# How to solve a polynomial homogeneous matrix

I am having trouble solving the solutions for this problem. It states:

The polynomials, $f_1(x)=x-2$, $f_2(x)=x^2-5x+4$, $f_3(x)=3x^2-4x$, $f_4(x)=x^2-1$ are linearly dependent since $f_1(x) + f_2(x) - f_3(x) + 2f_4(x) = 0$.

But how did they get coefficents?

My attempt:

$x^2(b+3c+d) +x(a-5b-4c)+(-2a+4b-1d) = 0$ but when I solve it, it doesn't give me the correct coefficents of the polynomial.

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## 1 Answer

You solve $b+3c+d=0, a-5b-4c=0, -2a+4b-d=0$. There are 3 equations in 4 unknowns so there are infinitely many solutions, $c$ is a free variable and $a=b=-c, d=-2c$ If you let $c=-1$ then you get the solution $a=1, b=1, d=2$ as desired.

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I understand that there is a free variable but how did you get that a=b=-c,d=-2c and also how did you get that c is the free variable, I got d to be the free variable? –  Q.matin Sep 22 '12 at 22:43
@queensmatin It is okay that $d$ is the free variable. Then a, b, c are expresed as multiples of $d$. If you put $d=2$ then you will get the solution $a=1, b=1, c=-1$ –  i. m. soloveichik Sep 22 '12 at 22:46
Oh okay, now I understand, thank you very much! –  Q.matin Sep 22 '12 at 23:00