Question

Prove that $\left|\begin{array}{cccc}1+a& 1& 1& 1\\ 1& 1+b& 1& 1\\ 1& 1& 1+c& 1\\ 1& 1& 1& 1+d\end{array}\right|=abcd(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$. Hence, find the value of the determinant if $a,b,c,d$ are the roots of the equation $p{x}^4+q{x}^3+r{x}^2+sx+t=0$.

Answer 1

$\left|\begin{array}{cccc}1+a& 1& 1& 1\\ 1& 1+b& 1& 1\\ 1& 1& 1+c& 1\\ 1& 1& 1& 1+d\end{array}\right|=\left|\begin{array}{cccc}a& -b& 0& 0\\ 1& 1+b& 1& 1\\ 1& 1& 1+c& 1\\ 1& 1& 1& 1+d\end{array}\right|=a\left|\begin{array}{ccc}1+b& 1& 1\\ 1& 1+c& 1\\ 1& 1& 1+d\end{array}\right|+b\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+c& 1\\ 1& 1& 1+d\end{array}\right|+0+0=a\left|\begin{array}{ccc}b& -c& 0\\ 1& 1+c& 1\\ 1& 1& 1+d\end{array}\right|+b\left|\begin{array}{ccc}0& -c& 0\\ 1& 1+c& 1\\ 1& 1& 1+d\end{array}\right|=a\left\{b\right((1+c)\left(1d\right)-1)+c(1+d-1)+b(c\left)\right(d)=abcd(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$

Now $a,b,c,d$ are the roots of the given equation

$=abcd(1+\frac{bcd+cda+dba+abc}{abcd})=abcd+bcd+cda+dba+abc=\frac{t}{p}-\frac{s}{p}$