Answer
Verified
423.9k+ views
Hint-In this question first analyse the question, then perform certain steps to simplify the determinant so that solving it is an easy task. Then by simplifying the determinant step by step we can reach the desired answer. Also study carefully the conditions given as they will be useful.
Complete step-by-step answer:
Considering the given condition in the question we can apply some operation on the determinant
Subtract third row from the first and save answer in first i.e.${R_1} \Rightarrow {R_1} - {R_3}$ and also apply ${R_2} \Rightarrow {R_2} - {R_3}{\text{ , where }}{R_1},{R_2}{\text{ and }}{R_3}{\text{ are first, second and third rows}}{\text{.}}$
Therefore, we get$\left| {\begin{array}{*{20}{c}}
{{\text{p - a}}}&{{\text{b - q}}}&{{\text{c - r}}} \\
0&{{\text{q - b}}}&{{\text{c - r}}} \\
0&{\text{b}}&{\text{r}}
\end{array}} \right| = 0$
As coefficient of ${\text{b - q }}$is $0$ .
Solving it further will generate equation
$\left( {{\text{p - a}}} \right)\left| {\begin{array}{*{20}{c}}
{{\text{q - b}}}&{{\text{c - r}}} \\
{\text{b}}&{\text{r}}
\end{array}} \right| + \left( {{\text{c - r}}} \right)\left| {\begin{array}{*{20}{c}}
0&{{\text{q - b}}} \\
{\text{a}}&{\text{b}}
\end{array}} \right| = 0$
Simplifying the above equation, $\left( {{\text{p - a}}} \right)\left[ {{\text{r}}\left( {{\text{q - b}}} \right) - {\text{b}}\left( {{\text{c - r}}} \right)} \right] + \left( {{\text{c - r}}} \right)\left[ { - {\text{a}}\left( {{\text{q - b}}} \right)} \right] = 0$
From given conditions we can deduce that ${\text{p - a}} \ne {\text{0,q - b}} \ne {\text{0,r - c}} \ne {\text{0}}$
Now, by dividing both sides by $\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]$ we get
$\dfrac{{{\text{r}}\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} -
\dfrac{{{\text{b}}\left( {{\text{p - a}}} \right)\left( {{\text{c - r}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} -
\dfrac{{{\text{a}}\left( {{\text{c - r}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} = 0$
By cancelling the same terms in numerator and denominator we get
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{b}}}{{\left( {{\text{q - b}}}
\right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$
Then we need to add and subtract ${\text{q}}$ from the numerator of term
$\dfrac{{\text{b}}}{{{\text{q - b}}}}$ and add and subtract ${\text{p}}$ from the numerator of term $\dfrac{{\text{a}}}{{{\text{p - a}}}}$
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{{\text{b + q - q}}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{{\text{a + p - p}}}}{{\left( {{\text{p - a}}} \right)}} = 0$
From this we get
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\left( {{\text{b - q}}} \right)}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\left( {{\text{a - p}}} \right)}}{{\left( {{\text{p - a}}} \right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$
This implies that
$
\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} - 1 + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} - 1 + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 0 \\
\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 2 \\
$
Hence the value of the question is 2.
Note- In getting the values of these types of questions we need to know the right steps in solving a determinant and then simplifying it because simplifying a determinant directly can sometimes be confusing and we may not get the desired result. Therefore, for these questions you need to practise solving the determinants.
Complete step-by-step answer:
Considering the given condition in the question we can apply some operation on the determinant
Subtract third row from the first and save answer in first i.e.${R_1} \Rightarrow {R_1} - {R_3}$ and also apply ${R_2} \Rightarrow {R_2} - {R_3}{\text{ , where }}{R_1},{R_2}{\text{ and }}{R_3}{\text{ are first, second and third rows}}{\text{.}}$
Therefore, we get$\left| {\begin{array}{*{20}{c}}
{{\text{p - a}}}&{{\text{b - q}}}&{{\text{c - r}}} \\
0&{{\text{q - b}}}&{{\text{c - r}}} \\
0&{\text{b}}&{\text{r}}
\end{array}} \right| = 0$
As coefficient of ${\text{b - q }}$is $0$ .
Solving it further will generate equation
$\left( {{\text{p - a}}} \right)\left| {\begin{array}{*{20}{c}}
{{\text{q - b}}}&{{\text{c - r}}} \\
{\text{b}}&{\text{r}}
\end{array}} \right| + \left( {{\text{c - r}}} \right)\left| {\begin{array}{*{20}{c}}
0&{{\text{q - b}}} \\
{\text{a}}&{\text{b}}
\end{array}} \right| = 0$
Simplifying the above equation, $\left( {{\text{p - a}}} \right)\left[ {{\text{r}}\left( {{\text{q - b}}} \right) - {\text{b}}\left( {{\text{c - r}}} \right)} \right] + \left( {{\text{c - r}}} \right)\left[ { - {\text{a}}\left( {{\text{q - b}}} \right)} \right] = 0$
From given conditions we can deduce that ${\text{p - a}} \ne {\text{0,q - b}} \ne {\text{0,r - c}} \ne {\text{0}}$
Now, by dividing both sides by $\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]$ we get
$\dfrac{{{\text{r}}\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} -
\dfrac{{{\text{b}}\left( {{\text{p - a}}} \right)\left( {{\text{c - r}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} -
\dfrac{{{\text{a}}\left( {{\text{c - r}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} = 0$
By cancelling the same terms in numerator and denominator we get
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{b}}}{{\left( {{\text{q - b}}}
\right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$
Then we need to add and subtract ${\text{q}}$ from the numerator of term
$\dfrac{{\text{b}}}{{{\text{q - b}}}}$ and add and subtract ${\text{p}}$ from the numerator of term $\dfrac{{\text{a}}}{{{\text{p - a}}}}$
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{{\text{b + q - q}}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{{\text{a + p - p}}}}{{\left( {{\text{p - a}}} \right)}} = 0$
From this we get
$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\left( {{\text{b - q}}} \right)}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\left( {{\text{a - p}}} \right)}}{{\left( {{\text{p - a}}} \right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$
This implies that
$
\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} - 1 + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} - 1 + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 0 \\
\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 2 \\
$
Hence the value of the question is 2.
Note- In getting the values of these types of questions we need to know the right steps in solving a determinant and then simplifying it because simplifying a determinant directly can sometimes be confusing and we may not get the desired result. Therefore, for these questions you need to practise solving the determinants.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Change the following sentences into negative and interrogative class 10 english CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What is a collective noun for bees class 10 english CBSE