# Purdue MA 26500 Fall 2022 Midterm II Solutions

Here comes the solution and analysis for Purdue MA 26500 Fall 2022 Midterm II. This second midterm covers topics in Chapter 4 (Vector Spaces) and Chapter 5 (Eigenvalues and Eigenvectors) of the textbook.

## Introduction

Purdue Department of Mathematics provides a linear algebra course MA 26500 every semester, which is mandatory for undergraduate students of almost all science and engineering majors.

### Textbook and Study Guide

:DisclosureThis blog site is reader-supported. When you buy through the affiliate links below, as an Amazon Associate, I earn a tiny commission from qualifying purchases. Thank you.

MA 26500 textbook is *Linear Algebra and its Applications* (6th Edition) by David C. Lay, Steven R. Lay, and Judi J. McDonald. The authors have also published a *student study guide* for it, which is available for purchase on Amazon as well.

### Exam Information

MA 26500 midterm II covers the topics of Sections 4.1 – 5.7 in the textbook. It is usually scheduled at the beginning of the thirteenth week. The exam format is a combination of multiple-choice questions and short-answer questions. Students are given one hour to finish answering the exam questions.

Based on the knowledge of linear equations and matrix algebra learned in the book chapters 1 and 2, Chapter 4 leads the student to a deep dive into the vector space framework. Chapter 5 introduces the important concepts of eigenvectors and eigenvalues. They are useful throughout pure and applied mathematics. Eigenvalues are also used to study differential equations and continuous dynamical systems, they provide critical information in engineering design,

### Reference Links

- Purdue Department of Mathematics Course Achive
- Purdue MA 26500 Spring 2024
- Purdue MA 26500 Exam Archive

## Fall 2022 Midterm II Solutions

### Problem 1 (10 points)

Let \[A=\begin{bmatrix}1 &0 &2 &0 &-1\\1 &2 &4 &-2 &-1\\2 &3 &7 &-3 &-2\end{bmatrix}\] Let \(a\) be the rank of \(A\) and \(b\) be the nullity of \(A\), find \(5b-3a\)

- A. 25
- B. 17
- C. 9
- D. 1
- E. 0

Problem 1 Solution

Do row reduction as follows:

- Add \(-1\) times row 1 to row 2
- Add \(-2\) times row 1 to row 2
- Scale row 2 by \(\frac{1}{2}\)
- Add \(-3\) times row 2 to row 3

\[ \begin{bmatrix}1 &0 &2 &0 &-1\\1 &2 &4 &-2 &-1\\2 &3 &7 &-3 &-2\end{bmatrix}\sim \begin{bmatrix}1 &0 &2 &0 &-1\\0 &2 &2 &-2 &0\\0 &3 &3 &-3 &0\end{bmatrix}\sim \begin{bmatrix}\color{fuchsia}1 &0 &2 &0 &-1\\0 &\color{fuchsia}1 &1 &-1 &0\\0 &0 &0 &0 &0\end{bmatrix} \]

So we have 2 pivots, the rank is 2 and the nullity is 3. This results in \(5b-3a=5\cdot 3-3\cdot 2=9\).

The answer is **C**.

### Problem 2 (10 points)

Let \(\pmb u=\begin{bmatrix}2\\0\\1\end{bmatrix}\), \(\pmb v=\begin{bmatrix}3\\1\\0\end{bmatrix}\), and \(\pmb w=\begin{bmatrix}1\\-1\\c\end{bmatrix}\) where \(c\) is a real number. The set \(\{\pmb u, \pmb v, \pmb w\}\) is a basis for \(\mathbb R^3\) provided that \(c\) is not equal

- A. \(-2\)
- B. \(2\)
- C. \(-3\)
- D. \(3\)
- E. \(-1\)

Problem 2 Solution

For set \(\{\pmb u, \pmb v, \pmb w\}\) to be a basis for \(\mathbb R^3\), the three vectors should be linearly independent. Let's create a matrix with these vectors as columns, then do row reduction like below \[ \begin{bmatrix}2 &3 &1\\0 &1 &-1\\1 &0 &c\end{bmatrix}\sim \begin{bmatrix}1 &0 &c\\0 &1 &-1\\2 &3 &1\end{bmatrix}\sim \begin{bmatrix}1 &0 &c\\0 &1 &-1\\0 &3 &1-2c\end{bmatrix}\sim \begin{bmatrix}1 &0 &c\\0 &1 &-1\\0 &0 &4-2c\end{bmatrix} \]

As can be seen, we need 3 pivots to make these column vectors linearly independent. If \(c\) is 2, the last row above has all-zero entries, there would be only 2 pivots. So C cannot be 2 for these three vectors to be linearly independent.

