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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?

$$-x_1 + x_2 + 4x_3 = 0$$ $$x_1 + 3x_2 + 8x_3 = 0$$ $$\frac 12 x_1 + x_2 + \frac 52 x_3 = 0$$

and

$$x_1 - x_3 = 0$$ $$x_2 + 3x_3 = 0$$

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.]

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You went wrong in that you solved the system and considered that the answer to the question. Both of these systems have infinite solutions but that is not what the question is asking. It is asking are the systems equivalent. One way to do that is solve them both and see that they give the same solution set. Another way to to transform one set of equations into the other using algebra.

$\begin{align} -x_1+x_2+4x_3 &= 0 \\ +x_1+3x_2+8x_3 &= 0 \\ \hline 4x_2+12x_3 &= 0 \end{align} $

Divide the last equation by $4$ and you get $x_2+3x_3=0$, which can also be viewed as $x_3=-3x_3$. Substituting this into one of the equations (say the second), we get $x_1+3x_2+8x_3=0 \iff x_1+3(-3x_3)+8x_3=0 \iff x_1-x_3=0$.

Sine we can transform the one system into the other, they are equivalent. This is often done using the augmented matrix version of an equation and Gaussian elimination (also called Gauss-Jordan elimination).


In response to the comment, label the first system's equations $a,b,c$, in order, and the second system's equations $d,e$, in order. The work above shows that $e=\frac{1}{4}a+\frac{1}{4}b$. Since we used $a$ and $b$ to get $e$, we have to pick a different pair to try and get $d$. It turns out that $d = \frac{2}{3}c-\frac{2}{3}a$ (though this is not the only possible answer).

The other direction is much simpler. Since $d$ has $x_1$ but not $x_2$, any $x_1$ terms in $a,b,c$ have to have come from $d$. The same is true for $e$ and $x_2$. Since $a$ has $-x_1$, we need a $-d$. Since $a$ has $x_2$, we need $e$. It turns out that this is enough to get the answer as the $x_3$ takes care of itself. So $a=e-d$. Similarly, $b=d+3e$ and $c=\frac{1}{2}d+e$.

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  • $\begingroup$ Aha, interesting, thanks. But I neglected to mention that the exercise also asks you to "express each equation in each system as a linear combination of the equations in the other system," which is why I was trying to find the co-efficients that would enable that. Where did I go wrong in that attempt? $\endgroup$ – user124384 Feb 18 '14 at 20:59
  • $\begingroup$ Cool, thanks. The book provided zero guidance on how to solve these (probably because I'm supposed to know how already. I'm reading this book way too far in advance, but I swear it's helping me understand matrices better than less advanced books). But if I may ask one quick follow-up question, why do you have to pick a different pair to get d, since you're not solving the system? The answer key I have just used the same pair. $\endgroup$ – user124384 Feb 18 '14 at 21:16
  • $\begingroup$ True enough. Since we are not solving, we don't really need a different pair in this case. Long years of habit in making sure things are independent lead me to pick different pairs. $\endgroup$ – John Habert Feb 19 '14 at 0:38
  • $\begingroup$ Ah, ok, just checking. Thanks so much for your help! $\endgroup$ – user124384 Feb 19 '14 at 0:47

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