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Section 2.1 #5: Linear or nonlinear?
$$3 \cos(x) - 4y + z =\sqrt{3}.$$ Solution: Nonlinear -- the variable "$x$" may not appear inside of a cosine function.

Section 2.1 #6: Linear or nonlinear?
$$(\cos(3))x - 4y + z = \sqrt{3}.$$ Solution: Linear. Although a cosine appears, $\cos(3)$ is simply a number. Although a $\sqrt{}$ appears, $\sqrt{3}$ is simply a number.

Section 2.1 #12: Find the solution set of $$2x_1 + 3x_2 =5.$$ Solution: Subtract $3x_2$ to get $$2x_1 = 5-3x_2.$$ Divide both sides by $2$ to get $$x_1 = \dfrac{5}{2} - \dfrac{3}{2}x_2.$$ Therefore the solution is $$\left( \dfrac{5}{2} - \dfrac{3}{2}x_2, x_2 \right).$$ (note: could also have solved for $x_2$ and got a slightly different looking solution)

Section 2.1 #28: Find the augmented matrix of the system $$\left\{ \begin{array}{llll} 2x_1 &+ 3x_2 &- x_3 &= 1 \\ x_1 & & +x_3 &= 0 \\ -x_1 &+ 2x_2 &-2x_3 &= 0. \end{array} \right.$$ Solution: $$\left[ \begin{array}{ll} 2 & 3 & -1 & 1 \\ 1 & 0 & 1 & 0 \\ -1 & 2 & -2 & 0 \end{array} \right]$$

Section 2.2 #1: Is the following matrix in row echelon form? Is it in reduced row echelon form?
$$\left[ \begin{array}{lll} 1 & 0 & 1 \\ 0 & 0 & 3 \\ 0 & 1 & 0 \end{array} \right]$$
Solution: No. Swapping row 2 and row 3 would put it into row echelon form. Furthermore, dividing the (new) third row by 3 would put it into reduced echelon form.

Section 2.2 #4: Is the following matrix in row echelon form? Is it in reduced row echelon form?
$$\left[ \begin{array}{ll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]$$
Solution: Yes! Echelon forms dictate properties that "the first nonzero entry in every row..." have. Since this matrix has no "nonzero entries" at all, all the properties for echelon form are vacuously true.

Section 2.2 #9: Use row operations to reduce the given matrix to reduced row echelon form: $$\left[ \begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right].$$
Solution: Calculate $$\begin{array}{ll} \left[ \begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right] &\stackrel{r_1 \leftrightarrow r_3}{\sim} \left[ \begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] \\ &\stackrel{r_1^*=r_1-r_2}{\sim} \left[ \begin{array}{ll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] \\ &\stackrel{r_2^*=r_2-r_3}{\sim} \left[ \begin{array}{ll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]. \end{array}$$

Section 2.2 #10: Use row operations to reduce the given matrix to reduced row echelon form: $$\left[ \begin{array}{ll} 4 & 3 \\ 2 & 1 \end{array} \right].$$
Solution: Calculate $$\begin{array}{ll} \left[ \begin{array}{ll} 4 & 3 \\ 2 & 1 \end{array} \right] & \stackrel{r_2^*=r_2-\frac{1}{2}r_1}{\sim} \left[ \begin{array}{lr} 4 & 3 \\ 0 & -\dfrac{1}{2} \end{array} \right] \\ &\stackrel{r_2^*=-2r_2}{\sim} \left[ \begin{array}{ll} 4 & 3 \\ 0 & 1 \end{array} \right] \\ &\stackrel{r_1^*=r_1-3r_2}{\sim} \left[ \begin{array}{ll} 4 & 0 \\ 0 & 1 \end{array} \right] \\ &\stackrel{r_1^*=\frac{1}{4}r_1}{\sim} \left[ \begin{array}{ll} 1 &0 \\ 0 & 1 \end{array} \right]. \end{array}$$

Section 2.2 #26: Solve the system of equations: $$\left\{ \begin{array}{llll} x& -y & +z &= 0 \\ -x &+ 3y & +z &= 5 \\ 3x &+ y &+ 7z &= 2. \end{array} \right.$$
Solution: First write the augmented matrix associated with this system: $$\left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ -1 & 3 & 1 & 5 \\ 3 & 1 & 7 & 2 \end{array} \right]$$ Now put this matrix into reduced echelon form using row operations: compute $$\begin{array}{ll} \left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ -1 & 3 & 1 & 5 \\ 3 & 1 & 7 & 2 \end{array} \right] & \stackrel{r_2^*=r_2+r_1, r_3^*=r_3-3r_1}{\sim} \left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ 0 & 2 & 2 & 5 \\ 0 & 4 & 4 & 2 \end{array} \right] \\ &\stackrel{r_3^*=r_3-2r_2}{\sim} \left[ \begin{array}{ll} 1 & -1 & 1 & 0 \\ 0 & 2 & 2 & 5 \\ 0 & 0 & 0 & -8 \end{array} \right] \\ &\stackrel{r_3^*=-\frac{1}{8}r_3}{\sim} \left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ 0 & 2 & 2 & 5 \\ 0 & 0 & 0 & 1 \end{array} \right] \\ &\stackrel{r_2^*=r_2-5r_3}{\sim} \left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \\ &\stackrel{r_2^* = \frac{1}{2}r_2}{\sim} \left[ \begin{array}{rrrr} 1 & -1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 &0&0&1 \end{array} \right] \\ &\stackrel{r_1^*=r_1+r_2}{\sim} \left[ \begin{array}{rrrr} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]. \end{array}$$ What system of equations is this (augmented) matrix encoding? Write the system: $$\left\{ \begin{array}{llll} x_1 + x_3 = 0 \\ x_2 + x_3 = 0 \\ 0 = 1. \end{array}\right.$$ Notice that the third equation is always false and so this system has no solution.
(note: this conclusion could have been reached once the row [0 0 0 -8] was found above (just write the system down) -- I went all the way to reduced echelon for practice).

