# Practice Problems for the Final Exam

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1 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 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 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 Find the subspace and bases: (a) Problem 2(a) in Midterm One. Copy right reserved by Yingwei Wang

2 (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 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

3 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 Higher order differential equations: (a) Find the null solutions to the homogeneous equations with constant coefficients: 3 Copy right reserved by Yingwei Wang

4 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 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 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

5 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 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 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

6 . 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] 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

7 Selected answers:. Omitted. 2. Omitted. 3. det(a) = t 2,det(A ) = t 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 Suppose t = 2, then 2/3 2/3 /3 A = /3 2/3 2/3. /3 /3 2/ 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 For t =, the rank of A is 2. 7 Copy right reserved by Yingwei Wang

8 2 0. For t =, the matrix A = 0. 0 (a) Col(A). Let us work on the A T. 0 A T = 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 = The basis set for Col(A T ) is 0,. And the dimension is dim(col(a T )) = dim(row(a)) = (3.) (3.2) (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

9 . 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 x 2 = x 3 0 Performing the Gaussian elimination for the coeffcient matrix yields Copy right reserved by Yingwei Wang

10 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 x 2 = x 3 0 Performing the Gaussian elimination for the coeffcient matrix yields 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

11 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 w 2 = w 3 Performing Gaussian elimination to the augment matrix leads to 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), = c e 5t +c 2 e.5t 2 +c 3 te.5t 2 +e.5t Copy right reserved by Yingwei Wang

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