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So I'm doing linear algebra right now and I have a question regarding addition of equations as part of Gauss' elimination algorithm. I understand why it's possible, as the LHS of one equation can be added/subtracted to another equation as long as its RHS is also added/subtracted. But I'm trying to better understand about the intersecting point of the equations.

Given two equations: $$x + 2y = 3$$ $$3x + 4y = 5$$

they intersect at point (-1, 2). When I add other equations which are combinations of these two equations (ie $-x - y = -1$), they also all pass through this point. Maybe I'm missing something, but could someone point out which property allows for this? Why does a new equation, composed of some combination of two others, resulting in an equation with a different slope and intercepts, still pass through that same point? What is so special about that point?

I'm a pretty visual learner, so any visual analogies would be much appreciated. Thanks for your time!

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Suppose lines $$L_1: a_1x+b_1y=c_1$$ and $$L_2:a_2x+b_2y=c_2$$ pass through the common point $(p,q)$

Then $$AL_1+BL_2: (Aa_1+Ba_2)x+(Ab_1+Bb_2)y=Ac_1+Bc_2$$ does too because $$(Aa_1+Ba_2)p+(Ab_1+Bb_2)q=A(a_1p+b_1q)+B(a_2p+b_2q)=Ac_1+Bc_2$$

It is perhaps easier to see writing the equations a bit differently (a form which makes it explicit that they pass through the point $(p,q)$ - equivalent to a change of variables to make $(p,q)$ the origin)

$$L_1:a_1(x-p)+b_1(x-q)=0$$$$L_2:a_2(x-p)+b_2(x-q)=0$$$$AL_1+BL_2: (Aa_1+Ba_2)(x-p)+(Ab_1+Bb_2)(x-q)=0$$

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This has nothing to do with linearity. It is simply an example of the age-old principal "equals added to equals form equals".

In this case, when you have a specific point $(x,y)=(a,b)$, and if that point is a solution of two equations with variables $x,y$, then substitution of $x=a$, $y=b$ gives two numerical equations. Adding those two equations gives another numerical equation.

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  • $\begingroup$ Yeah this explains it a lot better than I did. I thought the OP may have wanted to see how to get the result in the process of row reducing. (+1) $\endgroup$ – Mr.Fry May 27 '14 at 14:41
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If you put the coefficients into a matrix and you have that one row is a linear combination of the other two then by some row operations you can get the row of coefficients for your new equation to be a row of all zeros. You will actually use the same values needed to produce the new equation (linear combination). For instance if you multiplied the first equation by 2 and added it to the third then in the matrix to get a row of zeros from the equation produced by the linear combination just undo this operation. Hence you will just be solving the regular system all over again thus getting the same solution set.

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Well, if $(a,b)$ is a solution to your equation, it is also a solution to $$ x+2y-3=0, \qquad 3x+4y-5=0. $$ It is then clear that it will also be a solution to any linear combination of these equations.

The only thing I've done is moving RHS to LHS; and by moving the constants back to RHS, you can recover your equations and linear combinations.

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