Math 310 Final Exam Solutions

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1 Math 3 Final Exam Solutions. ( pts) Consider the system of equations Ax = b where: A, b (a) Compute deta. Is A singular or nonsingular? (b) Compute A, if possible. (c) Write the row reduced echelon form of A. (d) Find all solutions to the system Ax = b. (a) deta = A is nonsingular (b) A (c) rref(a) (d) x =, x =, x 3 =. omitted 3. ( pts) Consider the following matrix A: (a) Find the nullspace of A. A = 3 (b) Do the columns of A form a spanning set for R? Clearly explain why or why not. (a) The row reduced echelon form of A is: rref(a) = There is no pivot in the third column. Therefore, x 3 is a free variable. Let x 3 = α. Then we have x + x 3 = and x x 3 = which give us x = α and x = α. The nullspace of A is: N(A) α α α α R = Span (b) The columns of A form a spanning set for R because there is a solution to Ax = b for every b R (there are no zero rows in the row reduced echelon form of A).

2 4. ( pts) Do the vectors below form a basis for R 3? If so, explain. If not, remove as many vectors as you need to form a basis and show that the resulting set of vectors form a basis for R 3. x, x, x 3, x 4, The vectors do not form a basis for R 3 because dim R 3 = 3 so there can only be 3 vectors in any basis for R 3. If we remove x, then consider the matrix X whose columns are x, x 3, and x 4 : X Since detx =, X is invertible and there is only one solution to Ax = b for every b R 3. Therefore, the columns are LI and form a spanning set for R 3. Thus, they form a basis for R (3 pts) Consider the following mapping L : R R 3 : L(x) x x x + x (a) Show that L is a linear transformation. (b) Find a matrix representation for L using the standard basis for R 3 and the following basis vectors for R : u =, u = (a) Let x = (b) x x and y = ( x + y L(x+y) = L x + y y y ) ( αx L(αx) = L αx. Then, (x + y ) (x + y ) (x + y ) + (x + y ) ) αx αx αx + αx ( L(u ) = L ( L(u ) = L ) = α ) x x x + x x x x + x + y y y + y = αl(x) = L(x)+L(y) The matrix representation of L is then: A

3 6. (5 pts) Let Y = Span {x,x } where: x Find Y, the orthogonal complement of Y. 3, x We use the fact that Y = R(A T ) = N(A) where A T = x x. The matrix A is then the same as in Problem 3. Since we already found the nullspace of A in Problem 3, the answer is: Y = Span 7. (5 pts) Use the Gram-Schmidt method to find an orthonormal basis for R 3 from the basis: x, x, x 3 u = x x = p = x,u u u = x p x p = p = x 3,u u + x 3,u u u 3 = x 3 p x 3 p = 8. ( pts) Find a matrix X and a diagonal matrix D such that A = XDX where A 3 3

4 The eigenvalues of A are found as follows: Plugging λ = into (A λi)x = we get: det(a λi) = λ 3 λ λ = (3 λ) λ λ = (3 λ)( λ) ( ) = (3 λ)( λ + λ 4) = (3 λ)(λ λ 3) = (3 λ)(λ 3)(λ + ) = 4 λ =, λ = 3 (repeated) (A + I)x = x x x 3 The first and third equations tell us x x 3 = and the second equation tells us x =. Since x 3 is a free variable, let x 3 = α. Then we have x = α. Setting α = we get the eigenvector: λ =, x Plugging λ = 3 into (A λi)x = we get: (A 3I)x = x x x 3 The first and third equations tell us x + x 3 = and the second equation tells us =. Since both x and x 3 are free variables, let x = α and x 3 = β. Then we have x = β. The set of solutions is then: x = α + β Letting α = β =, we get the eigenvectors: λ,3 = 3, x, x 3 The matrix X has the eigenvectors as its columns and the diagonal matrix D has the corresponding eigenvalues along the main diagonal: X, D 3 3 4

