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).
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.
Let z1, z2 ∈ ℝ and let S = {(1, z1, z2), (2, 1, 0), (1, 0, − 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).
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.
Come up with a inconsistent linear system whose associated homogenous linear system is consistent.