

hints and selected answers to homework sets 9 the signed elementary products, all of which are zero, the determinant will also be zero. iii. This is a determinant of a lower triangular matrix. Use that observation to compute the determinant. iv. A column of zeros is analogous to a row of zeros. Use this fact to compute the determinant. v. Observe rows and. Is it possible to use a row operation that does not change the value of the determinant to get a row of zeros? vi. * vii. * viii. This matrix is one operation away from the identity matrix I 4. Also, note that it is a diagonal matrix, which is just a matrix that is both upper and lower triangular. ix. What row operations were applied to I 4 to get to this matrix? Use the answer to this question to compute the determinant. x. * xi. * a b c (b) Given that d e f 5, find the following determinants: g h i d e f g h i a b c R i. R 2 R g h i R d e f d e f 5 a b c a b c g h i a b c a b c ii. d e f ()( )(4) d e f ()( )(4)( 5) 6 4g 4h 4i g h i a + g b + h c + i a b c R +R iii. R d e f d e f 5 g h i g h i iv. *

hints and selected answers to homework sets 2 2.(a) i. 6 9 2 2 5 2 2 2 5 2 2 5 2 5 2 ( )()(5)( 2) ii. * iii. * 4 4 2 2 4 2 iv. 4 4 2 2 9 2 4 4 2 2 29 9 4 4 4 2 2 ( )( ) 4 29 9 4 4 2 2 4 5 5 v. * 5 (b) Start with the left side and use row-reduction. See if you can

hints and selected answers to homework sets 2 get to the following: (b a) c a b a. Continue b 2 a 2 c 2 a 2 to use row reduction to produce an upper triangular matrix to establish the identity: a b c (b a)(c a)(c b) a 2 b 2 c 2 The Goobbledygook Principle, that groups of objects often within parentheses can sometimes be treated a single object, might be of use in simplifying the above. Consider the following factorization using the Goobbledygook Principle: (x 2 y 2 ) (z x)(x y) (x + y)(x y) (z x)(x y) (x y)(x + y) (z x) (x y)(2x + y z) Lesson.(a) Suppose that A and B are matrices with det (A) 5 and det (B) 2. i. det (A ) det (AAA) det (A) det (A) det (A) (5)(5)(5) 25 ii. Trick question! Why? iii. Remember that 2B means that 2 is multiplied by every row of B so it needs to be pulled out three times, one for each row of B. det (2B) 2 det (B) 8( 2) 6 iv. * v. * (b) Suppose that C and D are 4 4 matrices with det (C) and det (D) 7. 2.(a) * i. det (C T C ) det (C T ) det (C ) det (C) det (C ) det (C) det (C) ii. Trick question, why? iii. Remember that 2 is being multiplied with each row of D2. So, det ( 2 D2 ) ( 2 )4 det (D 2 ) 6 49 det (D) det (D) 6 iv. * v. *

hints and selected answers to homework sets 22 4 2 8 4 2 8 (b) 2 4 96. Therefore, 2 4 is invertible 6 6 by the Determinant Test for Invertibility Theorem. 2 7 (c) Because 2 2 7 7 has a column of zeros, 2 7. 5 9 5 9 2 7 Thus, 2 7 is not invertible. 5 9 (d) *.(a) Look for proportional rows or columns after substituting x and x 2. (b) Can you think of a single row operation that will not change the value of the determinant that will yield proportional rows? Lesson 4.(a) * (c) * 2.(a) (i) All the components of u (,, 4) double: 2u (2, 2, 2 ( 4)) (2, 6, 8). Also, multiplying by k 2, doubles the length of the matrix. (ii) * (iii) * (iv) Multiplying by the scalar k produces the zero vector (,, ). (v) * (vi) Multiplying a vector u by the scalar k, reflects the vector through the origin: u (,, 4). (vii) * (b) i. * ii. * iii. Multiplying any vector u by < k < shrinks the size of the vector by a factor of k; however, it still points in the same direction. iv. * v. * vi. * vii. Multiplying any vector u by k <, reflects the vector through the origin and increases its length by the factor k.

