# Math Help - Systems of equations with arithmetic progression in coefficients

1. ## Systems of equations with arithmetic progression in coefficients

Hello guys,

I am analyzing system of linear equations with coefficients that are part of arithmetic progression.

Example of such system is

$\begin{array}{rcl}x+2y & = & 3 \\ x-2y & = & -5\end{array}$

I should draw conclusions and prove them for 2x2 and 3x3 systems.

2x2

By plotting equation lines it is easy to conclude that all such 2x2 systems have unique solution [-1;2].

By looking at a single equation
$ax+(a+k)y = a+2k \Leftrightarrow a(x+y)+ky=a+2k$
I have found that it will evaluate to true if
$ky=2k \Leftrightarrow y=2 \land x+y=1 \Leftrightarrow x=-1$.

Does this holds? I would appreciate very much if someone could provide real (rigor) proof.

3x3

I have tried few combinations of coefficients in matlab and got [0;-1;2] solution column vector few times, but I also got [NaN; Inf ; -Inf] in some cases. I do not have skills to prove this algebraically, any help?

Thanks!

2. ## Re: Systems of equations with arithmetic progression in coefficients

Originally Posted by losm1
Hello guys,

I am analyzing system of linear equations with coefficients that are part of arithmetic progression.

Example of such system is

$\begin{array}{rcl}x+2y & = & 3 \\ x-2y & = & -5\end{array}$

I should draw conclusions and prove them for 2x2 and 3x3 systems.

2x2

By plotting equation lines it is easy to conclude that all such 2x2 systems have unique solution [-1;2].

By looking at a single equation
$ax+(a+k)y = a+2k \Leftrightarrow a(x+y)+ky=a+2k$
I have found that it will evaluate to true if
$ky=2k \Leftrightarrow y=2 \land x+y=1 \Leftrightarrow x=-1$.

Does this holds? I would appreciate very much if someone could provide real (rigor) proof.

3x3

I have tried few combinations of coefficients in matlab and got [0;-1;2] solution column vector few times, but I also got [NaN; Inf ; -Inf] in some cases. I do not have skills to prove this algebraically, any help?

Thanks!
In the 2x2 case, it is correct that every line with equation of the form $ax + (a+k)y = a+2k$ passes through the point (–1,2). All you have to do to prove that is to plug the values x=–1 and y=2 into the equation and check that the equation is satisfied. If you have two equations of that form, they will usually have that as their unique solution. However, there is an exceptional case, namely when the two lines coincide. For example, the equations

$x+2y=3$
$2x+4y=6$

have infinitely many solutions, because every point on the first line is also on the second line.

In the 3x3 case, an equation of the form $ax + (a+k)y + (a+2k)z = a+3k$ represents a plane in three-dimensional space. It will always contain the whole of the line $(-2,3,0) + t(1,-2,1).$ Again, you can verify that by plugging $x=-2+t,$ $y=3-2t$ and $z=t$ into the equation and checking that it is satisfied. In particular, if you put t=2 then you see that the point $(0,-1,2)$ lies in the plane.

So if you take any number of planes of that form, their intersection will always contain that line. For two or more planes, the intersection will normally be exactly equal to that line. But as in the 2x2 case, there is an exceptional case that arises if all the equations are multiples of each other and therefore represent the same plane. In that case, every point of the plane will give a solution to the equations.

3. ## Re: Systems of equations with arithmetic progression in coefficients

Can someone help me prove that plane $ax + (a+k)y + (a+2k)z = a+3k$ always contains line that Opalg described?

Thanks