5 9. ( pts) Find the solution to the system of first order ODEs: dy dt = y 4y, y () = 3 dy dt = y, y () = Writing this system in matrix-vector form we have: y = Ay y 4 y = y y Since A is upper triangular, the eigenvalues are on the main diagonal: λ =,. Plugging λ = into (A λi)x = we get: 4 (A I)x = x = Both equations tell us that x =. However, x is free so we let x = α. Setting α = we get the eigenvector: λ =, x = Plugging λ = into (A λi)x = we get: 4 x (A + I)x = x = The first equation tells us that x x =. Since x is free we let x = α, which gives us x = α. Setting α = we get the eigenvector: λ =, x = The general solution to the system is: x y(t) = c e λt x + c e λt x y(t) = c e t + c e t Plugging in the initial conditions we get: 3 y() = = c + c The solution to this system of algebraic equations is c = and c =. Therefore, the solution is: y(t) = e t + e t. (a) ( pts) Let z = + i and w = i i. Compute z, z,w, and w,z. 5

6 (b) ( pts) Consider the following matrix: M i i Show that M is Hermitian and find a unitary matrix U that diagonalizes M. (a) z = z H z = i + i z,w = w H + i z = i + i = ( i)( + i) + ()() = i + = 3 = ( i)( + i) + ( + i)() = i i + + i = 3 w,z = z,w = 3 (b) To show that M is Hermitian, we must show that M = M H : M H = M T i i T i i = M To find the unitary matrix that diagonalizes M we must find the eigenvalues and eigenvectors for M. The eigenvalues are found as follows: det(a λi) = λ i i λ λ = ( λ) λ i i λ = ( λ)( λ) (i)( i) = ( λ)(4 4λ + λ ) = ( λ)(λ 4λ + 3) = ( λ)(λ )(λ 3) = The corresponding eigenvectors are: i λ =, x ; λ =, x λ =,,3 ; λ 3 = 3, x 3 Since M is Hermitian, the eigenvectors are orthogonal. Therefore, in order to construct the unitary matrix U we simply need to normalize the eigenvectors. The norms of both x and x 3 are. Therefore, the unitary matrix is: U i i i 6

7 Bonus: ( pts) Consider the set S which consists of all cubic polynomials p(t) = a + a t + a t + a 3 t 3 that satisfy the equation p () + 4p() =. That is, (a) Show that S is a subspace of P 4. (b) Find a basis for S. S = {p(t) p(t) P 4, p () + 4p() = }. (a) p(t) = is certainly in S because it satisfies the condition: (b) Let p(t), q(t) S. Therefore, we have: p () + 4p() = + 4() = p () + 4p() = q () + 4q() = Let r(t) = p(t) + q(t). Then we have r (t) = p (t) + q (t) and: Therefore, r(t) = p(t) + q(t) S. r + 4r() = p () + q () + 4(p() + q()) r + 4r() = = p () + 4p() + q () + 4q() = + (c) Let p(t) S and α R. Since p(t) S we have: p () + 4p() = Let r(t) = αp(t). Then we have r (t) = αp (t) and: Therefore, r(t) = αp(t) S. r + 4r() = αp () + 4αp() = α(p () + 4p()) = α() r + 4r() = Since the above three conditions are satisfied, S is a subspace of P 4.. Let p(t) = a + a t + a t + a 3 t 3. Then p (t) = a + 6a 3 t. The condition then tells us that: p () + 4p() = a + 6a 3 () + 4(a + a () + a () + a 3 () 3 ) = Therefore, the set S can be rewritten as follows: a + 4a = S = { a + a t a t + a 3 t 3 a,a,a 3 R } a = a S = { a ( t ) + a t + a 3 t 3 a,a,a 3 R } S = Span{ t,t,t 3 } The functions t,t,t 3 are LI and span S. Therefore, they form a basis for S. 7