hints and selected answers to homework sets 2.(a) * (c) u + v (, 4) + (2, ) ( + 2, 4 + ) (, 7) 7 2 2 4 7 4.(a) + 2 + 2 4 (c) * (d) kv + lw b k(2, ) + l((, 2) (, ) (2k l, k + 2l) (, ) 2k l Solving the linear system k + 2l Jordan Elimination Method: (e) * (f) * 2 2 G-J 2 2 (2, ) + (, 2) (, ) k 2 l with the Gauss- 5.(a) {k(, ) : k IR} is the 45 line passing through the origin. Its equation is y x. (b) {a(, ) : a Z} is the set of integer lattice points (points that lie on the intersection of integer grid lines) along the line x-axis, e.g.,... (, ), ( 2, ), (, ), (, ), (, ), (2, ), (, ),... (c) * (d) * (e) The entire xy-plane. The set of vectors {(, ), (, )} spans IR 2. (f) The linear combination of the two vectors will be a plane in -space. For those of you who have had Calc III, the equation of a plane can be written if a point on the plane is given along with a normal vector to that plane. The linear combination includes (,, ), so (,, ) is a point on the plane and a normal vector can be found by taking the cross-product of the two vectors: (,, ) (,, ) (,, ). Thus, the locus of the points represented by the linear combination of the two vectors is (x ) (y ) + (z ), which simplifies to x y + z. 6. For the vectors u and v below, calculate the dot product u v and the outer product u v.

hints and selected answers to homework sets 24 (a) * 2 (b) u v u T v 6 5 ()(2) + ()(5) + ( 6)( ) 29 ()(2) ()(5) ()( ) u v uv T 2 5 ()(2) ()(5) ()( ) 6 ( 6)(2) ( 6)(5) ( 6)( ) 6 5 9 2 5 2 8 (c) * 7.(a) * (b) w w w 2 + 2 2 + 5 2 + 4 + 25 (c) x x x ( 4) 2 + ( ) 2 25 5 (d) * 8.(a) Think of the two vectors as hands of an analog (not digital) clock. When are the tips of the hands as close as possible and when are they farthest apart? (b) Use the Magnitude of a Scalar-Vector Product Theorem to separate k and w. Then compute the magnitude of w and solve for k. (c) Try taking the magnitude of ŵ w and applying the w Magnitude of a Scalar-Vector Product Theorem. (d) ŵ w w 2 5 (2, 5) (2, 5) (, ). (2, 5) 29 29 29 (e) * 9.(a) i. u v u v cos (θ) (2, ) (, 7) (2, ) (, 7) cos (θ) (2)() + ( )( 7) 5 58 ( cos (θ) ) θ cos 29 ii. * θ 4.24 (b) Without calculating the exact value of the angle, determine whether the vectors u and v make an acute angle, make an obtuse angle, or are orthogonal.

hints and selected answers to homework sets 25 i. Since u v, then u v. ii. Since u v <, then the angle θ between the vectors u and v is obtuse. iii. Since u v 6 >, then the angle θ between the vectors u and v is acute. iv. * (c) i. a and b are parallel provided that they are scalar multiples of one another. So, n(2, k) (, 5) yields the following system of equations (note that the system is nonlinear but still solvable): 2n with solultions n 2 nk 5 and k. Alternatively, and perhaps more simply, another way to think of scalar multiples is that the two vectors are proportional. Then, their respective components must be proportional. 2 k 5 which has the same solution k. ii. By the Orthogonal Vectors Theorem, two vectors are orthogonal provided that their dot product is equal to zero. Use this fact to set up an appropriate equation and then solve. iii. Use the Angle Between Vectors Theorem to set up a trigonometric equation involving cos θ; then solve for k. iv. *.(a) proj a u u a (6, ) (, 9) a a 2 (, 9) 2 (, 9) 9 (, 9) ( 9 ), 9. ( 8 (b) proj b w, 2 ) (c) proj y x x y (,, 7) (,, 5) y y 2 (,, 5) 2 (,, 5) 2 (,, 5) ( 26 6 ) 8,, where x (,, 7) and y (,, 5). ( (d) proj x y 96 59, 2 59, 224 ) 59.(a) x proj a x x a (2, 5) (, ) a a 2 (, ) 2 (, ) (, ) (, ) ( x x x (2, 5), ) (, 9 ) (b) x (,, ) and x (2,, ) ( 5 (c) x 7, 7, 5 ) ( 6 and x 7 7, 8 7, 2 ) 7