The answer is **B**.

### Problem 3 (10 points)

Which of the following statements is always TRUE?

- A. If \(A\pmb x=\lambda\pmb x\) for some vector \(\pmb x\), then \(\lambda\) is an eigenvalue of \(A\).
- B. If \(\pmb v\) is an eigenvector corresponding to eigenvalue 2, then \(-\pmb v\) is an eigenvector corresonding to eigenvalue \(-2\).
- C. If \(B\) is invertible, then matrix \(A\) and \(B^{-1}AB\) could have different sets of eigenvalues.
- D. If \(\lambda\) is an eigenvalue of matrix \(A\), then \(\lambda^2\) is an eigenvalue of matrix \(A^2\).
- E. If \(-5\) is an eigenvalue of matrix \(B\), then matrix \(B-5I\) is not invertible.

Problem 3 Solution

Per definitions in 5.1 "Eigenvectors and Eigenvalues":

An eigenvector of an \(n\times n\) matrix \(A\) is a

nonzero vector\(\pmb x\) such that \(A\pmb x=\lambda\pmb x\) for some scalar \(\lambda\). A scalar \(\lambda\) is called an eigenvalue of \(A\) if there is anontrivial solution\(\pmb x\) of \(A\pmb x=\lambda\pmb x\); such an \(\pmb x\) is called an eigenvector corresponding to \(\lambda\).

Statement A is missing the "nonzero" keyword, so it is NOT always TRUE.

For Statement B, given \(A\pmb v=2\pmb v\), we can obtain \(A(\pmb{-v})=2(\pmb{-v})\). The eigenvalue is still 2, not \(-2\). This statement is FALSE.

Statement C involves the definition of Similarity. Denote \(P=B^{-1}AB\), we have \[BPB^{-1}=BB^{-1}ABB^{-1}=A\] So \(A\) and \(P\) **are similar**. __Similar matrices have the same eigenvalues__ (Theorem 4 in Section 5.2 "The Characteristic Equation"). Statement C is FALSE

This can be proved easily, as seen below \[\begin{align} \det (A-\lambda I)&=\det (BPB^{-1}-\lambda I)=\det (BPB^{-1}-\lambda BB^{-1})\\ &=\det(B)\det(P-\lambda I)\det(B^{-1})\\ &=\det(B)\det(B^{-1})\det(P-\lambda I) \end{align}\] Since \(\det(B)\det(B^{-1})=\det(BB^{-1})=\det I=1\), we see that \(\det (A-\lambda I)=\det(P-\lambda I)\). ■

For Statement D, given \(A\pmb x=\lambda\pmb x\), we can do the following deduction \[A^2\pmb x=AA\pmb x=A\lambda\pmb x=\lambda A\pmb x=\lambda^2\pmb x\] So it is always TRUE that \(\lambda^2\) is an eigenvalue of matrix \(A^2\).

Statement E is FALSE. An eigenvalue \(-5\) means matrix \(B-(-5)I\) is not invertible since \(\det(B-(-5)I)=\det(B+5I)=0\). But the statement refers to a different matrix \(B-5I\).

The answer is **D**.

### Problem 4 (10 points)

Let \(\mathbb P_3\) be the vector space of all polynomials of degree at most 3. Which of the following subsets are subspaces of \(\mathbb P_3\)?

- A set of polynomials in \(\mathbb P_3\) satisfying \(p(0)=p(1)\).

- A set of polynomials in \(\mathbb P_3\) satisfying \(p(0)p(1)=0\).

- A set of polynomials in \(\mathbb P_3\) with integer coefficients.

