Math 51, Homework-2 Solutions

Transcription

1 SSEA Summer 27 Math 5, Homework-2 Solutions Write the parametric equation of the plane that contains the following point and line: 3 2, t 3 t R 5 4 By substituting t = and t =, we get two points A and B on the line: A = 4 2, B = 7 2 AB = = If we call the given point as C, then: AC = = 4 4 Therefore, a parametric equation of the plane is: P = {OC + tab + sac s, t R} = s 3 + t 4 s, t R Consider the equation ax + by + cz = d, where a, b, c, d R are known constants (a) Write x in terms of y and z

2 SSEA Math Module Math 5 Homework-2 Solutions, Page 2 of 7 August 2, 27 Assuming a, x = b a y c a z + d a (b) Fill in the blank boxes in the following equation: x d/a -b/a -c/a y = + y + z z y, z R (c) Using the result from the previous part, explain the geometry described by the equation ax + by + cz = d The parametric equation above represents a plane in R 3 3 Recall that for nonzero vectors u, v R n : u v = u v cos(θ), where θ is the angle between the vectors Use this to find the angle between the diagonal of the unit square in: (a) R 2 and one of the axes: Let the diagonal point be P and consider the x axis Then, we have the following dot product: [ ] [ ] }{{} OP }{{} vector along x axis = cos(θ) cos(θ) = ( ) θ = cos 2 2 (b) R 3 and one of the axes: Let the diagonal point be P and consider the x axis Then, we have the following dot product: = cos(θ) }{{}}{{} OP vector along x axis cos(θ) = ( ) θ = cos 3 3

3 SSEA Math Module Math 5 Homework-2 Solutions, Page 3 of 7 August 2, 27 (c) R n and one of the axes: Let the diagonal point be P and consider the x axis Then, we have the following dot product: = } {{ + } cos(θ) n terms }{{} OP }{{} vector along x axis cos(θ) = ( ) θ = cos n n (d) What happens to this angle if n? lim θ = lim n n cos ( ) ( ) n = cos = cos () = π 2 4 Consider the matrix: (a) What is the size of A? A = , that is, 3 rows and 4 columns (b) What is element in the 2nd row, 3rd column of A? [A] 23 = 8 (c) The transpose of a matrix is another matrix obtained by reversing the rows and columns of the original matrix Write down the transpose of A which is represented as A T 6 3 A T =

4 SSEA Math Module Math 5 Homework-2 Solutions, Page 4 of 7 August 2, 27 5 The solution to the system of equations: 2x + 6x 2 + x 3 + 2x 4 = 5, 3x 2 + x 3 + 4x 4 =, 3x 2 + x 3 + 2x 4 = 5 is given as follows: x 9/2 /2 x 2 x 3 = 3 + x /3 3 x 4 2 x 3 R Describe the geometry of the solution set The solution represents a line in R 4 that passes through the point /2 and parallel to the vector /3 9/ Write down all possible 3 2 reduced row echelon form matrices,,, 7 Consider two planes in R 3 described parametrically as follows: 2 t + s 3 s, t R, + 3 k + m 4 k, m R (a) Which plane does not pass through the origin?

5 SSEA Math Module Math 5 Homework-2 Solutions, Page 5 of 7 August 2, 27 Choosing s = t = in the first parametric equation results in x = y = z = Therefore, the first plane contains the origin From the second parametric equation, we have: x = + 2m, y = + 3k, and z = 4m Choosing x = y = z = + 2m = + 3k =, 4m = From the last equation m =, and using this in the first equation results in =, which is not possible Therefore, no values for k and m will result in x = y = z = Therefore, the second plane does not contain the origin (b) Do these planes intersect at all? If yes, where do they intersect? If the planes intersect, then: 2 t = + 2m t 2m = 3 t + s = + 3k, s + t 3k =, 3 4 t + 3s = 4m 3s + t 4m = }{{} Augmented matrix 3 row row row 3 =row 3 3 row row =row row 2 row 3 =row 3 +2 row row 3 =(/9) row 3 2 8/9 /9 2/3 /3 row =row +3 row 3 2 8/9 /9 }{{} rref The rref above corresponds to the following equations with the variables k, s, t (pivot) and m (free): s 2 3 m = 3,

