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Worksheet 2

Due 10/13

1. We know how to obtain the general solution from a linear system. Let's try to reverse it. Find a linear system who's general solution is $$(x_1,x_2,x_3,x_4) = (1,2,3,4) + s_1(5,6,7,8) + s_2(9,0,1,2).$$

ANSWER: This question turned out to be easier and harder than I thought.

The easier approach is to write out the 4 equations, solve for $s_1$ and to get rid of it, then do the same for $s_2$.

Here's the harder approach I had in mind. We have 2 free variables. I will choose $x_2$ and $x_3$ to be the free variables. The goal is to manipulate the form of the general solution until it looks like $$(x_1,x_2,x_3,x_4)=(X,0,0,X)+s_1(X,1,0,X)+s_2(X,0,1,X),$$ where $X$ is any number.

Let $u=(1,2,3,4), v=(5,6,7,8), w=(9,0,1,2)$. The general solution is equivalent to $\{u+x: x\in \text{span}(v,w)\}$. We have some freedom. We can replace $u$ with any particular solution and we can replace $v,w$ with any other 2 vectors with the same span.

Let $v_2=(v-7w)/6$. Then $v_2$ and $w$ has the same span as $v,w$ and $v_2=(-29/3,1,0,-1)$.

Let $u_2=u-2v_2-3w$. So $u_2=(-20/3,0,0,0)$.

We now have that $$(x_1,x_2,x_3,x_4)=(-20/3,0,0,0)+s_1(-29/3,1,0,-1)+s_2(9,0,1,2)$$ is of the desired form. By setting $x_2=s_1$ and $x_3=s_2$, we see that $x_1=-20/3-29/3x_2$ and $x_4=-x_2+2x_3$.

2. Suppose $A$ is a matrix. Let $v,w$ be distinct (meaning $x\neq y$) vectors that solve $Ax=0$ so $Av=0$ and $Aw=0$ ($0$ here of course means the zero vector!). Let $L$ be the line that passes through $v$ and $w$. If $u$ is on $L$, then $Au=0$. Why? This exercise suggests that solution spaces are convex.

ANSWER:

If $u$ is on the line that passes through $v$ and $w$, then $u$ is of the form $sv+(1-s)w$. Then $A(u)=sA(v)+(1-s)A(w)=s0+(1-s)0=0$.

3. Let $z_1,z_2\in \mathbb{R}$ and let $S=\{(1,z_1,z_2)$, $(2,1,0)$, $(1,0,-1)\}$.

   • Find some values for $z_1$ and $z_2$ such that $S$ spans $\mathbb{R}^3$.

   • Find some values for $z_1$ and $z_2$ such that $S$ does not span $\mathbb{R}^3$.

   • Find all values for $z_1$ and $z_2$ such that $S$ spans $\mathbb{R}^3$. (In the process of solving this problem, some of you will be tempted to divide by zero. Resist that temptation.)

ANSWER: The problem becomes much easier once you reorder the vectors. This does not affect linear indepedence! The matrix use to determined linear independence is $$\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & z_1 \\ -1 & 0 & z_2 \end{bmatrix}$$ This matrix is much easier to use than the order the vectors were presented in. The nasty $(1,z_1,z_2)$ is in the last column and the topright entry is a 1. This matrix is equivalent to $$\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & z_1 \\ 0 & 2 & 1+z_2 \end{bmatrix}$$ by adding the first row to the last. We can then see that this matrix is has a pivot in each column whenever $(1,z_1)$ is not parallel to $(2,1+z_2)$. So the vectors are linearly independent whenever $2z_1\neq 1+z_2$.

4. Consider the following linear system that came from the book and the lecture. $$\begin{align} 2x_1-6x_2-x_3+8x_4 &= 0 \\ x_1 - 3x_2 - x_3 + 6x_4 &= 0 \\ -x_1+3x_2-x_3 +2x_4 &= 0. \end{align}$$ Using row reduction, we see that a general solution is of the form $x=s_1(3,1,0,0)+s_2(-2,0,4,1)$. Let $v_1=(2,1,-1), v_2=(-6,-3,3), v_3=(-1,-1,2), v_4=(8,6,2)$.

   • Is $\{v_1,v_2,v_3,v_4\}$ is linearly independent set? The answer should be no.

   • Express $v_1$ as a linear combination of $v_2,v_3,v_4$.

   • Express $v_2$ as a linear combination of $v_1,v_3,v_4$.

   • Express $v_3$ as a linear combination of $v_1,v_2,v_4$.

   • Express $v_4$ as a linear combination of $v_1,v_2,v_3$.

ANSWER:

By setting $s_1=1$ and $s_2=2$, we obtain a nontrivial solution to the system which implies $v_1+v_2+4v_3+v_4=0$. By solving for $v_1$ here, we can write $v_1$ as a linear combination of $v_2,v_3,v_4$. Same for the others.

5. Suppose $\{v_1,v_2,v_3\}$ is a linearly dependent set. Is it always the case that we can write $v_1$ as a linear combination of $v_2$ and $v_3$? If not, come up with a counterexample.

ANSWER:

No. Take $v_1=(1,0), v_2=(0,1), v_3=(0,2)$.

6. Come up with a inconsistent linear system whose associated homogenous linear system is consistent.

ANSWER:

A homogeneous linear system is always consistent so any inconsistent linear system is an example.