- A. (i) only
- B. (i) and (ii) only
- C. (i) and (iii) only
- D. (ii) only
- E. (ii) and (iii) only

Problem 4 Solution

Per the definition of Subspace in Section 4.1 "Vector Spaces and Subspaces"

A

subspaceof a vector space \(V\) is a subset \(H\) of \(V\) that has three properties:

a. The zero vector of \(V\) is in \(H\).

b. \(H\) is closed under vector addition. That is, for each \(\pmb u\) and \(\pmb v\) in \(H\), the sum \(\pmb u + \pmb v\) is in \(H\).

c. \(H\) is closed under multiplication by scalars. That is, for each \(\pmb u\) in \(H\) and each scalar \(c\), the vector \(c\pmb u\) is in \(H\).

So to be qualified as the subspace, the subset should have all the above three properties. Denote the polynomials as \(p(x)=a_0+a_1x+a_2x^2+a_3x^3\).

**(i)**Since \(p(0)=p(1)\), we have \(a_0=a_0+a_1+a_2+a_3\), so \(a_1+a_2+a_3=0\).- Obviously, it satisfies the first property as if \(a_i=0\) for all \(i\), \(a_1+a_2+a_3=0\) is true as well.
- Now assume \(p_1(x)\) and \(p_2(x)\) are two polynomials in this set and \[ p_1(x)=a_0+a_1x+a_2x^2+a_3x^3\\ p_2(x)=b_0+b_1x+b_2x^2+b_3x^3 \] So we have \(a_1+a_2+a_3=0\) and \(b_1+b_2+b_3=0\). Then define a third polynomial \[\begin{align} p_3(x)&=p_1(x)+p_2(x)\\ &=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+(a_3+b_3)x^3\\ &=c_0+c_1x+c_2x^2+c_3x^3 \end{align}\] It is true that \(c_1+c_2+c_3=0\) as well. So the set has the second property.
- This set does have the third property since \(cp(x)\) has \(ca_1+ca_2+ca_3=0\) and it is also in the same set.

This proves that set (i) is a subspace of \(\mathbb P_3\).

**(ii)**From \(p(0)p(1)=0\), we can deduce that \(a_0(a_0+a_1+a_2+a_3)=0\). So any polynomial in this set should satisfy this condition.- Obviously, it satisfies the first property as if \(a_i=0\) for all \(i\), \(a_0(a_0+a_1+a_2+a_3)=0\) is true as well.
- With the same notation of \(p_1(x)\), \(p_2(x)\) and \(p_3(x)\). We have \[\begin{align} c_0(c_0+c_1+c_2+c_3)&=(a_0+b_0)(a_0+b_0+a_1+b_1+a_2+b_2+a_3+b_3)\\ &=(a_0+b_0)((a_0+a_1+a_2+a_3)+(b_0+b_1++b_2+b_3))\\ &=a_0(a_0+a_1+a_2+a_3)+a_0(b_0+b_1++b_2+b_3)+b_0(a_0+a_1+a_2+a_3)+b_0(b_0+b_1++b_2+b_3)\\ &=a_0(b_0+b_1+b_2+b_3)+b_0(a_0+a_1+a_2+a_3) \end{align}\] If \(a_0=0\) and \(b_0\ne 0\), the above ends up with \(b_0(a_1+a_2+a_3)\), which is not necessary equal 0. So this polynomial in this set does NOT have the second property.

This proves that set (ii) is NOT a subspace of \(\mathbb P_3\).

**(iii)**It is easy to tell that this set is NOT a subspace of \(\mathbb P_3\). If we do multiplication by floating-point scalars, the new polynomial does not necessarily have an integer coefficient for each term and might not be in the same set.

So the answer is **A**.

### Problem 5 (10 points)

Consider the differential equation \[ \begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}= \begin{bmatrix}1 &3\\-2 &2\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatrix} \].

Then the origin is

- A. an attractor
- B. a repeller
- C. a saddle point
- D. a spiral point
- E. none of the above

Problem 5 Solution

First, write the system as a matrix differential equation \(\pmb x'(t)=A\pmb x(t)\). We learn from Section 5.7 "Applications to Differential Equations" that each eigenvalue–eigenvector pair provides a solution.