6 SSEA Math Module Math 5 Homework-2 Solutions, Page 6 of 7 August 2, 27 t 2m =, k 8 9 m = 9 We can assign an arbitrary value for m, and solve for the remaining variables in terms of m: s = 2m 3, t = 2m +, k = 8m, m = m 9 We can then solve for the intersection coordinates as x = + 2m, y = + 3k, and z = 4m: x + 2m + 2m y = + 3k = + 3 ( ) 2 8m = 2/3 + m 8/3, m R 9 z 4m 4m 4 which is a parametrization of a line in R 3 Therefore, the two planes intersect at a line 8 For each system of equations below: (a) Express the system as an augmented matrix Use Gaussian elimination to put the system in the reduced row echelon form Identify the pivot and free variables Determine whether the system has one of the following: No solutions One solution (find it) Infinitely many solutions (express the solution form) Describe the geometry of the solution: x + 2x 2 + 3x 3 =, 2x + x 2 2x 3 = [ ] [ ] 2 3 7/3 / /3 /3 }{{}}{{} Augmented matrix rref

7 SSEA Math Module Math 5 Homework-2 Solutions, Page 7 of 7 August 2, 27 The pivot variables are x, x 2 and the free variable is x 3 Since there is a free variable and there is no equation of the form = resulting from the rref, the system has infinitely many solutions The rref corresponds to the following equations: x 7 3 x 3 = 3, x x 3 = 3 We can solve for x and x 2 in terms of x 3 as follows: x = + 7x 3 3, x 2 = 8x 3, x 3 = x 3 3 We can write the solution as: x /3 7/3 x 2 = /3 + x 3 8/3, x 3 R, x 3 which is a line through the point with coordinates (/3, /3, ) and parallel to the vector with components (7/3, 8/3, ) (b) x + 2x 2 x 3 x 4 =, 2x + 4x 2 2x 3 + 3x 4 = 3, x + x 2 2x 3 + 4x 4 = 2 2 2/ / /5 }{{}}{{} Augmented matrix rref The pivot variables are x, x 2, x 4 and the free variable is x 3 Since there is a free variable and there is no equation of the form = resulting from the rref, the system has infinitely many solutions The rref corresponds to the following equations: x + x 3 = 2 5, x 2 x 3 = 4 5,

8 SSEA Math Module Math 5 Homework-2 Solutions, Page 8 of 7 August 2, 27 x 4 = 5 We can solve for x and x 2 in terms of x 3 as follows: x = x 3 (2/5), x 2 = x 3 + (4/5), x 3 = x 3, x 4 = /5 We can write the solution as: x 2/5 x 2 x 3 = 4/5 + x 3, x 3 R, x 4 /5 which is a line through the point with coordinates ( 2/5, 4/5,, /5) and parallel to the vector with components (,,, ) 9 Consider the equation: x + x 2 + x x n = b, where b R is a constant (a) Write the augmented matrix corresponding to the system [ ] b (b) Identify the pivot and free variables x is the pivot variable and the rest are free variables Exercise 6 Writing v as a linear combination of the vectors: = c 5 + c c + 2c 2 = 2, 5c 3c 2 = 2, 2c + c 2 =

9 SSEA Math Module Math 5 Homework-2 Solutions, Page 9 of 7 August 2, 27 The augmented matrix and rref for the system of equations in the unknowns c, c 2, c 3 are: , 2 } {{ } } {{ } Augmented matrix rref Since the third row of the augmented matrix corresponds to the equation =, there is no solution to the system and hence vector v cannot be written as a linear combination of the given vectors For the matrix A and vector x given below, compute the matrix-vector product Ax in two ways: Writing it as a linear combination of the columns of A, Writing it as dot product of rows of A (expressed as vectors) and x 5 5 [ ] A = , x = Using columns : Ax = = = 9, 5 5 [ ] [ ] [ ] [ ] using rows : Ax = [ ] [ 5 ] 2 2 = = 9 [ 3 ] [ 5 ] Graph the result of multiplying the vector x = [ cos(θ) sin(θ) A = sin(θ) cos(θ) [ ] by the following matrices: ], θ = π/4, and B = [ ] 4 4

