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Question: If \(a\ne p\) , \(b\ne q\) and \(c\ne r\),and \(\left| \begin{matrix} p & b & c \\\ a & q ...

If a≠pa\ne p , b≠qb\ne q and c≠rc\ne r,and ∣pbc aqc abr ∣=0\left| \begin{matrix} p & b & c \\\ a & q & c \\\ a & b & r \\\ \end{matrix} \right|=\text{0}, then find the value of pp−a+bq−b+rr−c\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}

Explanation

Solution

To solve this question, firstly we will use row operations R1→R1−R2{{R}_{1}}\to {{R}_{1}}-{{R}_{2}} and R2→R2−R3{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}. After that, we will expand the determinant along row R1{{R}_{1}}. After that, we will divide the whole equation by (p-a)(q–b)(c–r), and by doing some simplification, we will find the value of pp−a+bq−b+rr−c\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}.

Complete step by step answer:
Now, before we start solving the questions, let us see how we calculate determinant and what are its various properties
Now , if we want to calculate the determinant of matrix A of order 3×33\times 3, then determinant of matrix A of 3×33\times 3 is evaluated as,
∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a32a23)−a21(a12a33−a32a13)+a31(a23a12−a22a13)\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})
Some of the properties of determinant are as follows,
( a ) Determinant evaluated across any row or column is the same.
( b ) If an element of a row or a column are zeros, then the value of the determinant is equal to zero.
( c ) If rows and columns are interchanged then the value of the determinant remains the same.
( d ) Determinant of an identity matrix is 1.
Now, let us move to question now, it is asked to find the value of pp−a+bq−b+rr−c\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c} and in question it is given that a≠pa\ne p , b≠qb\ne q and c≠rc\ne r,and ∣pbc aqc abr ∣=0\left| \begin{matrix} p & b & c \\\ a & q & c \\\ a & b & r \\\ \end{matrix} \right|=\text{0}.
Now, using elementary row operation R1→R1−R2{{R}_{1}}\to {{R}_{1}}-{{R}_{2}} , we get
∣p−ab−qc−c aqc abr ∣=0\left| \begin{matrix} p-a & b-q & c-c \\\ a & q & c \\\ a & b & r \\\ \end{matrix} \right|=\text{0}
using elementary row operation R2→R2−R3{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} , we get
∣p−ab−q0 0q−bc−r abr ∣=0\left| \begin{matrix} p-a & b-q & 0 \\\ 0 & q-b & c-r \\\ a & b & r \\\ \end{matrix} \right|=\text{0}
Now, expanding determinant along R1{{R}_{1}}, we get
(p−a)(r(q−b)−b(c−r))−(b−q)(0−a(c−r))+0=0(p-a)(r(q-b)-b(c-r))-(b-q)(0-a(c-r))+0=0
On simplifying, we get
(p−a)(r(q−b)−b(c−r))+(b−q)a(c−r)=0(p-a)(r(q-b)-b(c-r))+(b-q)a(c-r)=0
⇒r(p−a)(q−b)−b(p−a)(c−r)+a(b−q)(c−r)=0\Rightarrow r(p-a)(q-b)-b(p-a)(c-r)+a(b-q)(c-r)=0
Now, dividing the whole equation by ( p –a )( q – b )( c – r ), we get
r(p−a)(q−b)−b(p−a)(c−r)+a(b−q)(c−r)(p−a)(q−b)(c−r)=0\dfrac{r(p-a)(q-b)-b(p-a)(c-r)+a(b-q)(c-r)}{(p-a)(q-b)(c-r)}=0
On simplifying, we get
r(p−a)(q−b)(p−a)(q−b)(c−r)−b(p−a)(c−r)(p−a)(q−b)(c−r)+a(b−q)(c−r)(p−a)(q−b)(c−r)=0\dfrac{r(p-a)(q-b)}{(p-a)(q-b)(c-r)}-\dfrac{b(p-a)(c-r)}{(p-a)(q-b)(c-r)}+\dfrac{a(b-q)(c-r)}{(p-a)(q-b)(c-r)}=0
Or, ⇒r(c−r)−b(q−b)−a(p−a)=0\Rightarrow \dfrac{r}{(c-r)}-\dfrac{b}{(q-b)}-\dfrac{a}{(p-a)}=0
Re – writing above equation, we get
−r(r−c)−b(q−b)−a(p−a)=0-\dfrac{r}{(r-c)}-\dfrac{b}{(q-b)}-\dfrac{a}{(p-a)}=0
Multiplying both sides by -1, we get
r(r−c)+b(q−b)+a(p−a)=0\dfrac{r}{(r-c)}+\dfrac{b}{(q-b)}+\dfrac{a}{(p-a)}=0
Now, adding and subtracting q in numerator of term b(q−b)\dfrac{b}{(q-b)} and adding and subtracting p in numerator of a(p−a)\dfrac{a}{(p-a)}, we get
r(r−c)+b+q−q(q−b)+a+p−p(p−a)=0\dfrac{r}{(r-c)}+\dfrac{b+q-q}{(q-b)}+\dfrac{a+p-p}{(p-a)}=0
⇒r(r−c)+b−q(q−b)+q(q−b)+a−p(p−a)+p(p−a)=0\Rightarrow \dfrac{r}{(r-c)}+\dfrac{b-q}{(q-b)}+\dfrac{q}{(q-b)}+\dfrac{a-p}{(p-a)}+\dfrac{p}{(p-a)}=0
On simplification, we get
r(r−c)−1+q(q−b)−1+p(p−a)=0\dfrac{r}{(r-c)}-1+\dfrac{q}{(q-b)}-1+\dfrac{p}{(p-a)}=0
On solving, we get
r(r−c)+q(q−b)+p(p−a)=2\dfrac{r}{(r-c)}+\dfrac{q}{(q-b)}+\dfrac{p}{(p-a)}=2

Note: It is very important to know how to solve determinant using it’s properties so knowledge of properties of determinant should be a priority. Always remember that ∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a32a23)−a21(a12a33−a32a13)+a31(a23a12−a22a13)\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}}). In determinant we can use both column and row elementary transformation. Calculation should be done carefully while solving determinant problems.