Now let's find out the eigenvalues of \(A\). From \(\det (A-\lambda I)=0\), we have \[\begin{vmatrix}1-\lambda &3\\-2 &2-\lambda\end{vmatrix}=\lambda^2-3\lambda+8=0\] This only gives two complex numbers as eigenvalues \[\lambda=\frac{3\pm\sqrt{23}i}{2}\]

Referring to the Complex Eigenvalues discussion at the end of this section, "the origin is called a spiral point of the dynamical system. The rotation is caused by the sine and cosine functions that arise from a complex eigenvalue". Because the complex eigenvalues have a positive real part, the trajectories spiral outward.

So the answer is **D**.

Refer to the following table for the mapping from \(2\times 2\) matrix eigenvalues to trajectories:

Eigenvalues Trajectories \(\lambda_1>0, \lambda_2>0\) Repeller/Source \(\lambda_1<0, \lambda_2<0\) Attactor/Sink \(\lambda_1<0, \lambda_2>0\) Saddle Point \(\lambda = a\pm bi, a>0\) Spiral (outward) Point \(\lambda = a\pm bi, a<0\) Spiral (inward) Point \(\lambda = \pm bi\) Ellipses (circles if \(b=1\))

### Problem 6 (10 points)

Which of the following matrices are diagonalizable over the real numbers?

- \(\begin{bmatrix}2 &-5\\3 &-6\end{bmatrix}\) (ii) \(\begin{bmatrix}4 &1\\0 &4\end{bmatrix}\) (iii) \(\begin{bmatrix}1 &-1 &3\\0 &5 &-2\\0 &0 &7\end{bmatrix}\) (iv) \(\begin{bmatrix}7 &1 &1\\0 &2 &2\\0 &1 &3\end{bmatrix}\)

- A. (i) and (iii) only
- B. (iii) and (iv) only
- C. (i), (iii) and (iv) only
- D. (i), (ii) and (iii) only
- E. (i), (ii) and (iv) only

Problem 6 Solution

This problem tests our knowledge of Theorem 6 of Section 5.3 "Diagonalization":

An \(n\times n\) matrix with \(n\) distinct eigenvalues is diagonalizable.

So let's find out the eigenvalues for each matrix:

- From the equation \(\det A-\lambda I=0\), we can obtain \[\begin{vmatrix}2-\lambda &-5\\3 &-6-\lambda\end{vmatrix}=(\lambda-2)(\lambda+6)+15=(\lambda+1)\lambda+3)=0\] This leads to two roots \(\lambda_1=-1\), \(\lambda_2=-3\).

- Since this is a triangular matrix, the eigenvalue is just 4, with multiplicity 2.

- For the same reason, this \(3\times 3\) matrix has eigenvalues 1, 5 and 7.

- Use cofactor expansion with \(C_{1,1}\), we have \[\begin{align} \begin{vmatrix}7-\lambda &1 &1\\0 &2-\lambda &2\\0 &1 &3-\lambda\end{vmatrix}&= (7-\lambda)(-1)^{1+1}\begin{vmatrix}2-\lambda &2\\1 &3-\lambda\end{vmatrix}\\ &=(7-\lambda)(\lambda^2-5\lambda+6-2)\\ &=(7-\lambda)(\lambda-4)(\lambda-1) \end{align}\] The eigenvalues are 7, 4, and 1.

Now we can see that (i), (iii), and (iv) have distinct eigenvalues, they are diagonalizable matrices.

So the answer is **C**.

### Problem 7 (10 points)

A real \(2\times 2\) matrix \(A\) has an eigenvalue \(\lambda_1=2+i\) with corresponding eigenvector \(\pmb v_1=\begin{bmatrix}3-i\\4+i\end{bmatrix}\). Which of the following is the general REAL solution to the system of differential equations \(\pmb x'(t)=A\pmb x(t)\)

