Question

1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} \right| =$$

• A. $xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$
• 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

Answer: (A)

$xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$

Answer 2

$\{\therefore C_{1} \equiv C_{2}\}$

= $xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$ $\left| \begin{matrix} 1 & 1 & 1 \\ \frac{1}{y} & 1 + \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & 1 + \frac{1}{z} \end{matrix} \right|$,

by $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$

=$x y z \left( 1 + \frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } \right)$ $\left| \begin{matrix} 1 & 0 & 0 \\ 1/y & 1 & 0 \\ 1/z & 0 & 1 \end{matrix} \right|$, by $C_{2} \rightarrow C_{2} - C_{1}$ $$C_{3} \rightarrow C_{3} - C_{1}$$

= $xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$ $\left| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right| = xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)$.

Trick: Put $x = 1,y = 2$ and $z = 3$, then

$$\left| \begin{matrix} 2 & 1 & 1 \ 1 & 3 & 1 \ 1 & 1 & 4 \end{matrix} \right| = 2(11) - 1(3) + 1(1 - 3) = 17$$

Option (1) gives, $1 \times 2 \times 3\left( 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \right) = 17$.