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Section 1.1 #12) Let $\vec{b}= \left[ \begin{array}{ll} 3 \\ 2 \\ 1 \end{array} \right]$, $\vec{c} = \left[ \begin{array}{ll} 1 \\ -2 \\ 1 \end{array} \right]$, and $\vec{d}=\left[ \begin{array}{ll} -1 \\ -1 \\ -2 \end{array} \right]$.
Compute $3\vec{b}-2\vec{c}+\vec{d}$.
Solution: Compute $$\begin{array}{ll} 3 \vec{b} - 2 \vec{c} + \vec{d} &= 3 \left[ \begin{array}{ll} 3 \\ 2 \\ 1 \end{array} \right] - 2 \left[ \begin{array}{ll} 1 \\ -2 \\ 1 \end{array} \right] + \left[ \begin{array}{ll} -1 \\ -1 \\ -2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 9 \\ 6 \\ 3 \end{array} \right] - \left[ \begin{array}{ll} 2 \\ -4 \\ 2 \end{array} \right] + \left[ \begin{array}{ll} -1 \\ -1 \\ -2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 9 - 2 + (-1) \\ 6 - (-4) + (-1) \\ 3 -2 + (-2) \end{array} \right] \\ &= \left[ \begin{array}{ll} 6 \\ 9 \\ -1 \end{array} \right]. \blacksquare \end{array}$$ Section 1.1 #18) Solve for the vector $\vec{x}$ in terms of the vectors $\vec{a}$ and $\vec{b}$: $$\vec{x} + 2\vec{a} - \vec{b} = 3(\vec{x}+\vec{a})-2(2\vec{a}-\vec{b}).$$ Solution: First distribute the coefficients on the right-hand side to get $$\vec{x} + 2\vec{a} - \vec{b} = 3\vec{x} + 3\vec{a} - 4\vec{a} + 2 \vec{b}=3\vec{x}-\vec{a}+2\vec{b}.$$ Now algebra shows us that $$(2\vec{a}-\vec{b})+\vec{a}-2\vec{b} = 3\vec{x} - \vec{x},$$ and simplifying yields $$3\vec{a}-3\vec{b} = 2\vec{x},$$ or equivalently, $$\dfrac{3}{2}\vec{a} - \dfrac{3}{2} \vec{b} = \vec{x}. \blacksquare$$ Section 1.1 #34) Calculate $3+1+2+3$ in $\mathbb{Z}_4$.
Solution: Since $3+1=0$ in $\mathbb{Z}_4$ and $2+3=1$ in $\mathbb{Z}_4$, we may calculate
$$3+1+2+3=(3+1)+(2+3)=0+1=1. \blacksquare$$ Section 1.1 #40) Calculate $2^{100}$ in $\mathbb{Z}_{11}$.
Solution: Since in this structure, $$2^2=4,$$ $$2^4=(2^2)^2=4^2=5,$$ $$2^8=(2^4)^2=5^2=3,$$ $$2^{16}=(2^8)^2=3^2=9,$$ $$2^{32}=(2^{16})^2=9^2=4,$$ $$2^{64}=(2^{32})^2=4^2=5,$$ $$5 \cdot 4 = 9,$$ and $$9 \cdot 5 = 1,$$ we see $$2^{100}=2^{64+32+4}=2^{64}2^{32}2^4=(5 \cdot 4) \cdot 5 = 9 \cdot 5 = 1.$$ Section 1.1 #43) Calculate $$2 \left( \left[ \begin{array}{ll} 3 \\ 1 \\ 1 \\ 2 \end{array} \right] + \left[ \begin{array}{ll} 3 \\ 3 \\ 2 \\ 1 \end{array} \right] \right)$$ in $\mathbb{Z}_4$ and $\mathbb{Z}_5$.