Problem A: Solve the system of equations (field: $\mathbb{C}$) $$\left\{ \begin{array}{llll} (2+i)z_1 & + 3z_2 &= 1 \\ - z_1 &+ iz_2 &= 0 \\ \end{array} \right.$$ Solution: Write the augmented matrix: $$\left[ \begin{array}{rrr} 2+i & 3 & 1 \\ -1 & i & 0 \end{array} \right].$$ Now put this matrix into reduced echelon form: compute $$\begin{array}{ll} \left[ \begin{array}{rrr} 2+i & 3 & 1 \\ -1 & i & 0 \end{array} \right] &\stackrel{r_2^*=r_2+\frac{1}{2+i} r_1}{\sim} \left[ \begin{array}{rrr} 2+i & 3 & 1 \\ 0 & i+\dfrac{3}{2+i} & \dfrac{1}{2+i} \end{array} \right] \\ &\stackrel{(2+i)r_2}{\sim} \left[ \begin{array}{rrr} 2+i & 3 & 1 \\ 0 & (2i-1)+3 & 1 \end{array} \right] \\ &= \left[ \begin{array}{rrr} 2+i & 3 & 1 \\ 0 & 2+2i & 1 \end{array} \right] \\ &\stackrel{r_2^*=\frac{1}{2+2i}r_2}{\sim} \left[ \begin{array}{rrr} 2+i & 3 & 1 \\ 0 & 1 & \dfrac{1}{2+2i} \end{array} \right] \\ &\stackrel{r_1^*=r_1-3r_2}{\sim} \left[ \begin{array}{rrr} 2+i & 0 & 1 - \dfrac{3}{2+2i} \\ 0 & 1 & \dfrac{1}{2+2i} \end{array} \right] \\ &\stackrel{1-\frac{3}{2+2i}=\frac{-1+2i}{2+2i}}{\sim} \left[ \begin{array}{rrr} 2+i & 0 & \dfrac{-1+2i}{2+2i} \\ 0 & 1 & \dfrac{1}{2+2i} \end{array} \right] \\ &\stackrel{\frac{1}{2+2i}=\frac{1}{2+2i} \frac{2-2i}{2-2i}=\frac{2-2i}{8}=\frac{1-i}{4}}{\sim} \left[ \begin{array}{rrr} 2+i & 0 & \dfrac{(-1+2i)(1-i)}{4} \\ 0 & 1 & \dfrac{1-i}{4} \end{array} \right] \\ &= \left[ \begin{array}{rrr} 2+i & 0 & \dfrac{1+3i}{4} \\ 0 & 1 & \dfrac{1-i}{4} \end{array} \right] \\ &\stackrel{r_1^*=\frac{1}{2+i}r_1; \frac{1}{2+i}=\frac{1}{2+i}\frac{2-i}{2-i}=\frac{2-i}{5}}{\sim} \left[ \begin{array}{rrr} 1 & 0 & \dfrac{(1+3i)(2-i)}{20} \\ 0 & 1 & \dfrac{1-i}{4} \end{array} \right] \\ &= \left[ \begin{array}{rrr} 1 & 0 & \dfrac{5+5i}{20} \\ 0 & 1 & \dfrac{1-i}{4} \end{array} \right] \\ &= \left[ \begin{array}{rrr} 1 & 0 & \dfrac{1+i}{4} \\ 0 & 1 & \dfrac{1-i}{4} \end{array} \right] \end{array}$$ Translate this back into a system to observe the original system is equivalent to $$\left\{ \begin{array}{lll} z_1 & &= \dfrac{1}{4} + \dfrac{1}{4}i \\ & z_2 &= \dfrac{1}{4} - \dfrac{1}{4}i \end{array} \right.$$ which could be written formally as a solution by writing $$\left( \dfrac{1}{4} + \dfrac{1}{4}i, \dfrac{1}{4} - \dfrac{1}{4}i \right).$$ Problem B: Solve the system of equations (field: $\mathbb{Z}_5$) $$\left\{ \begin{array}{lll} 3x_1 &+ 4x_2 &= 1 \\ x_1 &- 2x_2 &= 0 \end{array} \right.$$ Solution: First write the augmented matrix associated with the system: $$\left[ \begin{array}{rrr} 3 & 4 & 1 \\ 1 & -2 & 0 \end{array} \right].$$ Now compute, noting that "$-2$" is $3$ in this structure: $$\begin{array}{ll} \left[ \begin{array}{rrr} 3 & 4 & 1 \\ 1 & -2 & 0 \end{array} \right] &\stackrel{r_1^*=2r_1}{\sim} \left[ \begin{array}{rrr} 1 & 3 & 2 \\ 1 & 3 & 0 \end{array} \right] \\ &\stackrel{r_2^*=r_2-r_1}{\sim} \left[ \begin{array}{rrr} 1 & 3 & 2 \\ 0 & 0 & 3 \end{array} \right] \\ &\stackrel{r_2^*=2r_2}{\sim} \left[ \begin{array}{rrr} 1 & 3 & 2 \\ 0 & 0 & 1 \end{array} \right] \\ &\stackrel{r_1^*=r_1-2r_2}{\sim} \left[ \begin{array}{rrr} 1 & 3 & 0 \\ 0 & 0 & 1 \end{array} \right]. \end{array}$$ Translate this matrix back into a system to see that the original system is equivalent to $$\left\{ \begin{array}{lll} x_1 &+ 3x_2 &= 0 \\ 0& &= 1, \end{array} \right.$$ but since the second equation is false, we observe that the original system has no solution.