hints and selected answers to homework sets 26 2.(a) Find the distance between the point (2, ) and the line 4x + 2y + 7. First, pick any point on the line. Letting x and solving for y, we get the point Q(, 7 2 ). Let x be the vector that begins at Q(, 2 7 ) and terminates at P(2, ): x (2, 2 9 ). Note that the vector x is not necessarily perpendicular to the line, but we know that n (4, 2) is perpendicular to given line. So, the projection of x onto n would be orthogonal to the line and the length of the proj n x will the distance from the point to the line: proj n x x n n 2 n (2, 2 9 ) (4, 2) (4, 2) 2 (4, 2) 2 7 (4, 2) The magnitude of proj n x 2 7 (4, 2) is the distance from the point P(2, ) to the line: proj n x 7 2 (4, 2) 7 2 7 (4, 2) 2 2 7 5.8 (b) Follow the procedure of the problem above but for the arbitrary point (s, t) and the arbitrary line ax + by + c..(a) Prove: Symmetry Property. If u IR 2 and v IR 2, then u v v u. i. Assume u IR 2 and v IR 2. ii. Let u (u, u 2 ) and v (v, v 2 ). iii. Then u v u v + u 2 v 2 Definition of Inner (Dot) Product iv. v u + v 2 u 2 Commutative Property of Addition v. v u Definition of Inner (Dot) Product. Q.E.D. How would you modify the proof above for IR? (c) * (d) * Note that u, u 2, v, and v 2 are all real-valued scalars. Lesson 5 ( ) 2.(a) v w v (, 2, ) (2, 5, 4) 5 4, 2 4, 2 2 5 (,, ). (c) * 2. *.(a) u v (,, 4) (2,, ) (7,, 5). u (u v) (,, 4) (7,, 5) 2 2

hints and selected answers to homework sets 27 u (u v) v (u v) (2,, ) (7,, 5) 4 + 5 v (u v) 4.(a) Area(parallelogram spanned by u & v) u v (,, 2) (,, 4) ( 2, 2, 2) 2 (b) First, find two vectors that span the parallelogram (e.g., PQ and PR), then compute the magnitude of the cross product of these two vectors. What other pairs of vectors can be used to find the area of the parallelogram? 5.(a) Area(triangle spanned by u & v) 2 u v 2 (,, ) (2, 2, 2) 2 ( 4, 2, 8) 2 4 (b) Use the given three points A, B, and C to find two vectors that span the triangle. Then take half of the magnitude of the cross product of these two vectors. 6.(a) Given P(, ) and n (5, ): n PX (5, ) (x, y ( )) 5(x ) (y + ) 5x y 6 7.(a) Given P(,, 2) and n ( 2,, ): n PX ( 2,, ) (x, y ( ), z 2) 2(x ) + (y + ) + (z 2) 2x + y + z + 2x y z (c) * 8.(a) First note that the three points P(5,, 4), Q(2,, 7), and R(, 4, 5) determine a plane. Using the given points, find two vectors in the plane. Take the cross product of the two vectors to find an orthogonal vector to the plane. Use the vector equation n PX to write the equation of the plane. 9.(a) i. X vt + P (x, y) (, )t + ( 2, ) (x, y) (t 2, t + ) x t 2 y t +

hints and selected answers to homework sets 28 ii. * iii. Find a direction vector parallel to line by computing v PQ. Then use either P or Q as the fixed point F in the formula X vt + F. Note that you have freedom to use PQ or PQ as the direction vector and either P or Q as the fixed point. Do all of the choices lead to the same set of parametric equations? If the same, show algebraically that they are equivalent. If different, explain how two different representations can yield the same line. (b) i. * X vt + P (x, y, z) (,, )t + (2, 2, 5) ii. P(2, 2, 5) and v (,, ) (x, y, z) (t + 2, t + 2, t + 5) x t + 2 y 2 z t + 5 iii. Find a direction vector parallel to line by computing v QP. Then use either P or Q as the fixed point F in the formula X vt + F..(a) i. The vector (5,, 7) is orthogonal to the plane 5x y + 7z 2 and and the vector (, 9, 4) is orthogonal to the plane x + 9y 4z. The planes can only be parallel to one another if their respective orthogonal vectors are parallel. Since (5,, 7) k((, 9, 4) for any real valued k, then the vectors are not parallel and their associated planes are not parallel. Another way to determine if two D vectors are parallel is to take their cross product. If the cross product is the zero vector, then the two vectors are parallel. In this case, (5,, 7) (, 9, 4) ( 5, 4, 54) (,, ). Since the cross product of the two vectors is not the zero vector, then the two vectors are not parallel and neither are their associated planes. ii. * (b) i. The vector ( 4,, 2) is parallel to the line x 5 4t, y t, z + 2t and the vector (, 2, ) is normal to the plane x + 2y + z 9. The given line and plane will only be parallel if the vector in the same direction as the line and the vector normal to the plane are orthogonal. Taking the dot product of the two vectors, we get ( 4,, 2) (, 2, ) ( 4)() + ( )(2) + (2)(). Thus, the direction vector

