Syllabus: [pdf]

Previous course materials
Fall 2017 (Fairmont)
Spring 2015 (Missouri S&T)
Fall 2014 (Missouri S&T)
Exams
Exam 1: [pdf] [tex]

Homework
Homework 1 (due 26 August) (solution: [pdf]): Section 1.1: #11, 12, 13, 15, 16, 17, 18, 19, 20; Section 2.1: 1, 4, 5, 6, 11, 13, 14, 25, 27, 28, 29, 30, 31, 32; Section 2.2 (note: on problems 9-11, (only do part b)): 1, 2, 3, 4, 9, 10, 11, 25, 26, 27, 28, 32 and the following problem
Problem A: Solve the system of equations, where $i$ denotes the "imaginary number" obeying the property $i^2=-1$: $$\left\{ \begin{array}{llll} iz_1 & + 2iz_2 &= 1 \\ - z_1 &+ iz_2 &= 0 \\ \end{array} \right.$$ note: it is helpful to remember that to "simplify" a complex number of the form $\dfrac{a+bi}{c+di}$ you may multiply by a "convenient form of $1$" defined from the "complex conjugate" of the denominator as follows: $$\dfrac{a+bi}{c+di} = \dfrac{a+bi}{c+di} \dfrac{c-di}{c-di} = \dfrac{(a+bi)(c-di)}{c^2+d^2} = \dfrac{ac + bd}{c^2+d^2} + \dfrac{bc-ad}{c^2+d^2}i.$$

Homework 2 (due 4 September) (solution: [pdf]): Section 2.2 (note: on problems 9-11, only do part b): #1, 2, 3, 4, 9, 10, 11, 25, 26, 27, 28, 32; Section 2.3: 1, 2, 3, 4, 7, 8, 9, 10, 22, 23, 24, 25 and the following problem:
Problem A: Solve the system of equations, where $i$ denotes the "imaginary number" obeying the property $i^2=-1$: $$\left\{ \begin{array}{llll} iz_1 & + 2iz_2 &= 1 \\ - z_1 &+ iz_2 &= 0 \\ \end{array} \right.$$ note: it is helpful to remember that to "simplify" a complex number of the form $\dfrac{a+bi}{c+di}$ you may multiply by a "convenient form of $1$" defined from the "complex conjugate" of the denominator as follows: $$\dfrac{a+bi}{c+di} = \dfrac{a+bi}{c+di} \dfrac{c-di}{c-di} = \dfrac{(a+bi)(c-di)}{c^2+d^2} = \dfrac{ac + bd}{c^2+d^2} + \dfrac{bc-ad}{c^2+d^2}i.$$
Homework 3 (due 9 September) (solution: [pdf]): Section 2.3: 9, 10, 22, 23, 24, 25, 29, 30; Section 3.1: #1, 2, 3, 4, 9, 10, 11, 12, 17, 21, 22, 39, 40
Homework 4 (due 16 September) (solution: [pdf]): Section 3.1: #1, 2, 3, 4, 9, 10, 11, 12, 17, 21, 22, 39; Section 3.2: #2, 4, 5, 7, 14, 15, 37; Section 3.3: #1, 2, 3, 4, 50, 52, 53
Homework 5 (due 2 October) (solution: [pdf]): Section 6.1 #24, 26, 27, 28, 34, 36, 46, 53, 62; Section 6.2: #5, 6, 8, 10, 11, 18, 19, 24, 25, 27, 28, 29, 35, 36, 39
Homework 6 (due 7 October) (solution: [pdf]): Section 6.2: #5, 6, 8, 10, 11, 18, 19, 24, 25, 27, 28, 29, 35, 36, 39; Section 6.3: #2, 3, 7, 8; Section 6.4: #1, 2, 3, 5, 6, 9, 10, 15, 16, 24, 27
Homework 7 (due 14 October) (solution: [pdf]): Section 6.3: #2, 3, 7, 8; Section 6.4: #1, 2, 3, 5, 6, 9, 10, 15, 16, 24, 27
Homework 8 (due 23 October): Section 6.4: #15, 16, 18, 27, 28

Quizzes
Quiz 1 (due 28 August) (solution: [pdf]): Put the matrix $\begin{bmatrix} 0&1&2 \\ 3&4&5 \end{bmatrix}$ into row echelon form.
Quiz 2 (due 11 September) (solution: [pdf]): Consider a matrix $A \in \mathbb{R}^{3 \times 3}$. What size does a matrix $B$ have to be so that the product $AB \in \mathbb{R}^{3 \times 7}$? What size does a matrix $C$ have to be to obtain the product $CA \in \mathbb{R}^{11 \times 3}$?
Quiz 3 (due 30 September) (solution: [pdf]): Is $\mathbb{R}^{2 \times 1}$ a subspace of $\mathbb{R}^{3 \times 1}$? If so, explain why. If not, explain why not.
Quiz 4 (due 7 October) (solution: [pdf]): Prove that if $\mathscr{B}=\{v_1,\ldots,v_n\}$ is a basis for a vector space $V$ and $\alpha \in \mathbb{R}$ is a scalar, then $[\alpha u]_{\mathscr{B}}=\alpha [u]_{\mathscr{B}}$.
Quiz 5 (due 9 October) (solution: [pdf]): In class, we found that the change-of-basis matrix from the basis $\mathscr{B}=\left\{ \begin{bmatrix} -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \end{bmatrix} \right\}$ to the basis $\mathscr{C} = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}$ was $P_{\mathscr{C} \leftarrow \mathscr{B}} = \begin{bmatrix} -3 & 3 \\ 2 & -1 \end{bmatrix}$. We showed directly that if $\vec{x}=\begin{bmatrix} 5 \\ -1 \end{bmatrix}$, then $\left[ \vec{x} \right]_{\mathscr{B}}=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ and that $\left[ \vec{x} \right]_{\mathscr{C}}=\begin{bmatrix} 6 \\ -1 \end{bmatrix}$ and verified that $P_{\mathscr{C} \leftarrow \mathscr{B}} \left[ \vec{x} \right]_{\mathscr{B}} = \left[ \vec{x} \right]_{\mathscr{C}}$.

Use that information to find $P_{\mathscr{C} \leftarrow \mathscr{B}}^{-1}$ and use it to show $[\vec{x}]_{\mathscr{B}}=P_{\mathscr{C}\leftarrow\mathscr{B}}^{-1} [\vec{x}]_{\mathscr{C}}$.