Math 601 Solutions to Homework 3

Transcription

1 Math 601 Solutions to Homework 3 1 Use Cramer s Rule to solve the following system of linear equations (Solve for x 1, x 2, and x 3 in terms of a, b, and c 2x 1 x 2 + 7x 3 = a 5x 1 2x 2 x 3 = b 3x 1 x 2 3x 3 = c Answer: Let A be the matrix corresponding to the left side of the system of linear equations, and let A i denote the matrix A with column a i replaced by the vector b Then, to use Cramer s Rule, we must c compute the following determinants: det(a = = 5 det(a 1 det(a 2 = det(a 3 = a 1 7 b 2 1 c a 7 5 b 1 3 c a 5 2 b 3 1 c = 5a 10b + 15c = a 27b + 37c = a b + c Now, we can use Cramer s rule to find the solution: x 1 = det(a 1 5a 10b + 15c = = a 2b + 3c det(a 5 x 2 = det(a 2 det(a = a 27b + 37c 5 = 5 a 27 5 b c x 3 = det(a 3 det(a = a b + c 5 = 1 5 a 1 5 b c 1

2 2 Determine whether the following sets are subspaces of R 2 2 Explain your answers (a S 1 = { A R 2 2 a a 22 = 0 } (b S 2 = { A R 2 2 a 11 + a 22 = 1 } (c S 3 = { A R 2 2 ( (d S 4 = { A R 2 2 det(a = 0 } Answer: ( 2 1 A = A 3 4 (a Yes, S 1 is a subspace of R 2 2 To show that S 1 is a subspace, we must show that it is closed under scalar multiplication and closed under addition (1 Closed under scalar multiplication: } Suppose that A is a matrix in S 1 Then, ( a a 22 = 0, so a11 a a = a 22 and we can write A as A = a 21 a ( αa11 αa Consider αa = The 1, 2 and 2, 2 entries αa 21 αa of αa are equal, so αa S 1, and S 1 is closed under scalar multiplication (2 Closed under addition: Suppose that A and B are both matrices in S 1 Then, ( a = a 22 a11 a and b = b 22, so we can write A and B as A = ( a 21 a b11 b and B = b 21 b ( a11 + b Consider A + B = 11 a + b The 1, 2 and 2, 2 a 21 + b 21 a + b entries of A + B are equal, so A + B S 1 2

3 We have shown that S 1 is closed under addition and scalar multiplication Thus, S 1 is a subspace of R 2 2 (b No, S 2 is not a subspace of R 2 2 To show that S 2 is not a subspace, we will show that it is not closed under scalar multiplication Consider the matrix A = ( , which is in S 2 If we multiply A by the scalar 2, we get 2A = not in S 2 Thus, S 2 is not a subspace of R 2 2 ( , which is (c Yes, S 3 is a subspace of R 2 2 To show that S 3 is a subspace, we must show that S 3 is closed under scalar multiplication and closed under addition (1 Closed under scalar multiplication: ( 2 1 Let M = 3 4 Suppose that A is a matrix in S 3 Then, MA = AM Consider αa We would like to show that M(αA = (αam, but this just follows from properties of matrix multiplication and the assumption that MA = AM: M(αA = α(ma = α(am = (αam Thus, αa S 3 Thus, S 3 is closed under scalar multiplication (2 Closed under addition: Suppose that A and B are matrices in S 3 Then, MA = AM and MB = BM Consider A + B We would like to show that M(A+B = (A+BM This follows from the properties of matrix multiplication and the assumption that MA = AM and MB = BM: M(A + B = MA + MB = AM + BM = (A + BM Thus, A + B S 3 Thus, S 3 is closed under addition 3