hints and selected answers to homework sets 29 ii. * and the normal vectors are orthogonal, so the given line and plane are parallel. (c) i. Two planes are perpendicular only if their respective normal vectors are orthogonal. The normal vector for the plane x y + z 4 is n (,, ) and the normal vector for the plane x + 2z is n 2 (,, 2). Taking the dot product, we get n n 2 (,, ) (,, 2) 5. Since the n n 2, then the normal vectors are not orthogonal and their respective planes are not perpendicular. ii. * (d) i. A given line and plane are perpendicular when the direction vector of the line is parallel to the normal vector of the plane. The line x 2 4t, y 2t, z + 2t has direction vector ( 4, 2, 2) (which is parallel to the line), and 2x + y z 5 has normal vector (2,, ) (which is normal to the plane). Since ( 4, 2, 2) and (2,, ) are scalar multiples of one another, the two vectors are parallel, meaning that their associated line and plane are perpendicular. ii. * (e) There are two ways to approach this problem. The two planes intersect in a line. The direction vector for that line will be in the same direction as the cross product of the normal vectors of the given planes, (7, 2, ) and (,, 2) respectively. Taking the cross product, we get (7, 2, ) (,, 2) ( 7, 2, ), which will be the direction vector for the line of intersection. Now, we need to find a point that lies on line. Both planes will have a point when x, so, substituting gives us the system of equations 2x + z 2 y + 2z 5, with solution y 7, z 2 7. Using (, 7, 2 7 ) as a point on the line of intersection and ( 7, 2, ) as the direction vector of the line of intersection, we can write the following equation for the line: X vt + P (x, y, z) ( 7, 2, )t + (, 7, 2 7 ) (x, y, z) ( 7t, 2t 7, t 2 7 ) x 7t y 2t 7 z t 2 7 As an alternate solution, the two planes can be thought of as forming a system of linear equations, the solution of which is

hints and selected answers to homework sets 7 2 2 7 2 the line of intersection: G-J 2 5 2 4 x 7s 2 which has parametric solutions: y 2s 4 z s Observe that the the first parametric solution can be transformed into the second by substituting t s + 2 7. (f) A plane will be parallel to another plane if their respective normal vectors are parallel. Since we want the new plane to be parallel to the old plane 5x 2y + z 5, we can just use its normal vector n (5, 2, ). With the given point (, 6, 7) and the normal vector n (5, 2, ), the equation of the new plane is: n PX (5, 2, ) (x, y ( 6), z 7) 5(x ) 2(y + 6) + (z 7) 5x 2y + z 4 (g) Find the point of intersection by substituting the parametric equations x 5t + 9, y t, z t + for x, y, z in the equation for the plane 2x y + 4z + 7 : 2x y + 4z + 7 x 5t + 9, y t, z t + 2( 5t + 9) ( t ) + 4(t + ) + 7 t 4 t 4 Substituting the solution t 4 into the parametric equations produces the point of intersection: x 5( 4 ) + 9 7 y 4 z 4 + 49 So, the point of intersection of the line and plane is ( 7, 4, 49 ). Lesson 6. * 2.(a) Find u (,,, 9) (,,, ). (b) Find 2v w 2(2,,,, 7) (4, 2,,, ) ( 8, 8,,, 5) (c) Find scalars k, k 2, k, and k 4 such that k (,, 2, ) + k 2 (2,, 4, ) + k (7,,, 4) + k 4 (6,,, 2) (, 5, 6, ). (d) Linear vector equation: k (,, 2, ) + k 2 (2,, 4, ) + k (7,,, 4) + k 4 (6,,, 2) (, 5, 6, )

hints and selected answers to homework sets k + 2k 2 + 7k + 6k 4 yields the following system of linear equations: k + k + k 4 5 2k + 4k 2 + k + k 4 6 k 2 + 4k + 2k 4 which can be written as an augmented matrix and solved with the Gauss-Jordan Elimination algorithm: 2 7 6 5 2 4 6 4 2 The solution to the linear vector equation is k k 2 k 4 and k..(a) Find u v (,,,, 2, ) (, 2, 4,, 2, ) 2. (b) Find x y (, 7a, 2b,, ) (, 2,,, 4c) 4a + 2b. 4.(a) u (, 4,, 5, 2, ) 4. k( 2,,, 6) 5 k ( 2,,, 6) 5 (b) 7 k 5. k 5 7 k ± 5 7 v (c) Take the magnitude of v. (d) The distance between u and v is the length of the vector u v: u v. u v (, 2,, ) (, 2, 4, 4) (, 4, 5, ) 59 (e) x y (,, 2,, ) (, 5,, 4, 2) (4, 8, 2, 5, 2) 5.(a) u v (2, 4,,, 2, ) (, 5, 4,,, 2) 4. Therefore, the vectors are not orthogonal; the angle between the two vectors is obtuse. 6.(a) Start by letting u be an arbitrary n-tuple: u (u, u 2,..., u n ). Then replace u on one side of the equality and then produce the other side.

