Cramer's Rule Computation Help

Let D(x,y) denote the determinant of the 2x2 matrix with rows x and y. Assume the vectors v1,v2, v3 in R2 are pairwise linearly independent. Prove that D(v2, v3)v1 + D(v3, v1) v2 + D(v1, v2)v3=0.

(Hint: write v1 as a linear combination of v2 and v3 and use Cramer's rule to solve for the coefficients.)

I'm not sure how to set this up as linear combination. I've been trying, but I must be doing something wrong because its not making much sense.

Re: Cramer's Rule Computation Help

I prefer to consider vectors as vertical and D(u, v) as the determinant of the matrix with columns u and v. Note that this does not really change the definition of D since determinants are invariant under transposition (this sounds so sciency!)

As the hint says, consider the system of equations , i.e., . Since , the system has a unique solution given by the Cramer'r rule. Find x and y, and then rewrite .

Re: Cramer's Rule Computation Help

I think I'm getting stuck in the actual use of Cramer's rule then, because I had it set up like this, but I'm not sure how to translate that into a 2x2 matrix, or do I need to. I split the v's into two components, and then it doesnt really simplify in a way that I can solve it.

Re: Cramer's Rule Computation Help

From using Cramer's rule we get that . Try to write what y is.

Re: Cramer's Rule Computation Help

so the y= det (v2,v1)/det (v2,v3)?

How do I then turn that into a 2x2 I can use to prove this?

Re: Cramer's Rule Computation Help

Quote:

Originally Posted by

**renolovexoxo** so the y= det (v2,v1)/det (v2,v3)?

Yes.

Quote:

Originally Posted by

**renolovexoxo** How do I then turn that into a 2x2 I can use to prove this?

Did not understand the question. What is "that" and "this"?

Substituting the values of x and y into gives you the equality that post #1 asks to prove (up to a few algebraic manipulations).

Re: Cramer's Rule Computation Help

I'm getting how they translate, I think it's the actual evaluation of the determinants that is giving me trouble. In class we usually worked with numbers, but i can't get it to simplify and I'm wondering if I'm doing it wrong. My matrix for one part looks something like

[v1 v1

v3 v3]

but i feel like i need to put the vectors into two components,something like

[v1,1 v1,2

v3,1 v3,3]

Re: Cramer's Rule Computation Help

You don't have to evaluate the determinants beyond D(v2, v3) and similar expressions. The problem is to prove

D(v2, v3)v1 + D(v3, v1) v2 + D(v1, v2)v3=0 (*)

and this equality contains things like D(v2, v3), so there is no reason to further simplify D(v2, v3).

So, to repeat, substitute x = D(v1, v3) / D(v2, v3) and y = D(v2, v1) / D(v2, v3) into x * v2 + y * v3 = v1 and you'll get almost (*).

Re: Cramer's Rule Computation Help

Oh okay. That would explain why I was so lost. Thank you so much for your help!

Re: Cramer's Rule Computation Help

Now suppose a1, a2, a3 in R2 and for i=1,2,3, and li be the line in the plane passing through ai with direction vector vi. Prove that the three lines have a point in common if and only if D(a1,v1)D(v2,v3)+D(a2,v2)D(v3,v1)+D(a3,v3)D(v1,v2) =0. I've gone about it by trying to write the lines l1 and l2 and finding their intersection on l3. so I have something like: l1=a1+sv1 and l2=a2+tv2. I know I want to solve for a1+sv1=a2+tv2 for some s,t. THis is part of the last problem, so I know it involves Cramer's rule, but I think the addition of the point is confusing me a bit in trying to set it up.

Re: Cramer's Rule Computation Help

I recommend posting this as a new question so that more people can see it. Include a link to this thread if the solution to the second exercise is supposed to use the first one.