- A. \(c_{1}e^{2t}\begin{bmatrix}3\cos t-\sin t\\4\cos t+\sin t\end{bmatrix}+c_{2}e^{2t}\begin{bmatrix}3\sin t+\cos t\\4\sin t-\cos t\end{bmatrix}\)
- B. \(c_{1}e^{2t}\begin{bmatrix}-3\cos t+\sin t\\4\cos t-\sin t\end{bmatrix}+c_{2}e^{2t}\begin{bmatrix}3\sin t-\cos t\\4\sin t-\cos t\end{bmatrix}\)
- C. \(c_{1}e^{2t}\begin{bmatrix}3\cos t-\sin t\\4\cos t+\sin t\end{bmatrix}+c_{2}e^{2t}\begin{bmatrix}3\sin t-\cos t\\4\sin t-\cos t\end{bmatrix}\)
- D. \(c_{1}e^{2t}\begin{bmatrix}3\cos t+\sin t\\4\cos t-\sin t\end{bmatrix}+c_{2}e^{2t}\begin{bmatrix}3\sin t+\cos t\\4\sin t-\cos t\end{bmatrix}\)
- E. \(c_{1}e^{2t}\begin{bmatrix}3\cos t+\sin t\\4\cos t-\sin t\end{bmatrix}+c_{2}e^{2t}\begin{bmatrix}3\sin t-\cos t\\4\sin t+\cos t\end{bmatrix}\)

Problem 7 Solution

From Section 5.7 "Applications to Differential Equations", we learn that the general solution to a matrix differential equation is \[\pmb x(t)=c_1\pmb{v}_1 e^{\lambda_1 t}+c_2\pmb{v}_2 e^{\lambda_2 t}\] For a real matrix, complex eigenvalues and associated eigenvectors come in conjugate pairs. Hence we know that \(\lambda_2=2-i\) and \(\pmb{v}_2=\begin{bmatrix}3+i\\4-i\end{bmatrix}\). However, we do not need these two to find our solution here. **The real and imaginary parts of \(\pmb{v}_1 e^{\lambda_1 t}\) are (real) solutions of \(\pmb x'(t)=A\pmb x(t)\), because they are linear combinations of \(\pmb{v}_1 e^{\lambda_1 t}\) and \(\pmb{v}_2 e^{\lambda_2 t}\).** (See the proof in "Complex Eigenvalues" of Section 5.7)

Now use Euler's formula (\(e^{ix}=\cos x+i\sin x\)), we have \[\begin{align} \pmb{v}_1 e^{\lambda_1 t} &=e^{(2+i)t}\begin{bmatrix}3-i\\4+i\end{bmatrix}\\ &=e^{2t}(\cos t+i\sin t)\begin{bmatrix}3-i\\4+i\end{bmatrix}\\ &=e^{2t}\begin{bmatrix}(3\cos t+\sin t)+(3\sin t-\cos t)i\\(4\cos t-\sin t)+(4\sin t+\cos t)i\end{bmatrix} \end{align}\] The general REAL solution is the linear combination of the REAL and IMAGINARY parts of the result above, it is \[c_1 e^{2t}\begin{bmatrix}3\cos t+\sin t\\4\cos t-\sin t\end{bmatrix}+ c_2 e^{2t}\begin{bmatrix}3\sin t-\cos t\\4\sin t+\cos t\end{bmatrix}\]

So the answer is **E**.

### Problem 8 (10 points)

Let \(T: M_{2\times 2}\to M_{2\times 2}\) be a linear map defined as \(A\mapsto A+A^T\).

(2 points) (1) Find \(T(\begin{bmatrix}1 &2\\3 &4\end{bmatrix})\)

(4 points) (2) Find a basis for the range of \(T\).

(4 points) (3) Find a basis for the kernel of \(T\).

Problem 8 Solution

As the mapping rule is \(A\mapsto A+A^T\), we can directly write down the transformation as below \[T(\begin{bmatrix}1 &2\\3 &4\end{bmatrix})=\begin{bmatrix}1 &2\\3 &4\end{bmatrix}+\begin{bmatrix}1 &2\\3 &4\end{bmatrix}^T=\begin{bmatrix}2 &5\\5 &8\end{bmatrix}\]

If we denote the 4 entries of a \(2\times 2\) matrix as \(\begin{bmatrix}a &b\\c &d\end{bmatrix}\), the transformation can be written as \[\begin{align} T(\begin{bmatrix}a &b\\c &d\end{bmatrix}) &=\begin{bmatrix}a &b\\c &d\end{bmatrix}+\begin{bmatrix}a &b\\c &d\end{bmatrix}^T=\begin{bmatrix}2a &b+c\\b+c &2d\end{bmatrix}\\ &=2a\begin{bmatrix}1 &0\\0 &0\end{bmatrix}+(b+c)\begin{bmatrix}0 &1\\1 &0\end{bmatrix}+2d\begin{bmatrix}0 &0\\0 &1\end{bmatrix} \end{align}\] So the basis can be the set of three \(3\times 3\) matrices like below \[ \begin{Bmatrix}\begin{bmatrix}1 &0\\0 &0\end{bmatrix},\begin{bmatrix}0 &1\\1 &0\end{bmatrix},\begin{bmatrix}0 &0\\0 &1\end{bmatrix}\end{Bmatrix} \]

