# Determinant

if a x x x
x b x x = f(x)-xf/(x).
x x c x
x x x d
then f(x)=?

263
Sushovan Halder ·

this is the determinant[4*4] in LHS.Dont confuse.

1708
man111 singh ·

\hspace{-20}$Given$\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} = xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ Now Using property of Determinant, Sub.$\bf{(2)}$Row from$\bf{(1)}$\\\\\\$\bf{\Rightarrow \begin{vmatrix} \bf{a-x} & \bf{x-b} & \bf{0} & \bf{0}\\ \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}\; = xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\$\bf{\Rightarrow (a-x)\begin{vmatrix} \bf{b} & \bf{x} & \bf{x}\\ \bf{x}& \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{d} \end{vmatrix}-(x-b)\begin{vmatrix} \bf{x} & \bf{x} & \bf{x}\\ \bf{x}& \bf{c} & \bf{x}\\ \bf{x}& \bf{x} & \bf{d} \end{vmatrix}=xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ Now after opening the above Determinants......., we get\\\\\\$\bf{\Rightarrow -3x^4+2x^3(a+b+c+d)-x^2(ab+ac+ad+bc+bd+cd)+abcd=xf(x)-f\quad \hspace{-10}'(x)}$\\\\\\ \hspace{-20}$Now Let we assume $\bf{f(x)=px^4+qx^3+rx^2+sx+t}$\\\\\\ Then $\bf{f\hspace{-10}\quad'(x)=4px^3+3px^2+2rx+s}$\\\\\\ Now put into equation... and Camparing...\\\\\\ $\bf{\Rightarrow p=1\;\;,q=-(a+b+c+d)\;\;,r=(ab+ac+ad+bc+bd+cd)}$\\\\\\ and $\bf{s=0\;\;,t=abcd}$\\\\\\ Using Similarity of $\bf{f(x)-xf\hspace{-10}\quad'(x)}$ to the Determinants, we take $\bf{t=0}$\\\\\\ So $\bf{f(x)=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2+abcd.}$

• Manish Shankar good one :)
• man111 singh Thanks Manish Sir.
• Sushovan Halder thanks a lot,But i did'nt understand why u took xf(x)-f'(x).And also give me some tips to solve these type of typical questions.I try but it don't come out from my head.