hints and selected answers to homework sets 2 Test III: General Vector Spaces Lesson 7. * 2.(a) Consider Axiom 8, the Scalar Distributive Property: (α + β)x αx + βx. Observe that the left hand side (LHS) simplifies as follows: (α + β)x (α + β)(x, y) (x, (α + β)y) But the right hand side (RHS) simplies this way: αx + βx α(x, y) + β(x, y) (x, αy) + (x, βy) (2x, (α + β)y) Since the LHS RHS except when x, then Axiom 8 is not true in general. The set of objects above with the two given operations is not a real vector space. (b) Note that V is the right half-plane and is the set of ordered pairs in which the first component of (x, y) is always positive or zero. What axiom, if any, might need negative numbers? (c) It s often a good idea to test the simplest axiom(s) first. (d) Consider Axiom, the Scalar Identity Property: x x x (x, y) (2x, 2y) (x, y) Axiom fails and V is not a real vector space. Challenge: Can you find the other axiom that fails? Look closely at how scalar multiplication is defined. Then look for axioms that involve scalar multiplication. (e) Note that for the set of 2 2 matrices defined here, the s along the anti-diagonal are invariant. That means to be in the set, the number must be in each of those corners under all operations. Look for axioms that might lead to a change in the corners. (f) V is a vector space. You ll need to show all axioms hold. Remember to produce the zero vector when testing Axiom 4 and the additive inverse (formula) when testing Axiom 5. Most other axioms are equality axioms in which it is necessary to show that the left hand and the right hand sides are equal. (g) V is not a vector space. Look for an axiom that might change the value of the +. (h) V is a vector space. You ll need to show all axioms hold. Remember to produce the zero vector when testing Axiom 4 and the additive inverse (formula) when testing Axiom 5. Most other axioms are equality axioms in which it is necessary to show that the left hand and the right hand sides are equal. To show that something is a real vector space all axioms must be satisfied. To show something is not a real vector space, find at least one axiom that fails to hold.

hints and selected answers to homework sets (i) Note that (, ) V since 2( ) + () and ( 4, ) V since 2( 4) + (). Can you use one or both of these specific vectors to generate a counterexample to one of the axioms? (j) Write the solution to 2x + y as parametric equations for x and y. Use the resulting general vector (x, y) to help you determine if the axioms are satisfied. (k) For what values of c, will V be a vector space?. Establish additional properties of real vector spaces using mathematical proof. (a) i. ii. VS Axiom 7: Vector Distributive Property iii. VS Axiom 4: Additive Identity Property iv. VS Axiom 6: Closure Property of Scalar Multiplication v. VS Axiom 5: Additive Inverse Property vi. vii. Substituion (Note: Earlier in proof we showed that k + ku ku.) viii. VS Axiom : Associative Property (c) * ix. VS Axiom 5: Additive Inverse Property (Note: The axiom is being applied two times.) x. VS Axiom 4: Additive Identity xi. Lesson 8.(a) Three of the subsets S, S 2, S, S 4, S 5 are also subspaces under the usual vector operations of IR. 2 (b) i. Consider the following counterexample: T but 4 2 2 2 2 2 4 2 4 / T because the entries 2, 2 2, 2 2 and 4 2 are not integers. ii. It is a subspace. Verify this fact by establishing closure under addition and scalar multiplication. iii. It is a subspace. Verify this fact by establishing closure under addition and scalar multiplication. iv. Look for a counterexample. Can you find two matrices with determinants equal to zero, but whose sum does not have a determinant of zero?

hints and selected answers to homework sets 4 v. * (c) Two of the subsets U, U 2, U, U 4 are also subspaces under the usual vector operations of P. (d) Three of the subsets G, G 2, G, G 4, G 5 are also subspaces under the usual vector operations of F(, ). 2.(a) * (b) First, add v + w and v w. Then subtract v + w and v w. 5 aa + bb + cc 7 (c) 4 2 a + b + c 2 2 2 4 5 7 4a + b b + 2c 5 2a + 2b + c 7 2a + b + 4c Converting to an augmented matrix and applying the Gauss- Jordan algorithm: 4 2 5 2 2 7 2 4 G-J The last row of the RREF matrix indicates that the system of linear equations has no solution. So there are no values a, b, and 5 c that can be used to write as a linear combination of 7 A, B, and C. (d) The polynomial can be written as a linear combination 6 + x + 6x 2 αp + βp 2 + γp where α 4, β 5, and γ..(a) Let (x, y) be any vector in IR 2. To show that IR 2 span {v, w} where v (2, ) and w (, 4), it suffices to show that, given any vector (x, y), a set of scalars m and n can be found such that the following equation is true: (x, y) m(2, ) + n(, 4). In algebraic terms, this means we need to find m and n in terms of x and y: (x, y) m(2, ) + n(, 4) x 2m + n y m + 4n Which can be solved as a matrix equation:

hints and selected answers to homework sets 5 2 m 4 n m n m n x y 2 x 4 y 45 5 x y 5 2 5 Thus, for any arbitrary vector (x, y), we can find m and n such that (x, y) m(2, ) + n(, 4) using the following formulas: m 5 4 x 5 y n 5 x + 2 5 y (b) Notice that v (2, 8) and w (, 4) are scalar multiples. What are the consequences for the span when two vectors are scalar multiples? (c) No, it doesn t span IR 2. Explain why one vector cannot be the span of IR 2. (d) Is IR span {x, y, z} where x (2, 2, 2), y (,, ), z (,, )? Let (a, b, c) be any vector in IR. To show that IR span {x, y, z} where x (2, 2, 2), y (,, ), z (,, ), it suffices to show that, given any vector (a, b, c), a set of scalars p, q, and r can be found such that the following equation is true: (a, b, c) p(2, 2, 2) + q(,, ) + r(,, ). In algebraic terms, this means we need to find p, q, and r in terms of a, b, and c: (a, b, c) p(2, 2, 2) + q(,, ) + r(,, ) a 2p b 2p + r c 2p + q + r Which can be solved as a matrix equation: 2 p a 2 q b 2 r c p 2 a q 2 b r 2 c p 2 a q b r c Thus, for any arbitrary vector (a, b, c), we can find p, q, and r such that (a, b, c) p(2, 2, 2) + q(,, ) + r(,, ) using the following formulas:

hints and selected answers to homework sets 6 (e) * (f) * p 2 a q b + c r a + b 4.(a) Let f cos 2 x and g sin 2 x. Which of the following vectors lie in span {f, g}? i. Yes, use the trigonometric double angle formula to rewrite cos 2x. ii. No, explain why not. iii. Yes, which trigonometric identity did you use? iv. Yes, which scalars did you multiply f and g by? v. No, explain why not. (b) Let x (2,,, ), y (,, 5, 2), and z (,, 2, ). Which of the following vectors lie in span {x, y, z}? i. (2,, 7, ) will be in the span span {x, y, z} if the following linear vector equation has at least one solution: a(2,,, ) + b(,, 5, 2) + c(,, 2, ) (2,, 7, ). Write the resulting system of linear equations as an augmented matrix and reduce by the Gauss-Jordan Elimination algorithm: 2 2 2 5 2 7 G-J 2 Thus, 2(2,,, ) (,, 5, 2) (,, 2, ) (2,, 7, ) and (2,, 7, ) span {x, y, z}. ii. Yes, observe that (,,, ) x + y + z. iii. No, it s not. Now, show why it is not. (c) What geometric shape is represented by spans below? Then give the shape s mathematical equation or set of equations. i. The line passing through the origin and in the direction of (2, 5): x 2t y 5t ii. Note that (, ) and (6, 2) are scalar multiples of one another. The span of the two vectors is the line passing through the origin and in the direction of (, ): x t y t

hints and selected answers to homework sets 7 iii. The line passing through the origin and in the direction of (2,, ): x 2t iv. IR 2 v. * y t z t vi. It s going to be a plane in IR. To find the equation of the plane find a vector n that is normal to both (,, ) and (4, 4, ). (What vector operation will generate such a vector?) Then, observe that the point (,, ) lies on the plane. (Why is this so?). Use the normal vector n, the origin as the fixed point P, and the equation n PX to find the equation for the span of the two vectors. (d) i. () There are three equations m. (2) There are four unknowns n 4. () Solving with Gauss-Jordan: t x x 5 t x 2 5 5 7 t. x 4 t This is the line passing through the origin in the direction of (, -, 7, 5). (Note that any scalar multiple of ( 5, 5, 5 7, ) will work here.) (4) The the solution space to the homogeneous system is described by {(,, 7, 5)t : t IR} and is a subspace of IR 4. ii. () There are four equations m 4. (2) There are five un- x s t x 2 s knowns n 5. () Solving with Gauss-Jordan: x t which can be written in column vector form: x x 2 x s + t. x 4 x 5 x 4 x 5 t (4) The the solution space to the homogeneous system is de-