The kernel (or null space) of such a \(T\) is the set of all \(\pmb u\) in vector space \(V\) such that \(T(\pmb u)=\pmb 0\). Write this as \[T(\begin{bmatrix}a &b\\c &d\end{bmatrix})=\begin{bmatrix}2a &b+c\\b+c &2d\end{bmatrix}=\begin{bmatrix}0 &0\\0 &0\end{bmatrix}\] This leads to \(a=d=0\) and \(c=-b\). So the original matrix \(A\) that satified this conditioncan be represented as \(c\begin{bmatrix}0 &1\\-1 &0\end{bmatrix}\). This shows that \(\begin{bmatrix}0 &1\\-1 &0\end{bmatrix}\) (or \(\begin{bmatrix}0 &-1\\1 &0\end{bmatrix}\)) is the basis for the null space of \(T\).

### Problem 9 (10 points)

(6 points) (1) Find all the eigenvalues of matrix \(A=\begin{bmatrix}4 &0 &0\\1 &2 &1\\-1 &2 &3\end{bmatrix}\), and find a basis for the eigenspace corresponding to each of the eigenvalues.

(4 points) (2) Find an invertible matrix \(P\) and a diagonal matrix \(D\) such that \[ \begin{bmatrix}4 &0 &0\\1 &2 &1\\-1 &2 &3\end{bmatrix}=PDP^{-1} \]

Problem 9 Solution

- Apply the equation \(\det A-\lambda I=0\), we have \[\begin{vmatrix}4-\lambda &0 &0\\1 &2-\lambda &1\\-1 &2 &3-\lambda\end{vmatrix}=(4-\lambda)\begin{vmatrix}2-\lambda &1\\2 &3-\lambda\end{vmatrix}=-(\lambda-4)^2(\lambda-1)=0\] So the eigenvalues are 4 an 1. Now to find eigenvector for each eigenvalue, we take the eigenvalue to the system \((A-\lambda I)\pmb x=\pmb 0\) and find the basis vector(s) which would be the eigenvector.
- For \(\lambda_1=\lambda_2=4\), we have the new matrix as \[\begin{bmatrix}0 &0 &0\\1 &-2 &1\\-1 &2 &-1\end{bmatrix}\sim \begin{bmatrix}0 &0 &0\\1 &-2 &1\\0 &0 &0\end{bmatrix}\] This gives \(x_1-2x_2+x_3=0\) with two free variables \(x_2\) and \(x_3\). Now in parametric vector form, we can obtain \[\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}2x_2-x_3\\x_2\\x_3\end{bmatrix}=x_2\begin{bmatrix}2\\1\\0\end{bmatrix}+x_3\begin{bmatrix}-1\\0\\1\end{bmatrix}\] A basis is \(\begin{Bmatrix}\begin{bmatrix}2\\1\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix}\end{Bmatrix}\).
- For \(\lambda_3=1\), the new matrix is \[\begin{bmatrix}3 &0 &0\\1 &1 &1\\-1 &2 &2\end{bmatrix}\sim \begin{bmatrix}1 &0 &0\\0 &1 &1\\0 &2 &2\end{bmatrix}\sim \begin{bmatrix}1 &0 &0\\0 &1 &1\\0 &0 &0\end{bmatrix}\] This gives \(x_1=0\) and \(x_2=-x_3\) with one free variable \(x_3\). Again in parametric vector form, we can obtain \[\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\-x_3\\x_3\end{bmatrix}=x_3\begin{bmatrix}0\\-1\\1\end{bmatrix}\] A basis is \(\begin{Bmatrix}\begin{bmatrix}0\\-1\\1\end{bmatrix}\end{Bmatrix}\).

