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
    (x1, x2, x3, x4) = (1, 2, 3, 4) + s1(5, 6, 7, 8) + s2(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 s1 and to get rid of it, then do the same for s2.

Here’s the harder approach I had in mind. We have 2 free variables. I will choose x2 and x3 to be the free variables. The goal is to manipulate the form of the general solution until it looks like
(x1, x2, x3, x4) = (X, 0, 0, X) + s1(X, 1, 0, X) + s2(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 ∈ 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 v2 = (v − 7w)/6. Then v2 and w has the same span as v, w and v2 = ( − 29/3, 1, 0,  − 1).

Let u2 = u − 2v2 − 3w. So u2 = ( − 20/3, 0, 0, 0).

We now have that
(x1, x2, x3, x4) = ( − 20/3, 0, 0, 0) + s1( − 29/3, 1, 0,  − 1) + s2(9, 0, 1, 2)
is of the desired form. By setting x2 = s1 and x3 = s2, we see that x1 =  − 20/3 − 29/3x2 and x4 =  − x2 + 2x3.

  1. Suppose A is a matrix. Let v, w be distinct (meaning x ≠ 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.

  1. Let z1, z2 ∈ ℝ and let S = {(1, z1, z2), (2, 1, 0), (1, 0,  − 1)}.

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, z1, z2) 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, z1) is not parallel to (2, 1 + z2). So the vectors are linearly independent whenever 2z1 ≠ 1 + z2.

  1. Consider the following linear system that came from the book and the lecture. Using row reduction, we see that a general solution is of the form x = s1(3, 1, 0, 0) + s2( − 2, 0, 4, 1). Let v1 = (2, 1,  − 1), v2 = ( − 6,  − 3, 3), v3 = ( − 1,  − 1, 2), v4 = (8, 6, 2).

ANSWER:

By setting s1 = 1 and s2 = 2, we obtain a nontrivial solution to the system which implies v1 + v2 + 4v3 + v4 = 0. By solving for v1 here, we can write v1 as a linear combination of v2, v3, v4. Same for the others.

  1. Suppose {v1, v2, v3} is a linearly dependent set. Is it always the case that we can write v1 as a linear combination of v2 and v3? If not, come up with a counterexample.

ANSWER:

No. Take v1 = (1, 0), v2 = (0, 1), v3 = (0, 2).

  1. 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.