Question

$\left|\begin{array}{ccc}1+x& 1& 1\\ 1& 1+y& 1\\ 1& 1& 1+z\end{array}\right|=$

• A. $xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
• B. $xyz$
• C. $1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
• D. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

Answer 1

The correct option is A $xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

$\Delta =xyz\left|\begin{array}{ccc}1+\frac{1}{x}& \frac{1}{x}& \frac{1}{x}\\ \frac{1}{y}& 1+\frac{1}{y}& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}& 1+\frac{1}{z}\end{array}\right|=xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{y}& 1+\frac{1}{y}& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}& 1+\frac{1}{z}\end{array}\right|,by{R}_1\to R_1+R_2+R_3=xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{y}& 1& 0\\ \frac{1}{z}& 0& 1\end{array}\right|,by\begin{array}{c}{C}_2\to C_2-C_1\\ C_3\to C_3-C_1\end{array}=xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=xyz(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$.
Trick : Put x=1, y=2 and z=3, then $\left|\begin{array}{ccc}2& 1& 1\\ 1& 3& 1\\ 1& 1& 4\end{array}\right|=2\left(11\right)-1\left(3\right)+1(1-3)=17$
Option (a) gives, $1\times 2\times 3(1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3})=17$.