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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$
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Answer

$\text{L.H.S}=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$
$=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\0&\text{q}-\text{b}&\text{c}-\text{r}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_3$]
$=\text{p}[\text{r}(\text{q}-\text{b})-\text{b}(\text{c}-\text{r})]+\text{a}[\text{b}(\text{c}-\text{r})-\text{c}(\text{q}-\text{b})]$
$=\text{pr}(\text{q}-\text{b})+\text{pb}(\text{r}-\text{c})-\text{ab}(\text{r}-\text{c})-\text{ac}(\text{q}-\text{b})$
$=(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})$
Since, $\triangle=0$
$\therefore(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})=0$
$\Rightarrow\frac{\text{pr}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{pr}-\text{ar}+\text{ar}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}(\text{p}-\text{a})+\text{a}(\text{r}-\text{c})}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}}{\text{r}-\text{c}}+\frac{\text{a}}{\text{p}-\text{a}}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}\\=\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}-\frac{\text{a}}{\text{p}-\text{a}}-\frac{\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=\frac{\text{p}-\text{a}}{\text{p}-\text{a}}+\frac{\text{q}-\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=2$

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