hints and selected answers to homework sets 8 scribed by s + t : s, t IR and is a subspace of IR 5. iii. * Lesson 9. * 2.(a) The following sets of vectors are taken from vector spaces IR 2, IR, IR 4, P 2, and M 2 2. In each case determine whether the set of vectors is linearly independent or dependent. Then explain why. i. Linearly dependent. Note that the two vectors are scalar multiples of one another. ii. The cardinality of the set is and the dimension of IR 2 is 2. Therefore, the set is linearly dependent by the Cardinality Exceeds Dimension Theorem. iii. * iv. * v. Linearly independent. Note that the only solution to the trivial equation k (4,, 2) + k 2 ( 4,, 2) (,, ) is the trivial solutionk k 2. vi. Compare the cardinality of the set with dimension of the vector space. vii. Investigate the associated trivial equation to determine whether the set of vectors is LI or LD. viii. Investigate the associated trivial equation to determine whether the set of vectors is LI or LD. ix. There are only two vectors in the set and clearly they are not scalar multiples of one another. (Why?). Therefore, the set is linearly independent. (b) For which real values of λ do the following vectors form a linearly dependent set in IR? x (λ, 2, 2 ), x 2 ( 2, λ, 2 ), x ( 2, 2, λ) Although it s clear that the set {x, x, x }, where x (λ, 2, 2 ), x 2 ( 2, λ, 2 ), x ( 2, 2, λ), is linearly dependent when

hints and selected answers to homework sets 9 λ 2 (why?), it is not at all clear if there are any other values for λ in which the set of vectors will be linearly dependent. Guess-and-check might lead to other values, but there would always be doubt whether all of the values had been found. Approach the question systematically by converting the trivial equation: a(λ, 2, 2 ) + b( 2, λ, 2 ) + c( 2, 2, λ) (,, ) into a system of linear equations in a, b, c. Notice that the coefficient matrix has numerical entries but also some entries in terms of λ. Review the theorems on determinants and homogeneous systems of linear equations in order to make a statement about the coefficient matrix and when it will have only one solution or when it will have an infinite number of solutions. Find all the values of λ in which the homogeneous system will have infinitely many solutions, indicating that the set will be linearly dependent.. Establish properties of linear dependent and independent sets using mathematical proof. (a) As for all proofs of conditional statements, assume the if part of the conditional. Use the definition of linear dependence to write a statement about the solutions to the trivial equation for the set {x, x 2, x }. Now explore the set {x, x 2, x, x 4 }. Write the trivial equation for this set and investigate its solution(s). Use the outcome of the investigation to determine whether the new set is LI or LD. (b) As for all proofs of conditional statements, assume the if part of the conditional. Use the definition of linear independence to write a statement about the solutions to the trivial equation for the set {x, x 2, x }. Now explore the set {x, x 2, x, }. Write the trivial equation for this set and investigate its solution(s). Use the outcome of the investigation to determine whether the new set is LI or LD. Lesson 2. * 2.(a) From inspection, explain why the following sets of vectors are not bases for the indicated vector spaces. i. Notice that S {(2, ), (, 5), (, 7)} consists of three vectors from IR 2. What does this tell you about linear independence/dependence?

hints and selected answers to homework sets 4 ii. In a previous homework problem, you proved that a set of vectors that contained a zero vector would be linearly dependent. Thus, S {(, ), (, 7)} is LD and cannot be a basis for IR 2. iii. Notice that the two vectors in S {(, 9), ( 5, 5)} are scalar multiples of one another. Can S be a basis? Why or why not? iv. T {( 2,, ), (5,, )} doesn t span IR. (Why?) Therefore, T is not a basis for IR. v. * vi. * (b) Using the definition of a basis or any theorems about bases, determine whether the following sets of vectors are bases for the indicated vector spaces. i. In order for J to be linearly independent, there can be only one solution to the trivial vector equation: a(,, ) + b(2, 2, ) + c(,, ) (,, ). (Can you write the system of three linear equations?) In order for J to span IR, there must be one solution to the following equation: a(,, ) + b(2, 2, ) + c(,, ) (x, y, z). (Can you write the system of three linear equations?) Both linear vector equations above, have the same coefficient matrix: 2 2 2 Since 2 6, then J is LI and J spans IR. Thus, J {(,, ), (2, 2, ), (,, )} is a basis for IR. ii. * iii. Both the trivial equation k( x + 2x 2 ) + l( + x + 4x 2 ) + m( 7x) + x + x 2 and the spanning equation k( x + 2x 2 ) + l( + x + 4x 2 ) + m( 7x) a + bx + cx 2 have the same coefficient matrix: 7. 2 4 2 Since 2, then L is LD and L doesn t span P 2. Thus, L { x + 2x 2, + x + 4x 2, 7x} is not a basis for P 2.