4 We have shown that S 3 is closed under scalar multiplication and closed under addition Thus, S 3 is a subspace of R 2 2 (d No, S 4 is not a subspace of R 2 2 To show that S 4 is not a subspace, we will show that S 4 is not closed under addition Consider the matrices A = ( and B = ( Since det(a = 0 and det(b ( = 0, both A and B are in the set S Now, consider A + B = Since, det(a + B = 1 0, 1 2 the matrix A + B is not in the set S 4 Thus, S 4 is not closed under addition, so it is not a subspace 3 Let P 4 denote the vector space of all polynomials with degree less than or equal to 3 Consider the subspace of P 4 consisting of polynomials p(x for which p( 1 = 0 Find a basis for this subspace } Answer: We are considering the subspace S = {p(x P 4 p( 1 = 0 A polynomial in P 4 is of the form p(x = a + bx + cx 2 + dx 3 If the polynomial is in the the subspace S, then p( 1 = 0, which means (plugging x = 1 into the polynomial p(x that a b + c d = 0 If we solve this equation for a, we get a = b c + d Thus, every polynomial in S can be written in the form p(x = (b c + d + bx + cx 2 + dx 3 We can rewrite this as p(x = b(1 + x + c( 1 + x 2 + d(1 + x 3 Thus, every polynomial in S can be written as a linear combination of the polynomials 1 + x, 1 + x 2, and 1 + x 3 This means that Span(1 + x, 1 + x 2, 1 + x 3 = S 4

5 All we need to do now is show that these three polynomials are linearly independent: Suppose there exists c 1, c 2, c 3 such that Rearranging, we get c 1 (1 + x + c 2 ( 1 + x 2 + c 3 (1 + x 3 = 0 (c 1 c 2 + c 3 + c 1 x + c 2 x 2 + c 3 x 3 = 0 Since this must be true for all values of x, the coefficients on the left side must all be 0 Thus, using the coefficients of the x, x 2, and x 3 terms, we get that c 1 = 0, c 2 = 0, and c 3 = 0 Thus, the polynomials 1+x, 1+x 2, and 1+x 3 are linearly independent We have shown that the polynomials 1 + x, 1 + x 2, and 1 + x 3 span S and are linearly independent Thus, they form a basis for S A basis for the subspace S is {1 + x, 1 + x 2, 1 + x 3 } 4 Let C[ π, π] denote the vector space of all real-valued functions that are defined and continuous on the closed interval [ π, π] (a Consider the subspace of C[ π, π] spanned by the vectors cos x, sin x, and sin(2x What is the dimension of this subspace? Explain your answer (b Consider the subspace of C[ π, π] spanned by the vectors cos x, cos(3x, and cos 3 x What is the dimension of this subspace? Explain your answer Answer: (a The main thing we need to do is determine whether the vectors are linearly independent Suppose there exists c 1, c 2, c 3 such that 5

6 c 1 cos x + c 2 sin x + c 3 sin(2x = 0 This must hold for all values of x, so we can plug in specific values of x and get linear equations involving c 1, c 2, and c 3 We will plug in x = 0, x = π 2, and x = π 4 : x = 0 c = 0 x = π 2 x = π c = 0 1 c c 2 + c 3 = Solving this system of linear equations, we get c 1 = 0, c 2 = 0, and c 3 = 0 Thus, the vectors cos x, sin x, and sin(2x are linearly independent Since these vectors also span the subspace, they form a basis for the subspace Thus, the dimension of the subspace is 3 (b Again, we need to determine if the vectors are linearly independent In this case, the vectors are linearly dependent due to the following trig identity: cos(3x = 3 cos x + 4 cos 3 x Here is one way to see that this trig identity holds: cos(3x = Re(e i3x = Re((cos x + i sin x 3 = Re(cos 3 x + 3i cos 2 x sin x 3 cos x sin 2 x i sin 3 x = cos 3 x 3 cos x sin 2 x = cos 3 x 3 cos x(1 cos 2 x = cos 3 x 3 cos x + 3 cos 3 x = 3 cos x + 4 cos 3 x 6

7 Next, we choose 2 of the vectors and see if they are linearly independent (it does not matter which 2 we choose We will choose cos x and cos 3x Suppose there exists c 1 and c 2 such that c 1 cos x + c 2 cos(3x = 0 Since this must hold for all values of x, we can plug in specific values of x and get linear equations involving c 1 and c 2 We will plug in x = 0 and x = π This gives us the equations 3 c 1 + c 2 = c 1 c 3 = 0 Solving, we get c 1 = 0 and c 2 = 0 Thus, the two vectors are linearly independent They also span the subspace, and so they form a basis for the subspace Thus, the dimension of the subspace is 2 7