Matematické Fórum

stuart clark · 21. 06. 2014 06:00

If $\bf{\begin{vmatrix} \bf{a} & \bf{x} & \bf{x} & \bf{x}\\ \bf{x} & \bf{b} & \bf{x} & \bf{x}\\ \bf{x} & \bf{x} & \bf{c} & \bf{x}\\ \bf{x} & \bf{x} & \bf{x} & \bf{d} \end{vmatrix} = f(x)-x\cdot f^{'}(x)}.$ Then value of $\bf{f(x)},$ is

sugyman

Substracting second row from the first and using Laplace expansion we get
$\begin{vmatrix} \bf{a} & \bf{x} & \bf{x} & \bf{x}\\ \bf{x} & \bf{b} & \bf{x} & \bf{x}\\ \bf{x} & \bf{x} & \bf{c} & \bf{x}\\ \bf{x} & \bf{x} & \bf{x} & \bf{d} \end{vmatrix} = (a-x)(-1)^{1+1} \begin{vmatrix} \bf{b} & \bf{x} & \bf{x}\\ \bf{x} & \bf{c} & \bf{x}\\ \bf{x} & \bf{x} & \bf{d} \end{vmatrix} + (x-b)(-1)^{2+1} \begin{vmatrix} \bf{x} & \bf{x} & \bf{x}\\ \bf{x} & \bf{c} & \bf{x}\\ \bf{x} & \bf{x} & \bf{d} \end{vmatrix} =$
and after some uncomfortable calculations: $= -3x^4+x^3(2a+2b+2c+2d)-x^2(ac+ab+ad+bc+bd+cd)+abcd$
If we now assume $f(x)=px^4+qx^3+rx^2+sx+t$
$xf^{'}(x)=4px^4+3qx^3+2rx^2+sx$
$f(x)-xf^{'}(x)=-3px^3-2qx^3-rx^2+t$
Seeing the similarity with the value of determinant let $p=1, q=-a-b-c-d, r=ac+ab+ad+bc+cd+bd, s=0, t=abcd$ in order to make the starting equation true. Naturally because of our assumptions we should look if our discovered function satisfy your equation, but it is pretty obvious.
To sum up: $f(x)=x^4-x^3(a+b+c+d)+x^2(ac+av+ad+bc+cd+bd)+abcd$ .

stuart clark · 21. 06. 2014 19:54

Thanks ↑ sugyman:

Matematické Fórum

#1 21. 06. 2014 06:00

value of f(x) in Determinant

#2 21. 06. 2014 15:54 — Editoval sugyman (21. 06. 2014 15:55)

Re: value of f(x) in Determinant

#3 21. 06. 2014 19:54

Re: value of f(x) in Determinant

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