Practice Problems for the Final Exam Linear Algebra. Matrix multiplication: (a) Problem 3 in Midterm One. (b) Problem 2 in Quiz. 2. Solve the linear system: (a) Problem 4 in Midterm One. (b) Problem in Quiz. (c) Problem in Quiz 2. (d) Problem 2 in Quiz 3. (e) Section 3 in Lecture note 3. (f) Section in Lecture note 0. (g) Problem 5.3A and 5.3B on pages 278-279 in the textbook. 3. Find the inverse of a matrix: (a) Problem 5 (a) in Midterm One. (b) Problem 2 in Quiz 2. (c) Problem 3 in Quiz 3. (d) Section 2 in Lecture note 5. 4. Find the determinant of a matrix: (a) Problem 5 (b) (c) in Midterm One. (b) Problem in Quiz 3. (c) Section.4 in Lecture note 6. 5. Find the subspace and bases: (a) Problem 2(a) in Midterm One. Copy right reserved by Yingwei Wang
(b) Problem 6 in Midterm One. (c) Quiz 5. (d) Section 3 in Lecture note 9. (e) Section 2 in Lecture note. (f) Section 4 in Lecture note 2. 6. Find the eigenvalue, eigenvectors and also the powers of a matrix: (a) Problem 2 in Quiz. (b) Problem in Quiz 2. (c) Examples -6 in Lecture note 30. (d) Examples -5 in Section 2 and examples -2 in Section 3 of Lecture note 3. (e) Examples -2 in Chapter 6.2 and examples in Chapter 6.4 of the textbook. 2 Differential Equations. First order differential equations: (a) Separable equations: i. Problem in Quiz 7. ii. Problem 5 (b) in Midterm Two. (b) Standard form and integrating factors: i. Problem 2(a) in Midterm Two. ii. Problem 2 in Quiz 6. iii. Problem 2 in Quiz 7. iv. Examples 2., 2.2, 2.3 in Lecture note 7. (c) Exact equations: i. Problem 2(b) in Midterm Two. ii. Problem 3 in Quiz 7. iii. Examples on pages 3-4 in Lecture note 8. (d) Homogeneous equations: page in Lecture note 8. 2 Copy right reserved by Yingwei Wang
2. Second order differential equations: (a) Find the null solutions to the homogeneous equations with constant coefficients: i. Problem 3(a) in Midterm Two. ii. Examples -4 in Lecture 9; iii. Problem 2. in Quiz 8. iv. Problem. in Quiz 9. (b) Find the particular solution to non-homogeneous equations by undetermined coefficients: i. Problem 2.2 in Quiz 8. ii. Problem.2 in Quiz 9. iii. Examples -4 in Chapter 2.6 in the textbook. iv. Examples -2 in Lecture note 22. (c) Find the particular solution to non-homogeneous equations by variation of parameters: i. Problem 3(b) in Midterm Two. ii. Problem.3 in Quiz 9. iii. Examples 5-6 in Chapter 2.6 in the textbook. iv. Eamples 2-3 in Lecture note 23. (d) Solve the problems with constant coefficients by Laplace transforms: i. Problem 3(c) in Midterm Two. ii. Problem 2 in Quiz 9. iii. Example in Lecture note 24. iv. Examples -3 in Chapter 2.7 in the textbook. (e) From solutions back to differential equations: i. Problem 3(d) in Midterm Two. ii. Example 3 in Lecture note 22. 3. Higher order differential equations: (a) Find the null solutions to the homogeneous equations with constant coefficients: 3 Copy right reserved by Yingwei Wang
i. Problem 4(a) in Midterm Two. ii. Problem 3. in Quiz 8. iii. Examples in Section of Lecture note 2. (b) Find the particular solution to non-homogeneous equations by undetermined coefficients: i. Problem 4(b) in Midterm Two. ii. Problem 3.2 in Quiz 8. iii. Examples in Section 2 of Lecture note 2. 4. Use graphical methods to sketch the solutions: (a) Slope field for first order: Examples -4 in Chapter 3. in the textbook; Problem 5 in Midterm Two. (b) Phase plane for second order: Examples on page 62-63 in the textbook. 5. Numerical methods for first order differential equations: (a) Forward Euler method: i. Examples -2 in Lecture note 28; ii. Problem. in Quiz. (b) Improved Euler (Simplified RK) method: i. Examples -2 in Lecture note 29; ii. Problem.2 in Quiz. 6. First order system: (a) How to find the null solution: i. Examples -4 in Section of Lecture note 32; ii. Problem 2 in Quiz 2. (b) How to find the particular solution: i. Examples -3 in Section 2 of Lecture note 32; ii. Examples -3 in Lecture note 33; iii. Problem 2 in Quiz 2. 4 Copy right reserved by Yingwei Wang