10 SSEA Math Module Math 5 Homework-2 Solutions, Page of 7 August 2, 27 Explain in words the effect of these matrix multiplications with the vector x [ ] [ ] cos(θ) sin(θ) Ax = = sin(θ) cos(θ) [ ] cos(θ) sin(θ) = cos(θ) + sin(θ) [ ] [ ] cos(π/4) sin(π/4) = 2 cos(π/4) + sin(π/4) The effect of multiplying the vector is to rotate it counterclockwise by the angle π/4 = 45 without changing its magnitude as shown in the figure below: [ v = ] 45 [ ] [ ] 4 4 For the matrix A =, Ax = The effect is stretch the vector to 4 times 4 4 its original length without changing its orientation 3 Describe in words the effect of multiplying a vector x R 5 by the matrix A: A = What happens when you multiply x by the transpose of A?

11 SSEA Math Module Math 5 Homework-2 Solutions, Page of 7 August 2, 27 x 5 x 2 x Ax = x 2 x 3, x 3 AT x = x 4 x 5 x 4 x The effect of multiplying by A is to shift the components of vector x downward and the effect of multiplying by A T is to shift the components of vector x upward 4 Exercise 84 Solving Ax = to find out the nullspace of A [ ] [ ] [ ] 2 4 x = 2 x 2 2x + 4x 2 =, x 2x 2 =, ([ ]) 2 which results in x = 2, x 2 = Therefore, the nullspace of A is the span [ ] 6 For the right hand side vector b = : 3 [ ] [ ] }{{}}{{} Augmented matrix rref x is the pivot variable and x 2 is the free variable that takes any arbitrary value The rref results in the equation x + 2x 2 = 3 x = 3 2x 2 The solution can therefore be represented as: [ ] [ ] [ ] x 3 2 = + x 2, x 2 R, x 2 which is a straight linethe nullspace of A and the solutions to the system Ax = b are shown in the following figure:

12 SSEA Math Module Math 5 Homework-2 Solutions, Page 2 of 7 August 2, 27 [ [ ] x ] 2, x 2 R ([ 2 N(A) = span ]) 5 Exercise 825 Let us consider the equation: 4 c + c 2 + c 3 = 3 4c + c 2 + c 3 =, c + c 2 c 3 =, 3c + c 2 + c 3 = 4 3 }{{}}{{} Augmented matrix rref Therefore, the vectors are linearly independent c =, c 2 =, c 3 = 6 Exercise 93 To be updated

13 SSEA Math Module Math 5 Homework-2 Solutions, Page 3 of 7 August 2, 27 Optional Challenge Problems 7 Am Math Mon, Feb 933: A ship is sailing with velocity vector v ; the wind blows apparently (judging by the vane on the mast) in the direction of a vector a; on changing the direction and speed of the ship from v to v 2, the apparent wind is in the direction of a vector b Find the velocity vector of the wind Referring to the figure below, we have the following vectors and scalars: v : initial ship velocity vector v 2 : new ship velocity vector w: absolute wind velocity vector a, b: apparent wind directions and these are unit vectors That is, a =, b = s, t: scalar constants From the geometry, w = v + sa = v 2 + tb Taking dot products with a and b v + sa = v 2 + tb a (v + sa) = a (v 2 + tb) a v + s (a a) = a v }{{} 2 + t (a b) = s (a b) t = a (v 2 v ) v + sa = v 2 + tb b (v + sa) = b (v 2 + tb) b v + s (b a) = b v }{{} 2 + t (b b) }{{} a b = (a b) s t = b (v 2 v ) We can solve for s and t from the two equations above to get: s = [a (a b) b] (v 2 v ) (a b) 2

