Question

If x > m, y > n, z > r (x, y, z > 0) such that \[\left| \begin{matrix}
   x & n & r \\
   m & y & r \\
   m & n & z \\
\end{matrix} \right|=0\]. The value of $\dfrac{x}{x-m}+\dfrac{y}{y-n}+\dfrac{z}{z-r}$ is?
(a) 1
(b) -1
(c) 2
(d) -2

Answer 1

Hint: Consider the given determinant and perform elementary row operations \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\]. Now, expand the determinant along the first row and form an expression with the given variables. Divide both the sides with $\left( x-m \right)\left( y-n \right)\left( z-r \right)$ and form a simplified relation. Further, write the terms $\dfrac{x}{x-m}=\left( 1+\dfrac{m}{x-m} \right)$ and$\dfrac{y}{y-n}=\left( 1+\dfrac{n}{y-n} \right)$, use the above obtained relation and substitute its value to get the answer.

Complete step by step solution:
Here we have been provided with the determinant \[\left| \begin{matrix}
   x & n & r \\
   m & y & r \\
   m & n & z \\
\end{matrix} \right|=0\] and we are asked to find the value of the expression $\dfrac{x}{x-m}+\dfrac{y}{y-n}+\dfrac{z}{z-r}$ with the given conditions that x > m, y > n, z > r (x, y, z > 0). Let us assume the expression as E, so we have,
$\Rightarrow E=\dfrac{x}{x-m}+\dfrac{y}{y-n}+\dfrac{z}{z-r}$
Let us simplify the given determinant, so performing the elementary row operations \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] we get,
\[\Rightarrow \left| \begin{matrix}
   x-m & n-y & 0 \\
   0 & y-n & r-z \\
   m & n & z \\
\end{matrix} \right|=0\]
On expanding the determinant along the first row we get,
\[\begin{align}
  & \Rightarrow \left( x-m \right)\left[ \left( y-n \right)z-n\left( r-z \right) \right]-\left( n-y \right)\left[ 0\left( z \right)-m\left( r-z \right) \right]+0\left[ 0\left( n \right)-m\left( r-z \right) \right]=0 \\
 & \Rightarrow z\left( x-m \right)\left( y-n \right)-n\left( x-m \right)\left( r-z \right)+m\left( n-y \right)\left( r-z \right)=0 \\
 & \Rightarrow z\left( x-m \right)\left( y-n \right)+n\left( x-m \right)\left( z-r \right)+m\left( y-n \right)\left( z-r \right)=0 \\
\end{align}\]
Dividing both the sides with $\left( x-m \right)\left( y-n \right)\left( z-r \right)$ we get,
\[\begin{align}
  & \Rightarrow \dfrac{z\left( x-m \right)\left( y-n \right)+n\left( x-m \right)\left( z-r \right)+m\left( y-n \right)\left( z-r \right)}{\left( x-m \right)\left( y-n \right)\left( z-r \right)}=0 \\
 & \Rightarrow \dfrac{z}{z-r}+\dfrac{n}{y-n}+\dfrac{m}{x-m}=0........\left( i \right) \\
\end{align}\]
Now, let us simplify the expression E, so we can write the terms $\dfrac{x}{x-m}=\left( 1+\dfrac{m}{x-m} \right)$ and$\dfrac{y}{y-n}=\left( 1+\dfrac{n}{y-n} \right)$, therefore we get,
$\begin{align}
  & \Rightarrow E=1+\dfrac{m}{x-m}+1+\dfrac{n}{y-n}+\dfrac{z}{z-r} \\
 & \Rightarrow E=2+\left( \dfrac{m}{x-m}+\dfrac{n}{y-n}+\dfrac{z}{z-r} \right) \\
\end{align}$
Substituting the value of equation (i) in the above expression we get,
$\begin{align}
  & \Rightarrow E=2+0 \\
 & \therefore E=2 \\
\end{align}$

So, the correct answer is “Option c”.

Note: Note that we haven’t directly expanded the determinant but performed some operations first because we can see that in the expression we required terms like $\left( x-m \right)$, $\left( y-n \right)$ and $\left( z-r \right)$. If we would have directly expanded the determinant then our calculations would have become difficult. If options are provided and time is less then you can assign some particular values to the variables x, y, z, m, n and r according to the conditions given. It will reduce the calculation.