Math 2 Homework #7 March 4, 2 7.3.3. Find the solution set of 2x 3y = 5. Answer: We solve for x = (5 + 3y/2. Hence the solution space consists of all vectors of the form ( ( ( ( x (5 + 3y/2 5/2 3/2 x = = = + y, y y where y is an arbitrary real number. 7.3.5. Find the solution set of 2x 3y = x 4y = 2 Answer: In matrix form the system is ( ( ( 2 3 x =. 4 y 2 The augmented matrix is ( 2 3 4 We reduce to row echelon form by first interchanging the rows and then subtracting 2 times the first row from the second: ( ( 4 2 4 2 M 2 3 5 4 The simplified system of equations is x 4y = 2 5y = 4. Backsolving we get y = 4/5 and then x = 2 + 4y = 2 6/5 = 6/5. Hence the solution is ( 6/5, 4/5 T. 7.3.7. Find the solution set of 2x 3y = 5 x + 4y = 7 Answer: In matrix form the system is ( ( ( 2 3 x 5 =. 4 y 7 The augmented matrix is ( 2 3 5 4 7
We reduce to row echelon form by first interchanging the rows and then subtracting 2 times the first row from the second: ( ( 4 7 4 7 M 2 3 5 9 The simplified system of equations is x + 4y = 7 y = 9. Backsolving we get y = 9/ and then x = 7 4y = 7 36/ = 4/. Hence the solution is (4/, 9/ T. 7.3.3. Find the solution set of Answer: In matrix form the system is The augmented matrix is 4x + 7y + 5z = 8 2x + y z = ( 4 7 5 2 ( x y = z ( 8. ( 4 7 5 8 2 We reduce to row echelon form by first interchanging the rows and then subtracting 2 times the first row from the second: ( ( 2 2 M 4 7 5 8 9 3 8 The simplified system is 2x + y z = 9y + 3z = 8 To backsolve we first set the free variable z = t. Next we solve for y = 2 t/3. Finally we solve for x = (y z/2 = 2t/3. Hence the solutions are the vectors of the form x = ( x y z = ( ( 2t/3 2 t/3 = 2 + t t ( 2/3 /3. 7.3.9. Find the solution set of the system Ay = b, where ( 4 2 5 A = 4 8 8 and b = 3 2 4 ( 5 6 3
Answer: In MATLAB we do the following A = [4 2-5; -4-8 8; -3-2 4]; b = [-5 6 3] ; M=[A b] 4 2-5 -5-4 -8 8 6-3 -2 4 3 M(2,: = M(2,: - M(2,*M(,:/M(,; M(3,: = M(3,: - M(3,*M(,:/M(,; M(3,: = M(3,: - M(3,2*M(2,:/M(2,2 4 2-5 -5 - /2-3/2 The simplified system is 4y + 2y 2 5y 3 = 5 y 2 + y 3 /2 = 3/2 To backsolve, we first set the free variable y 3 = t. Next we solve for y 2 = 3/2 + y 3 /2 = (3 + t/2. Finally, we solve for y = ( 5 2y 2 + 5y 3 /4 = [ 5 (3 + t+ 5t]/4 = 2 + t. Hence our solutions are all vectors of the form y = ( y y 2 y 3 = ( 2 + t (3 + t/2 t = ( 2 3/2 + t ( /2. 7.3.2. Find the solution set of the system Ay = b, where ( ( 3 3 4 A = 8 7 2 and b = 8 8 6 5 Answer: In MATLAB we do the following A=[-3-3 ;8 7-2;8 6 -]; b=[4-8 -5] ;
[A b] -3-3 4 8 7-2 -8 8 6 - -5 M(2,: = M(2,: - M(2,*M(,:/M(,; M(3,: = M(3,: - M(3,*M(,:/M(,; M(3,: = M(3,: - M(3,2*M(2,:/M(2,2-3 -3 4-2/3 8/3 /3 /3 The simplified system is 3y 3y 2 + y 3 = 4 y 2 + 2y 3 /3 = 8/3 y 3 /3 = /3 Backsolving we get y 3 =. Next y 2 = 8/3 + 2/3 = 2. Finally, y = ( 4 3y 2 + y 3 /3 =. Hence the only solution is y = (, 2, T 7.3.27. Find the solution set of the system Ay = b, where ( 7 7 8 3 A = 9 5 8 2 and b = 5 2 8 Answer: Using MATLAB we do the following ( 37 35 9 A=[-7 7-8 -3; 9-5 8-2 ; 5 2 8]; b=[37-35 -9] ; [A b] -7 7-8 -3 37 9-5 8-2 -35 5 2 8-9 M(2,: = M(2,: - M(2,*M(,:/M(,; M(3,: = M(3,: - M(3,*M(,:/M(,;
