I'm going through Hoffman and Kunze's Linear Algebra on my own and can't for the life of me figure out part of Exercise 3 of Section 1.2:
Are the following two systems of linear equations equivalent?
$$-x_1 + x_2 + 4x_3 = 0$$ $$x_1 + 3x_2 + 8x_3 = 0$$ $$\frac 12 x_1 + x_2 + \frac 52 x_3 = 0$$
$$x_1 - x_3 = 0$$ $$x_2 + 3x_3 = 0$$
I've figured it out for the first system, but have not been able to solve the resulting systems of equations for the second system. My own calculations resulted in infinite solutions, but according to an answer key I found, there is definitely one solution (i.e. the systems are equivalent).
For the linear combination of the first equation of the second system, I got:
$$x_1 - x_3 = x_1(-a+b+ \frac 12 c) + x_2(a + 3b + c) + x_3(4a + 8b + \frac 52 c)$$
which results in this system of equations in order to solve for a, b and c:
$$(1)\; -a+b+ \frac 12 c = 1$$ $$(2)\; a + 3b + c = 0$$ $$(3)\; 4a + 8b + \frac 52 c = -1$$
Simplifying (1) and (2) gives you $4b+ \frac 32 c = 1$, and simplifying (2) and (3) gives you the same equation, which is why I figured there were infinite solutions. I used a different technique and different combination of equations, with the same result. (And the same thing happened with the second equation of the second system.)
Where did I go wrong???
[I know similar questions have been asked, but the ones I found weren't quite right in helping me solve this particular problem. Hope that's okay.]