3 Additional problems Consider the following 3-by-3 matrix t 2 A = 0, 0 t where t is a real parameter. Questions:. Find A 2. 2. Find A T and verify that both A T +A and A T A are symmetric matrices. 3. Find det(a) and det(a ). 4. For what values of t, so that A is nonsingular; and find A in terms of t by Gaussian-Jordan elimination. 5. For what values of t, so that the columns of A form a basis set for R 3. 6. For t = 2, find A using adjugate method. 7. For t = 0, solve the linear system Ax = b by any legal way, where b = (3,2,) T. 8. For what values of t, so that Ax = 0 has infinitely many solutions; and find the complete solution to Ax = b, where b = (t+3,2,t+) T, for these values of t. 9. Let t = and find rank(a). 0. Let t = and find dimensions and bases for four fundamental subspaces of A: (a) Col(A); (b) Col(A T ) or Row(A); (c) Null(A); (d) Null(A T ). 5 Copy right reserved by Yingwei Wang
. For what values of k such that the vectors t 0,, 0 2 t are linearly dependent? 2. For what values of t such that the vector is in the space 0 span t 2 0,. t 3. Suppose after several steps of Gauss-Jordan elimination for the matrix A, we arrive at 0 0 Gauss-Jordan elimination [A I] 0 0 0. 0 0 0 Can you determine the value of t? 4. For t = 3.5, find the eigenvalues and eigenvectors. Is A diagonalizable in this case? Why? 5. For t = 3.5, find the general solution to the problem y = Ay. 6 Copy right reserved by Yingwei Wang
Selected answers:. Omitted. 2. Omitted. 3. det(a) = t 2,det(A ) = t 2. 4. If t ±, then A is nonsingular and A = t t t 2 2 t. t 2 t 5. If t ±, then columns of A form a basis set for R 3. 6. Suppose t = 2, then 2/3 2/3 /3 A = /3 2/3 2/3. /3 /3 2/3 3 7. Suppose t = 0, then the solution to Ax = 2 is x =. 8. If t = or t =, then Ax = 0 has infinitely many solutions; Furthermore, we have (a) If t =, then the complete solution to Ax = b, where b = (t+3,2,t+) T, 2 is x = c + 2 ; 0 (b) If t =, then the complete solution to Ax = b, where b = (t+3,2,t+) T, 0 is x = c + 2. 0 9. For t =, the rank of A is 2. 7 Copy right reserved by Yingwei Wang
2 0. For t =, the matrix A = 0. 0 (a) Col(A). Let us work on the A T. 0 A T = 0 2 0 0 0 0 The basis set for Col(A) is 0,. And the dimension is dim(col(a)) = 2. (b) Col(A T ). Let us work on A. 2 A = 0 0 0 0 0 0 The basis set for Col(A T ) is 0,. And the dimension is dim(col(a T )) = dim(row(a)) = 2. 0 0. (3.) 0 0 0 0 0. (3.2) 0 0 0 (c) Null(A). By (3.2), it is easy to know that the basis set for Null(A) is. And the dimension is dim(null(a)) =. (d) Null(A T ). By (3.), it is easy to know that the basis set for Null(A) is. And the dimension is dim(null(a T )) =. 8 Copy right reserved by Yingwei Wang
. Inorderto have linearlydependent columns ofthematrixa, we needdet(a) = 0. It implies that t = ±. 2. Bythedefinitionofspan, weknow that[,,0] T isalinear combinationof[t,0,] T and [2,,t] T. In other word, there exist x,y such that t 2 = x 0 +y, 0 t = xt+2y, = 0x+y, 0 = x+ty t 2 =, t = ±. 3. t =. Note that for each step of Gauss-Jordan elimination, [A I] [B C] [I A], we always have the following relation for the intermediate step: C B = A. 4. The characteristic polynomial of A in this case is p(λ) = det(a λi) = (λ 5)(λ.5) 2. (a) For λ = 5, the eigenvector x satisfies: (A 5I)x = 0,.5 2 x 0 0 4 x 2 = 0. 0.5 x 3 0 Performing the Gaussian elimination for the coeffcient matrix yields.5 2 0.5 0 4 0 0.25. 0.5 0 0 0 9 Copy right reserved by Yingwei Wang
It follows that the eigenvector is 6 x =. 4 (b) For λ 2 =.5, the eigenvector x satisfies: (A.5I)x = 0, 2 2 x 0 0 0.5 x 2 = 0. 0 2 x 3 0 Performing the Gaussian elimination for the coeffcient matrix yields 2 2 0 2 0 0.5 0 2. 0 2 0 0 0 It follows that there is only one linearly independent eigenvector 2 x 2 = 2. In other word, for λ 2 =.5, its algebraic multiplicity is 2 while its geometric multiplicity is. So the matrix A is not diagonalizable in this case. 5. The null solution corresponding to λ,x is 6 y (t) = e λt x = e 5t. 4 The null solutions corresponding to λ 2,x 2 are 2 y 2 (t) = e λ2t x 2 = e.5t 2, 0 Copy right reserved by Yingwei Wang
y 3 (t) = te λ 2t x 2 +e λ 2t w, where w satisfies (A λ 2 I)w = x 2, 2 2 w 2 0 0.5 w 2 = 2. 0 2 w 3 Performing Gaussian elimination to the augment matrix leads to 2 2 2 0 0 0.5 2 0 2 4 0 2 0 0 0 0 One of the solutions is So the y 3 (t) is w = 4. 0 Finally, the null solution is y 3 (t) = te λ2t x 2 +e λ2t w, 2 = te.5t 2 +e.5t 4. 0 y(t) = c y (t)+c 2 y 2 (t)+c 3 y 3 (t), 6 2 2 = c e 5t +c 2 e.5t 2 +c 3 te.5t 2 +e.5t 4. 4 0 Copy right reserved by Yingwei Wang