Solution:
$\underline{\mathrm{In} \hspace{2pt} \mathbb{Z}_4}$
Calculate $$\begin{array}{ll} 2 \left( \left[ \begin{array}{ll} 3 \\ 1 \\ 1 \\ 2 \end{array} \right] + \left[ \begin{array}{ll} 3 \\ 3 \\ 2 \\ 1 \end{array} \right] \right) &= \left[ \begin{array}{ll} 2 \\ 2 \\ 2 \\ 0 \end{array} \right] + \left[ \begin{array}{ll} 2 \\ 2 \\ 0 \\ 2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 2+2 \\ 2+2 \\ 2+0 \\ 0+2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 0 \\ 0 \\ 2 \\ 2 \end{array} \right] \end{array}$$ $\underline{\mathrm{In} \hspace{2pt} \mathbb{Z}_5}$
Calculate $$\begin{array}{ll} 2 \left( \left[ \begin{array}{ll} 3 \\ 1 \\ 1 \\ 2 \end{array} \right] + \left[ \begin{array}{ll} 3 \\ 3 \\ 2 \\ 1 \end{array} \right] \right) &= \left[ \begin{array}{ll} 1 \\ 2 \\ 2 \\ 4 \end{array} \right] + \left[ \begin{array}{ll} 1 \\ 1 \\ 4 \\ 2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 1+1 \\ 2+1 \\ 2+4 \\ 4+2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 2 \\ 3 \\ 1 \\ 1 \end{array} \right]. \blacksquare \end{array}$$ Section 1.1: #52) Solve or indicate that there is no solution: $8x=9$ in $\mathbb{Z}_{11}$.
Solution: To solve this, note that $8 \cdot 7 = 1$ in $\mathbb{Z}_{11}$ and $9 \cdot 7 = 8$ in $\mathbb{Z}_{11}$. Therefore we multply both sides of the equation by $7$ to get $$8 \cdot 7 x = 9 \cdot 7,$$ or equivalently $$x = 8.$$ Section 1.1: #55) Solve or indicate that there is no solution: $6x+3=1$ in $\mathbb{Z}_8$.
Solution: This equation has solution $x=1$ in $\mathbb{Z}_8$. To see this, first subtract $3$ from both sides (noting that "$-2$" is $6$ in $\mathbb{Z}_8$) to get $$6x = 6.$$ Clearly $x=1$ is a solution. But it is not the only one! Any solution to this eqution would be an $x \in \mathbb{Z}_8$ with the property that $6x=6$. Consider the following (exhaustive) table to see that a second solution also exists: $$\begin{array}{|l|l|} \hline x & 6x \\ \hline 0 & 0 \\ 1 & 6 \\ 2 & 4 \\ 3 & 2 \\ 4 & 0 \\ 5 & 6 \\ 6 & 4 \\ 7 & 2 \\ \hline \end{array}$$ so we see that there are two solutions: $x=1$ and $x=5$. $\blacksquare$

Section 1.1 #57)
(a) For which values of $a$ does $ax=1$ have a solution in $\mathbb{Z}_5$?
(b) For which values of $a$ does $ax=1$ have a solution in $\mathbb{Z}_6$?
Solution:
(a): The only way that we may solve $ax=1$ in $\mathbb{Z}_5$ is to find some number $a^{-1} \in \mathbb{Z}_5$ with the property that $a a^{-1}=1$. Once such a number is found, the solution of $ax=1$ can be obtained by multiplying both sides by $a^{-1}$ to get $x=a^{-1}$. The table below shows that every nonzero $a \in \mathbb{Z}_5$ has an associated $a^{-1} \in \mathbb{Z}_5$ (and hence may be solved): $$\begin{array}{|l|l|} \hline a & a^{-1} \\ \hline 0 & \mathrm{n/a} \\ 1 & 1 \\ 2 & 3 \\ 3 & 2 \\ 4 & 4 \\ \hline \end{array}$$ (b): Here we seek to find an $a^{-1}$ for every $a \in \mathbb{Z}_6$. From the table we will see that it is not always possible to find an $a^{-1}$:
$$\begin{array}{|l|l|} \hline a & a^{-1} \\ \hline 0 & \mathrm{n/a} \\ 1 & 1 \\ 2 & DNE \\ 3 & DNE \\ 4 & DNE \\ 5 & 5 \\ \hline \end{array}$$ Since $a^{-1}$ exists only when $a \in \{1,5\}$, the only equations of the form $ax=1$ that may be solved in $\mathbb{Z}_6$ are $x=1$ and $5x=1$. $\blacksquare$