# quadratic equations in matrix form.

• Feb 12th 2011, 01:30 AM
mgrafton
Hi

I'm sure this question will seem seem very trivial, but I was wondering if you could help me out.

Although I can see that quadratic equations such as

5x^2 -6xy+5y^2=8 can be shown as the matrix (xy)(5 -3/-3 5)=8 (Using the divide symbol to indicate row down.)

Similarly

7x^2-12xy-2y^2=10 would be (x y)(7 -6 /-6 -2)

but I was stuck when trying to convert the quadratic equation x^2 + x-8 +5xy -6y +2y^2=0 into matrix form.

If someone could shed some knowledge on the subject I would be deeply appreciative.

Thanks

Matt
• Feb 12th 2011, 02:48 AM
Moo
Hello,

That's something about quadratic forms. If I'm not mistaking, the third one isn't one, so you can't find the matrix form of this equation...

Also, don't forget $\begin{pmatrix} x\\ y\end{pmatrix}$ on the right side of the 2x2 matrix...
• Feb 12th 2011, 03:28 AM
mgrafton
Hi there, thanks for the response.

Then reason I asked was because I am trying to transform the last equation so that it can be expressed in the form
AX^2 + BY^2=1 and was trying to do this using eigen analysis, and I'm not sure if this is possible without matrices?
Do you know of any other ways I could set about doing this?

Thanks
Matt
• Feb 12th 2011, 03:38 AM
Ackbeet
Actually, you can do it, but it's a bit tricky, and it has to be with shifted variables. You'd like to write your equation this way:

$(x+a)^{2}+5(x+a)(y+b)+2(y+b)^{2}=C.$

Multiplying this out and comparing coefficients with your original equation lead to the following three equations:

$2a+5b=1$

$5a+4b=-6$

$C=8+a^{2}+5ab+2b^{2}.$

You can solve this system rather straight-forwardly. Then you can write your quadratic form as the following:

$\begin{bmatrix}x+a&y+b\end{bmatrix}\begin{bmatrix} 1 &5/2\\
5/2 &\sqrt{2}\end{bmatrix}\begin{bmatrix}x+a\\ y+b\end{bmatrix}=C.$

This may or may not be allowed, depending on your definitions, as Moo pointed out. But this computation can still be done.