14 SSEA Math Module Math 5 Homework-2 Solutions, Page 4 of 7 August 2, 27 We can use this in the original vector equation for the absolute wind velocity vector w to get: ( ) [a (a b) b] (v2 v ) w = v + sa = v + (a b) 2 a 8 Consider three vectors x, x 2, and x 3 Suppose x perpendicular to x 2, and x 2 is not parallel to x 3, then can the vectors x, x 2, and x 3 form a linearly independent set? Give an example or a counterexample To be updated 9 Consider vector x R n whose components x k represent the value of a signal at time stamp k =, 2,, n Assuming n is even, we want to construct a 2 downsampled signal y of size n/2 by multiplying x by an appropriate matrix A such that: y = Ax, where y k = x 2k, k =, 2,, n/2 Using n = 6, write the elements of matrix A y x 2 y 2 = x 3 y 3 x 4 }{{} x 5 A x 6 x 2 Consider an m n matrix A and an n p matrix B The product of the two matrices is an m p matrix C that is defined as follows: C = AB = [ Ab, Ab 2,, Ab p ] (a) Write the elements of the matrix C in terms of dot products rows of matrix A (written in a vertical fashion) and columns of matrix B

15 SSEA Math Module Math 5 Homework-2 Solutions, Page 5 of 7 August 2, 27 Using the notation a b = a T b for the dot product of vectors a and b, we have: AB = A [ b, b 2,, b p ] = [ Ab, Ab 2,, Ab p ], ã T ã T ã T B = 2 B = ã T 2 B, ã T mb = ã T m ã T ã T 2 ã T m [ ] [ b, b 2,, b p ] b, b 2,, b p ã T b ã T b p [ = ] ã b, b 2,, b T mb ã T mb p p b T b T 2 b T n, = [ ] a a 2 a n = a b T + a 2 bt a n bt n (b) Two vectors x and y are said to be orthogonal if x y = ; that is the angle between them is 9 Suppose both matrices A and B are of size n n Comment on the nature of orthogonality between the rows and columns of the matrices A and B under each one of the following conditions: AB =, A T B =, AB T =, A T B T = We are looking for a statement like: the columns of A are orthogonal to the columns of B etc AB = rows of A are orthogonal to the columns of B A T B = columns of A are orthogonal to the columns of B AB T = rows of A are orthogonal to the rows of B A T B T = columns of A are orthogonal to the rows of B 2 The system shown in the figure below consists of n linear springs that support n masses:

16 SSEA Math Module Math 5 Homework-2 Solutions, Page 6 of 7 August 2, 27 k W x k 2 W 2 x 2 k 3 k n W n x n For i =, 2,, n, the spring stiffnesses are denoted by k i, the weights of the masses are W i, and x i are the unknown displacements of the masses (measured from the positions where the springs are undeformed) The so-called displacement formulation is obtained by writing the equilibrium equation of each mass and substituting F i = k i (x i+ x i ) for the spring forces The result is the following system of equations: (k + k 2 )x k 2 x 2 = W k i x i + (k i + k i+ )x i k i+ x i+ = W i, i = 2, 3,, n k n x n + k n x n = W n (a) Write the above set of equations in the matrix vector form Ax = b k + k 2 k 2 k 2 k 2 + k 3 k 3 k i k i + k i+ k i+ k n k n + k n k n } {{ k n k n } A x x 2 x i x n x n }{{} x = W W 2 W i W n W n } {{ } b

17 SSEA Math Module Math 5 Homework-2 Solutions, Page 7 of 7 August 2, 27 (b) Is there something unique about the symmetric structure of the matrix A if the spring constants are all equal If the spring constants are all equal to k,we have the following structure for the matrix A: 2k k k 2k k A = k 2k k k 2k k k k