- Home
- Standard 12
- Mathematics
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
$2$
$3$
$0$
$1$
Solution
$\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3$
$\left|\begin{array}{ccc}\alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma\end{array}\right|=0$
$ (\alpha-\mathrm{a})(\gamma(\beta-\mathrm{b})-\mathrm{b}(\mathrm{c}-\gamma))-(\mathrm{b}-\beta)(-\mathrm{a}(\mathrm{c}-\gamma))=0 $
$ \gamma(\alpha-\mathrm{a})(\beta-\mathrm{b})-\mathrm{b}(\alpha-\mathrm{a})(\mathrm{c}-\gamma)+\mathrm{a}(\mathrm{b}-\beta)(\mathrm{c}-\gamma) $
$ \frac{\gamma}{\gamma-\mathrm{c}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\mathrm{a}}{\alpha-\mathrm{a}}=0$