M(3,: = M(3,: - M(3,2*M(2,:/M(2,2-7 7-8 -3 37 4-6/7-4/7 88/7-6/7 369/28 2/7 The simplified system of equations is 7y + 7y 2 8y 3 3y 4 = 37 4y 2 6y 3 /7 4y 4 /7 = 88/7 6y 3 /7 + 369y 4 /28 = 2/7 To backsolve, we first set the free variable y 4 = t. Then we solve for y 3 = ( 2 + 369t/4/6 = 2 + 23t/8. Next, Finally y 2 = (88/7 + 6y 3 /7 + 4y 4 /7/4 = [88 + 6( 2 + 23t/8 + 4t]/28 = [56 + 287t]/28 = 2 + 4t/4. y = [ 37 + 7y 2 8y 3 3y 4 ]/7 = [ 37 + 7(2 + 4t/4 8( 2 + 23t/8 3t]/7 = 3t/4. Hence the solutions are the vectors y = 3t/4 2 + 4t/4 2 + 23t/8 t = 2 2 + t 3/4 4/4. 23/8 7.3.29. Find the solution set of the system Ay = b, where A = ( 8 6 9 8 9 5 7 9 4 3 7 and b = ( 5 3 9
Answer: Using MATLAB we do the following A=[8-6 9 8 - ;-9 5-7 9 ; -4-3 -7]; b=[5-3 9] ; M=[A b] 8-6 9 8-5 -9 5-7 9-3 -4-3 -7 9 M(2,: = M(2,: - M(2,*M(,:/M(,; M(3,: = M(3,: - M(3,*M(,:/M(,; M(3,: = M(3,: - M(3,2*M(2,:/M(2,2 8-6 9 8-5 -7/4 25/8 8-9/8-5/8-83/4-262/7-67/4 63/2 M(2,: = 8*M(2,:; M(3,: = 4*M(3,: 8-6 9 8-5 -4 25 44-9 -5-83 -524-67 44 The simplified system of equation is 8y 6y 2 + 9y 3 + 8y 4 y 5 = 5 4y 2 + 25y 3 + 44y 4 9y 5 = 5 83y 3 524y 4 67y 5 = 44 To backsolve, we first set the free variables y 4 = s and y 5 = t. The we solve for y 3 = [ 44 524s 67t]/83. Next, y 2 = [5 + 25y 3 + 44y 4 9y 5 ]/4 = [5 + 25( 44 524s 67t/83 = 44s 9t]/4 = [ 65 82s 73t]/83.
and finally, y = [5 + 6y 2 9y 3 8y 4 + y 5 ]/8 = [5 + 6( 65 + 82s + 73t/83 9( 44 524s 67t/83 8s + t]/8 = [528 + 445s 44t]/83. Hence the solutions are the vectors [528 + 445s 44t]/83 [ 65 82s 73t]/83 y = [ 44 524s 67t]/83 s t 528/83 445/83 44/83 65/83 82/83 73/83 = 44/83 + s 524/83 + t 67/83. 7.3.3. Find the solution set of the system Ay = b, where A = ( 2 3 6 7 6 6 8 9 5 9 ( 3 and b = 6 Answer: Using MATLAB we do the following: A=[-2 3 6-7 -;- -6-6 -8; -9-5 - 9]; b = [3 6] ; [A b] -2 3 6-7 - 3 - -6-6 -8-9 -5-9 6 M(2,: = M(2,: - M(2,*M(,:/M(,; M(3,: = M(3,: - M(3,2*M(2,:/M(2,2-2 3 6-7 - 3-5/2-2 -5/2-5/2 -/2-3/5 2 8 83/5
M(2,: = -2*M(2,:; M(3,: = -5*M(3,: -2 3 6-7 - 3 5 4 5 5 3 - -9-83 The simplified system of equations is 2y + 3y 2 + 6y 3 7y 4 y 5 = 3 5y 2 + 4y 3 + 5y 4 + 5y 5 = 3y 3 y 4 9y 5 = 83 To backsolve, we first set the free variables y 4 = s and y 5 = t. Then we solve for y 3 = [ 83 + s + 9t]/3. Next, y 2 = [ 4y 3 5y 4 5y 5 ]/5 = [ 4( 83 + s + 9t/3 5s 5t]/5 = [23 7s 37t]/3. and finally, y = [3 3y 2 6y 3 + 7y 4 + y 5 ]/2 = [3 3(23 7s 37t/3 6( 83 + s + 9t/3 + 7s + t]/2 = 8 2s + 6t. Hence the solutions are the vectors 8 2s + 6t [23 7s 37t]/3 y = [ 83 + s + 9t]/3 s t 8 2 6 23/3 7/3 37/3 = 83/3 + s +/3 + t 9/3. 7.4.. The matrix ( 2 3 is the reduced row echelon form of the augmented matrix [A, b] representing the system Ax = b. Is the system inconsistent? Answer: Consistent.
