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I am tutoring a high school student in Algebra 2. Her class has just covered systems of linear equations, mainly via the "substitution" and "elimination" methods. Her textbook concludes that unit with Cramer's rule (which is only used for 2 x 2 and 3 x 3 systems).

My student likes using Cramer's rule, and asks me if there is any easy way to explain why it works. I'm having a hard time coming up with an explanation that doesn't require undergraduate-level linear algebra. Is there a simple way to explain it?

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  • $\begingroup$ In 2D, you are looking to scale two vectors forming a parallelogram to reach a given point in the plane. The area of that parallelogram is given by the determinant. I realize this is hastily (and inaccurately!) stated, but I believe one could fashion an intuitive explanation based on the volume of a parallelopiped. $\endgroup$ – Joseph O'Rourke Oct 25 '15 at 17:56
  • $\begingroup$ @JosephO'Rourke That's a nice beginning, but I'm not sure it's appropriate for the context of high school Algebra 2. I'll have to give some thought as to whether there's a way to translate that into more elementary language. $\endgroup$ – mweiss Oct 25 '15 at 18:48
  • $\begingroup$ For 2x2 systems it is certainly straightforward to check that the general solution given by Cramer's rule is in fact the solution. So the "proof is in the pudding." This does not explain how one might derive the rule, but this might be the best approach at the level of algebra 2. $\endgroup$ – user52817 Oct 27 '15 at 0:49
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    $\begingroup$ I agree with the others that in the $2 \times 2$ case, probably showing the student that it works through direct calculation is best. In some sense, you should just try to get her through this topic, because Cramer's rule is inappropriate in a high school algebra course for a whole host of reasons. Presumably the teacher is covering it because "it's in the book"... $\endgroup$ – Michael Joyce Oct 30 '15 at 12:52
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    $\begingroup$ ... And presumably it's in the book because the publisher, while no doubt releasing many various editions with different numberings of the exercises, has no interest or incentive to adapt to the computer age which has made Cramer's rule obsolete as a computational tool. $\endgroup$ – Michael Joyce Oct 30 '15 at 12:53
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Just solve a general $2$ by $2$ linear system using substitution, and out pops Cramer's rule.

Start with: \begin{align} ax + by = u ,\\ cx + dy = v. \end{align} Multiplying both sides of the first equation by $c$, and both sides of the second equation by $a$, then subtracting, we find that \begin{equation} (ad - bc)y = av - uc. \end{equation} Assuming that the determinant $ad - bc$ is not $0$, we find that \begin{equation} y = \frac{av - uc}{ad - bc} = \frac{ \begin{vmatrix} a & u \\ c & v \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}. \end{equation} The formula for $x$ can be derived similarly.

I think this is the easiest way to discover the determinant in the first place, and to recognize its relevance to solving linear systems.

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    $\begingroup$ It's true that the rule appears when you solve the general 2 by 2 system by hand, but this doesn't really explain why the rule should work in general or how to interpret what it does. $\endgroup$ – Joonas Ilmavirta Oct 30 '15 at 10:07
  • $\begingroup$ Yes, I agree with @JoonasIlmavirta. Even for the 3 x 3 case, an explicit calculation like this seems unwieldy to the point of being unreasonable. $\endgroup$ – mweiss Oct 30 '15 at 13:20
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    $\begingroup$ I'd be interested in learning a more enlightening viewpoint. But, I do think it's helpful to know that there is nothing difficult about discovering Cramer's rule. Anyone can easily work out the $2 \times 2$ and $3 \times 3$ cases by hand, and that's enough to guess the general rule. $\endgroup$ – littleO Oct 30 '15 at 13:30
  • $\begingroup$ @mweiss I am a fairly unreasonable person, if you look at page 194 of supermath.info/LinearNotes2015.pdf you'll see where I derive the formula for the 3x3 determinant from essentially the same line of thought as here ( I admit there is an ambiguity which we must fix by the consideration of volume...) this question ultimately comes back to the what is a determinant really discussion... $\endgroup$ – James S. Cook Nov 1 '15 at 1:14
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You can look at this short proof. Basically, all you need to know is that $\det$ is unchanged by adding any multiple of any row to any other row, and then use that to manipulate the original system of equations to get row $k$ being all zero except for the $k$-th entry, which is the trivial 1-dimensional case. But the big question is how students at that level be possibly expected to know and understand the determinant!

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