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

Last updated date: 14th Jun 2024
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Hint: To solve this question, firstly we will use row operations ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ and ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$. After that, we will expand the determinant along row ${{R}_{1}}$. 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 $\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}$.

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 $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
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 $\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}$ and in question it is given that $a\ne p$ , $b\ne q$ and $c\ne r$,and $\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \\ \end{matrix} \right|=\text{0}$.
Now, using elementary row operation ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ , we get
$\left| \begin{matrix} p-a & b-q & c-c \\ a & q & c \\ a & b & r \\ \end{matrix} \right|=\text{0}$
using elementary row operation ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ , we get
$\left| \begin{matrix} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \\ \end{matrix} \right|=\text{0}$
Now, expanding determinant along ${{R}_{1}}$, we get
$(p-a)(r(q-b)-b(c-r))-(b-q)(0-a(c-r))+0=0$
On simplifying, we get
$(p-a)(r(q-b)-b(c-r))+(b-q)a(c-r)=0$
$\Rightarrow r(p-a)(q-b)-b(p-a)(c-r)+a(b-q)(c-r)=0$
Now, dividing the whole equation by ( p –a )( q – b )( c – r ), we get
$\dfrac{r(p-a)(q-b)-b(p-a)(c-r)+a(b-q)(c-r)}{(p-a)(q-b)(c-r)}=0$
On simplifying, we get
$\dfrac{r(p-a)(q-b)}{(p-a)(q-b)(c-r)}-\dfrac{b(p-a)(c-r)}{(p-a)(q-b)(c-r)}+\dfrac{a(b-q)(c-r)}{(p-a)(q-b)(c-r)}=0$
Or, $\Rightarrow \dfrac{r}{(c-r)}-\dfrac{b}{(q-b)}-\dfrac{a}{(p-a)}=0$
Re – writing above equation, we get
$-\dfrac{r}{(r-c)}-\dfrac{b}{(q-b)}-\dfrac{a}{(p-a)}=0$
Multiplying both sides by -1, we get
$\dfrac{r}{(r-c)}+\dfrac{b}{(q-b)}+\dfrac{a}{(p-a)}=0$
Now, adding and subtracting q in numerator of term $\dfrac{b}{(q-b)}$ and adding and subtracting p in numerator of $\dfrac{a}{(p-a)}$, we get
$\dfrac{r}{(r-c)}+\dfrac{b+q-q}{(q-b)}+\dfrac{a+p-p}{(p-a)}=0$
$\Rightarrow \dfrac{r}{(r-c)}+\dfrac{b-q}{(q-b)}+\dfrac{q}{(q-b)}+\dfrac{a-p}{(p-a)}+\dfrac{p}{(p-a)}=0$
On simplification, we get
$\dfrac{r}{(r-c)}-1+\dfrac{q}{(q-b)}-1+\dfrac{p}{(p-a)}=0$
On solving, we get
$\dfrac{r}{(r-c)}+\dfrac{q}{(q-b)}+\dfrac{p}{(p-a)}=2$

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 $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$. In determinant we can use both column and row elementary transformation. Calculation should be done carefully while solving determinant problems.