7.4.3. The matrix ( 2 is the reduced row echelon form of the augmented matrix [A, b] representing the system Ax = b. Is the system inconsistent? Answer: Inconsistent. 7.4.7. Find all solutions of the homogeneous system Ax = where ( 2 A =. Are there solutions other than the zero vector? Is the coefficient matrix singular? Answer: Using row operations we reduce A to row echelon form. ( 2 A. We now see that the only solution of the homogeneous system is the zero vector (, T. Hence the matrix is nonsingular. 7.4.. Find all solutions of the homogeneous system Ax = where ( A =. Are there solutions other than the zero vector? Is the coefficient matrix singular? Answer: Using row operations we reduce A to row echelon form ( ( A. We now see that the only solution of the homogeneous system is the zero vector (, T. Hence the matrix is nonsingular. 7.4.8. Is the matrix A = ( 4 2 singular? If it is nonsingular find A. Answer: We augment A with the identity and perform row operations: ( ( ( 4 2 /2 2 /4 /4 Hence A is nonsingular and A = ( /2. /4
7.4.2. Is the matrix A = ( 2 2 singular? If it is nonsingular find A. Answer: We augment A with the identity and perform row operations: ( [A, I] /2 /2. Hence A is nonsingular and ( A = /2 /2. 7.4.23. Does the system a x + a 2 x 2 + a 3 x 3 = b a 2 x + a 22 x 2 + a 23 x 3 = b 2 have a unique solution? Answer: No. There are fewer equations than unknowns, so there are free variables. There is either no solution or there are many solutions. 7.4.3. For which values of a and b do the following equations have a solution? ( ( ( 2 4 u 3 5 v = a 2 4 8 w b Answer: The augmented matrix is ( 2 4 3 5 a. 2 4 8 b We use row operations to reduce this to row echelon form ( 2 4 b/2 2 (2a 3b/2. (3b 2a/2 This system will be consistent if and only if 3b 2a =. 7.4.35. List as many properties as you can of an invertible matrix. Answer: Suppose that A is invertible. Then: ( A is nonsingular. (2 The only solution of the homogeneous system Ay = is the zero vector. (3 The equation Ax = b has a unique solution for any right hand side b. (4 If A is put into row echelon form then the diagonal entries of he result are nonzero. (5 If A is put into reduced row echelon form then the result is the identity matrix.
7.4.36. List as many properties as you can of a nonsingular matrix. Answer: Suppose that A is nonsingular. Then: ( A is invertible. (2 The only solution of the homogeneous system Ay = is the zero vector. (3 The equation Ax = b has a unique solution for any right hand side b. (4 If A is put into row echelon form then the diagonal entries of he result are nonzero. (5 If A is put into reduced row echelon form then the result is the identity matrix. 7.5.3. Find the nullspace of the matrix ( 4 4. 2 2 Answer: The nullspace consists of solutions of ( ( ( 4 4 x =. 2 2 y Set up the augmented matrix and reduce to row echelon form. ( ( 4 4 2 2 Hence, y is free and x + y = x = y. Therefore, all solutions have the form ( ( ( x y x = = = y, y y where y is free. 7.5.5. Find the nullspace of the matrix ( 5 2 5. Answer: We use row operations to reduce A to row echelon form ( 3. Hence z is a free variable. We backsolve to find that y = and x = y z = z. Hence the nullspace consists of all vectors of the form ( ( ( x z v = y = = z. z z
7.5.7. Find the nullspace of the matrix 2 2 2 2 4 4. 2 4 3 Answer: We use row operations to reduce A to row echelon form 2 2 4 4 2 2. We set the free variables y 3 = s and y 4 = t. Then we backsolve to find that y 2 = y 4 = t, and then that y = [2y 2 + 4y 3 + 4y 4 ]/2 = 2s + t. Hence the nullspace consists of all vectors of the form y y 2 y = = y 3 y 4 2s + t t s t 2 = s + t 7.5.9. Either show that the following vectors are linearly independent or find a nontrivial linear combination that is equal to. ( ( v = and v 2 2 = 3 Answer: We must examine the nullspace of the matrix V = [v, v 2 ]. Using row operations we get ( ( V =. 2 3 5 Since the diagonal entries are nonzero, V is nonsingular, and its nullspace is trivial (i.e., it contains only the zero vector. Hence v and v 2 are linearly independent. 7.5.4. Either show that the following vectors are linearly independent or find a nontrivial linear combination that is equal to. ( 8 ( 2 ( 8 v = 9 6, v 2 = 7 and v 3 = 8 4 Answer: We must examine the nullspace of the matrix V = [v, v 2, v 3 ]. Using row operations we transform V to row echelon form ( 8 2 8 9 36. Thus we see that the third component is free or y 3 = t. We backsolve for y 2 = 4y 3 = 4t, and y = [8y 3 2y 2 ]/8 = 2t. Hence the nullspace of V consists of all vectors of the form t(2, 4, T. Taking t = we have the vector (2, 4, T. Hence we have 2v 4v 2 + v 3 =,.