hints and selected answers to homework sets 4 iv. * 6 8 v. Both the trivial equation a( + b + c + d 6 2 4 2 6 8 x y and the spanning equation a( + b + c + d 6 2 4 2 z w have the same coefficient matrix: 6 8 2. 6 4 2 6 8 Since 48, then N is LI and N 2 6 4 2 { } 6 8 spans M 2 2. Thus, N,,, 6 2 4 2 is a basis for M 2 2..(a) u 2(, ) ( 2, ) (, + 2 ) (c) (v) P 2 2 2 2 (d) To find the coordinate vector (w) S, we must write w (, ) as a linear combination of the vectors in the basis S {(2, 4), (, 8)}: (e) * c (2, 4) + c 2 (, 8) (, ) 2c + c 2 4c + 8c 2 Thus, (w) S ( 5 28, 4 ) c 5 28 c 2 4 c (, ) + c 2 (, 2) (a, b) (f) c a c + 2c 2 b c a c 2 b a 2

hints and selected answers to homework sets 42 (g) * Thus, (x) U ( a, b a ) 2 c + c 2 + c + c 4 2 (h) c + c 2 2 c + c 2 c c 4 c c 2 c c 4 Thus, (u) J (,,, ) 4.(a) The augmented matrix can be reduced as follows with the Gauss-Jordan Elimination algorithm: 4 2 G-J 5 2 5 The RREF form of the matrix translates as the following parametric equations: x θ + 5 2 λ x 2 θ 5 λ x θ x 4 λ Which can be written in column vector format: 2 x 5 2 x 2 θ + λ 5 θ + ρ 5 x x 4 2 From the above, S, clearly spans the solution space. The set S is also LI (from the pattern of zeros in the 5 two vectors, they cannot be scalar multiples of one another.). Therefore S is a basis of the solution space of the homogeneous

hints and selected answers to homework sets 4 system of equations. Since there are two basis vectors in the solution space, its dimension is 2. (b) The augmented matrix can be reduced as follows with the Gauss-Jordan Elimination algorithm: 2 2 4 G-J 6 5 2 The solution of which is the trivial space which consists only of the zero vector:. Remember that we define the dimension of the trivial space to be. Since the dimension is the number of basis vectors and the dimension in this case is, this means that the basis for the trivial space contains no vectors. Thus, the basis of the solution space of this set of homogeneous equations is and its dimension is. (c) Note that the second and third equations are simply scalar multiples of the first equation: x x 2 + x After parameterization, the solution set can be written in vector form: x x 2 q + r x Since T, spans the solution set and is linearly independent, then T is a basis for the solution space of the homogeneous system and its dimension is 2. (d) * 5.(a) i. The equation of the plane 2x y + 7z can be parameterized as: x 2 α 7 2 β y α z β

hints and selected answers to homework sets 44 The parameterized equations in vector format yield: x 2 7 2 7 y α + β γ 2 + δ (result after z 2 subbing α 2γ and β 2δ) 7 The set 2, spans the plane and is linearly independent; thus, it is the basis for the plane with dimension 2 2. ii. Remember that, although the equation has only two unknowns x and y, it is still a plane in IR. So, you ll also need to parameterize z. iii. In vector form the parametric equations of the line x 2t, y t, z 5t can be written as follows: x 2 y t z 5 Use the equation above to find a basis for the line and its dimension. iv. * v. * (b) i. Since (w, x, y, ) w(,,, ) + x(,,, ) + y(,,, ), it s clear that the set {(,,, ), (,,, ), (,,, )} spans the subspace. From the pattern of s and s in the vectors, it is also clear that none of the vectors can be written as linear combinations of the others so the set is also linearly independent. Since the set spans the subspace and is linearly independent, the set {(,,, ), (,,, ), (,,, )} is a basis and dimension of the space is. ii. Explain why the basis is {(,,, )} and the dimension is. iii. To find the basis and dimension of {(w, x, y, z) IR 4 : y w x, z w + x}, use the observation below: (w, x, y, z) (w, x, w x, w + x) w(,,, ) + x(,,, ) (c) The set {x 4, x, x 2, x} is a basis for the subspace because it spans the subspace and it is also linearly independent. The dimension is 4. Lesson 2. The row vectors are r 2 5 4 6, r 2 2 5, and r 4 2 2. The column vectors are c 2, 4