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If $a\ne p,b\ne q$ and $c\ne r$ and also, $\left| \begin{matrix}
   p & b & c \\
   a & q & c \\
   a & b & r \\
\end{matrix} \right|=0$ then, find the value of :
$\dfrac{p}{p-a}+\dfrac{q}{q-b}+\dfrac{r}{r-c}=?$


Last updated date: 20th Mar 2023
Total views: 306.3k
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Answer
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306.3k+ views
Hint: To solve this question, use the properties of determinants. In a determinant, if we subtract a row from another row, the value of the determinant remains the same. Use this property of determinants to solve this question.

Before proceeding with the question, we must know all the properties that will be required to solve this question.
In determinants, the determinant of any matrix $A=\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)$ is evaluated by the formula,
\[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-fh \right)-b\left( di-fg \right)+c\left( dh-eg \right)...................\left( 1 \right)\]
Also, there is a property in determinants which states that subtracting any row of a determinant from another row of the same determinant doesn’t change the value of the determinant. $...........\left( 2 \right)$
In the question, we are given that $\left| \begin{matrix}
   p & b & c \\
   a & q & c \\
   a & b & r \\
\end{matrix} \right|=0$ and we have to find the value of $\dfrac{p}{p-a}+\dfrac{q}{q-b}+\dfrac{r}{r-c}$.
$\left| \begin{matrix}
   p & b & c \\
   a & q & c \\
   a & b & r \\
\end{matrix} \right|=0$
Using property $\left( 2 \right)$, subtracting row 3 from row 2, we get,
$\begin{align}
  & \left| \begin{matrix}
   p & b & c \\
   a-a & q-b & c-r \\
   a & b & r \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   p & b & c \\
   0 & q-b & c-r \\
   a & b & r \\
\end{matrix} \right|=0 \\
\end{align}$
Using property $\left( 2 \right)$, subtracting row 3 from row 1, we get,
$\begin{align}
  & \left| \begin{matrix}
   p-a & b-b & c-r \\
   0 & q-b & c-r \\
   a & b & r \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   p-a & 0 & c-r \\
   0 & q-b & c-r \\
   a & b & r \\
\end{matrix} \right|=0 \\
\end{align}$
Using formula $\left( 1 \right)$ in the above equation, we get,
$\begin{align}
  & \left( p-a \right)\left[ r\left( q-b \right)-b\left( c-r \right) \right]-0\left[ 0.b-a\left( q-b \right) \right]+\left( c-r \right)\left[ 0.b-a\left( q-b \right) \right]=0 \\
 & \Rightarrow \left( p-a \right)\left[ r\left( q-b \right)-b\left( c-r \right) \right]-a\left( c-r \right)\left( q-b \right)=0 \\
 & \Rightarrow r\left( p-a \right)\left( q-b \right)+b\left( p-a \right)\left( r-c \right)+a\left( r-c \right)\left( q-b \right)=0 \\
\end{align}$
Dividing the above equation by $\left( p-a \right)\left( q-b \right)\left( r-c \right)$, we get,
\[\begin{align}
  & \dfrac{r\left( p-a \right)\left( q-b \right)}{\left( p-a \right)\left( q-b \right)\left( r-c \right)}+\dfrac{b\left( p-a \right)\left( r-c \right)}{\left( p-a \right)\left( q-b \right)\left( r-c \right)}+\dfrac{a\left( r-c \right)\left( q-b \right)}{\left( p-a \right)\left( q-b \right)\left( r-c \right)}=0 \\
 & \Rightarrow \dfrac{r}{\left( r-c \right)}+\dfrac{b}{\left( q-b \right)}+\dfrac{a}{\left( p-a \right)}=0 \\
\end{align}\]
The above equation can also be written as,
\[\begin{align}
  & \dfrac{r}{\left( r-c \right)}+\dfrac{b-q+q}{\left( q-b \right)}+\dfrac{a-p+p}{\left( p-a \right)}=0 \\
 & \Rightarrow \dfrac{r}{\left( r-c \right)}+\dfrac{b-q}{\left( q-b \right)}+\dfrac{q}{\left( q-b \right)}+\dfrac{a-p}{\left( p-a \right)}+\dfrac{p}{\left( p-a \right)}=0 \\
 & \Rightarrow \dfrac{p}{\left( p-a \right)}+\dfrac{q}{\left( q-b \right)}+\dfrac{r}{\left( r-c \right)}+\left( -1 \right)+\left( -1 \right)=0 \\
 & \Rightarrow \dfrac{p}{\left( p-a \right)}+\dfrac{q}{\left( q-b \right)}+\dfrac{r}{\left( r-c \right)}=2 \\
\end{align}\]
Hence, the answer is 2.

Note: One can also do this question without using the property $\left( 2 \right)$ i.e. subtracting one row from another row of the determinant does not affect the value of the determinant. If one evaluates the determinant by using this property of determinants, he/she has to factorise the final expression which he/she will get after evaluating the determinant without using that property. This process of factorisation will take a lot of time and thus, this method will take a much longer time to solve this question.