so v, v 2, and v 3 are linearly dependent. 7.5.7. Find a basis for the span of the vectors v and v 2 in Exercise 7.5.9. What is the dimension of the span? Answer: Since the vectors v and v 2 are linearly independent, they form a basis for their span. Since there are two vectors in the basis, the dimension is 2. 7.5.22. Find a basis for the span of the vectors v, v 2, and v 3 in Exercise 7.5.4. What is the dimension of the span? Answer: Since v, v 2, and v 3 are linearly dependent, they do not form a basis. From Exercise 7.5.4 we know that 2v 4v 2 + v 3 =, or v 3 = 4v 2 2v. Hence any vector in the spaesn of v, v 2, and v 3 can be written as a v + a 2 v 2 + a 3 v 3 = a v + a 2 v 2 + a 3 (4v 2 2v = (a 2a 3 v + (a 2 + 4a 3 v 2. Thus every vector in the span of v, v 2, and v 3 is also in the span of v and v 2. Furthermore, since the vectors v and v 2 are not multiples of each other, they are linearly independent. Hence they form a basis of the span of v, v 2, and v 3. The dimension of the span is 2. 7.5.3. Find a basis for the nullspace of the matrix in Exercise 7.5.7. Answer: In Exercise 7.5.7 we found that the nullspace consists of all vectors of the form 2 y = s + t. Let v = (2,,, T and v 2 = (,,, T. We see that v and v 2 span the nullspace. In addition they are not multiples of each other, so they form a basis. 7.5.35. Describe the solution space of the system Ax = b, where A is the matrix in Exercise 7.5.3, and b = (4, 2 T. Answer: By observation we find that v p = (, T is a particular solution to the system. In Exercise 7.5.3 we found that the nullspace is generated by v h = (, T. Hence the solution space to the system Ax = b consists of all vectors of the form x = v p + tv h, where t R. 7.5.39. Describe the solution space of the system Ax = b, where A is the matrix in Exercise 7.5.7 and b = (, 6,, 6 T. Answer: By observation we find that v p = (3,,, T is a particular solution. In Exercise 7.5.3 we discovered that the nullspace of A has as a basis the vectors v = (2,,, T and v 2 = (,,, T. Hence every solution to the system Ax = b is of the form x = v p + sv + tv 2, where s, t R.
M9.2c. Answer: Using MATLAB we do the following: >> A=[-9, -28, 8, 38; 8, 6, -27, -6; 2, 4, 2, -4; -8, -6, 2, 6]; >> b = [3; 5; -2; -4]; >> [A, b] -9-28 8 38 3 8 6-27 -6 5 2 4 2-4 -2-8 -6 2 6-4 >> rref(m ans= -2 Since the right hand column contains a pivot, we know that the system is inconsistent. There are no solutions. M9.2f. Answer: Using MATLAB we do the following: >> A = [-23, 26, -42, -32, -9; -2,,, -3, -4; -7, 9, -28, -22, -63; -4, 4, -24, -6, -52; 8, -2, 32, 23, 69]; >> b = [-6; -2; -3; -2; 3];
>> [A, b] -23 26-42 -32-9 -6-2 -3-4 -2-7 9-28 -22-63 -3-4 4-24 -6-52 -2 8-2 32 23 69 3 >> rref(m ans= Now we backsolve and discover that the solution space is the single point x = (,,,, T. This problem can also be done with the single command >> A\b ans= -.. -... M9.3f. Answer: Using MATLAB we do the following: >> A = [6, -4, 2, -2, -8; 6, -4, 2, -2, -8; 29, -4,, -54, -36; 3, -6, 5, -24, -6; -, 4, -4, 8, 2]
A = 6-4 2-2 -8 6-4 2-2 -8 29-4 -54-36 3-6 5-24 -6-4 -4 8 2 >> null(a, r ans= -.5.5. -.25 -.75 -.5... M9.4g. Thus the three vectors /2 3/2 /4 3/4 /2 v =, v 2 = and v 3 = are a basis of the nullspace of A. The dimension is 3. Answer: >> v = [; ; ; ]; >> v2 = [; ; ; ]; >> v3 = [5; -6; ; -6]; >> V = [v, v2, v3] V = 5-6 -6 >> null(v, r ans= Empty matrix: 3-by- This means that the nullspace is trivial, so the vectors are linearly independent.