*Solutions to [[Algebra2Sheet1.pdf]].* ### Section A: Preliminary warmup on simultaneous linear equations 1. (a) We have $\begin{align}x+ y &= 0 \tag{1} \\ 2x + 4y &= 0 \tag{2} \end{align}$$(2)-2\times(1)$ gives $2y = 0$ hence $(x,y) = (0,0)$. (b) We have $\begin{align}x+ y &= 1 \tag{1} \\ 2x + 4y &= 6 \tag{2} \end{align}$$(2)-2\times(1)$ gives $2y =4 \implies y=2$. Since $x+y=1$, we have $(x,y)=(-1,2).$ (c) ### Section B: Exercises on §1.1 of the lecture notes 4. We're given $\underline{v} = (1,2,3)^{T}$ and $\underline{w}=(2,-1,0)^{T} \in \mathbb{R}^{3}$. (a) $3\underline{v}-2\underline{w}=(-1,8,9)^{T}$. (b) Considering the components separately, this is the same as solving the following equations simultaneously $\begin{align}3+2\mu &= -5 \tag{1} \\ 6 - \mu & = 10 \tag{2} \end{align} $ rearranging $(1)$ gives $\mu=-4$ which satisfies $(2)$ so we've found a solution. (c) Let $\lambda\in \mathbb{R}$. Considering the components separately, this is the same as solving the following equations simultaneously $\begin{align} -2\mu+3 \lambda & = 3 \tag{1} \\ \mu + \lambda &= 6 \tag{2} \\ 3\lambda &= 9 \tag{3} \end{align}$$(3)$ give $\lambda=3$. Substituting $\lambda$ in $(2)$ gives $\mu =3$. $(\lambda,\mu)=(3,3)$ indeed satisfies $(1),(2)$ and $(3)$. 5. We have $\underline{v} = (2,1)^{T}$ and $\underline{w}=(1,3)^{T}$. (a) Sufficient to solve the following equations simultaneously $\begin{align}2 + \lambda &= \mu \tag{1} \\ 1 + 3\lambda &= 2\mu \tag{2}\end{align}$for some $\mu\in \mathbb{R}$. $(2)-2\times(1)$ gives $-3+\lambda=0 \implies \lambda=3$. (b) Sufficient to solve the following equations simultaneously $\begin{align}2 + \lambda &= 0 \tag{1} \\ 1 + 3\lambda &= \mu \tag{2}\end{align}$for some $\mu \in \mathbb{R}$. $(1)$ gives $\lambda=-2$. (c) $\underline{v}+\lambda \underline{w}$ describes a strip of length $\underline{v}$ ? 7. We have $v_{1}= (1,-1,2)^{T}$ and $v_{2}=(2,1,1)^{T} \in \mathbb{R}^{3}$ (a) By considering component separately, sufficient to solve the following equations simultaneously: $\begin{align} \lambda_{1} + 2 \lambda_{2} = 0 \tag{1} \\ -\lambda_{1} + \lambda_{2} = 0 \tag{2} \\ 2\lambda_{1} + \lambda_{2} = 0 \tag{3} \end{align}$$(2)$ gives $\lambda_{1}=\lambda_{2}$. Substituting in $(1)$ gives $\lambda_{1}=\lambda_{2}= 0$ which satisfies $(3)$. Hence $v_{1},v_{2}$ are linearly independent. (b) By considering component separately, sufficient to solve the following equations simultaneously: $\begin{align} \lambda_{1} + 2 \lambda_{2} = -1 \tag{1} \\ -\lambda_{1} + \lambda_{2} = -8 \tag{2} \\ 2\lambda_{1} + \lambda_{2} = 7 \tag{3} \end{align}$$(1) +(2)$ gives $3\lambda_{2}=-9 \implies \lambda_{2}=-3$. Substituting in $(2)$ gives $(\lambda_{1},\lambda_{2})=(5,-3)$ which indeed satisfies $(1),(2),\&,(3).$ (c) By considering component separately, sufficient to solve the following equations simultaneously: $\begin{align} \lambda_{1} + 2 \lambda_{2} = 4 \tag{1} \\ -\lambda_{1} + \lambda_{2} = 5 \tag{2} \\ 2\lambda_{1} + \lambda_{2} = -3 \tag{3} \end{align}$$(1)+(2)$ gives $3\lambda_{2}=9 \implies \lambda_{2}=3$. Substituting in $(2)$ gives $(\lambda_{1},\lambda_{2})= (-2,3)$ which however does not satisfy $(3)$ hence no $\lambda_{i}$ exists. 8. (c) $\underline{r} = (1,2,3)^{T} + \lambda \, (2, 2,1)^{T}$ or $\underline{r} = (-1,0,2)^{T} + \lambda \, (2, 2,1)^{T}$ with $\lambda \in \mathbb{R}$.