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Previous Year Question (PYQs)

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




Solution

Solution: Given $\left|\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}\right| = 0,$ the rows are linearly dependent. Using the determinant identity, we get $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 1.$

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