- From the above solution we can directly write \(P\) and \(D\) below \[P=\begin{bmatrix}2 &-1 &0\\1 &0 &-1\\0 &1 &1\end{bmatrix}\quad D=\begin{bmatrix}4 &0 &0\\0 &4 &0\\0 &0 &1\end{bmatrix}\]

### Problem 10 (10 points)

(4 points) (1) Find the eigenvalues and corresponding eigenvectors of the matrix \[\begin{bmatrix}-5 &1\\4 &-2\end{bmatrix}\]

(2 points) (2) Find a general solution to the system of differential equations \[ \begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}= \begin{bmatrix}-5 &1\\4 &-2\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatrix} \]

(4 points) (3) Let \(\begin{bmatrix}x(t)\\y(t)\end{bmatrix}\) be a particular soilution to the initial value problem \[ \begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}= \begin{bmatrix}-5 &1\\4 &-2\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatrix}, \begin{bmatrix}x(0)\\y(0)\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}. \] Find \(x(1)+y(1)\).

Problem 10 Solution

- To find eigenvalues, write down the determinant as \[\begin{vmatrix}-5-\lambda &1\\4 &-2-\lambda\end{vmatrix}=(\lambda+6)(\lambda+1)=0\] So the eigenvalues are \(\lambda_1=-6\) and \(\lambda_2=-1\). Now follow the same method as Problem 9 solution to get eigenvectors for them.
- For \(\lambda_1=-6\), the new matrix is \[\begin{bmatrix}1 &1\\4 &4\end{bmatrix}\sim \begin{bmatrix}1 &1\\0 &0\end{bmatrix}\] The eigenvector is \(\begin{bmatrix}1\\-1\end{bmatrix}\).
- For \(\lambda_1=-1\), the new matrix is \[\begin{bmatrix}-4 &1\\4 &-1\end{bmatrix}\sim \begin{bmatrix}-4 &1\\0 &0\end{bmatrix}\] The eigenvector is \(\begin{bmatrix}1\\4\end{bmatrix}\).

- The general solution to a matrix differential equation is \[\pmb x(t)=c_1\pmb{v}_1 e^{\lambda_1 t}+c_2\pmb{v}_2 e^{\lambda_2 t}\] So from this, since we already found out the eigenvalues and the corresponding eigenvectors, we can write down \[ \begin{bmatrix}x(t)\\y(t)\end{bmatrix}=c_1\begin{bmatrix}1\\-1\end{bmatrix}e^{-6t}+c_2\begin{bmatrix}1\\4\end{bmatrix}e^{-t} \]
- Now apply the initial values of \(x(0)\) and \(y(0)\), here comes the following equations: \[\begin{align} c_1+c_2&=3\\ -c_1+4c_2&=7 \end{align}\] This gives \(c_1=1\) and \(c_2=2\). So \(x(1)+y(1)=e^{-6}+2e^{-1}-e^{-6}+8e^{-1}=10e^{-1}\).

## Summary

Here is the table listing the key knowledge points for each problem in this exam:

Problem # | Points of Knowledge | Book Sections |
---|---|---|

1 | The Rank Theorem | 4.6 "Rank" |

2 | Linear dependence, Invertible Matrix Theorem | 4.3 "Linearly Independent Sets; Bases", 4.6 "Rank" |

3 | Eigenvectors and Eigenvalues | 5.1 "Eigenvectors and Eigenvalues" |

4 | Vector Spaces and Subspaces | 4.1 "Vector Spaces and Subspaces" |

5 | Eigenfunctions of the Differential Equation | 5.7 "Applications to Differential Equations" |

6 | The Diagonalization Theorem, Diagonalizing Matrices | 5.3 "Diagonalization" |

7 | Complex Eigenvalues and Eigenvectors | 5.5 "Complex Eigenvalues" |

8 | Kernel and Range of a Linear Transformation | 4.2 "Null Spaces, Column Spaces, and Linear Transformations" |

9 | Eigenvalues, Basis for Eigenspace, Diagonalizing Matrices | 5.1 "Eigenvectors and Eigenvalues", 5.3 "Diagonalization" |

10 | Eigenvectors and Eigenvalues | 5.1 "Eigenvectors and Eigenvalues", 5.7 "Applications to Differential Equations" |