Question:
If \(\left|\begin{matrix}p& q-b& r-cp-a& q& r-cp-a& q-b& r\end{matrix}\right|=0\), then the value of \(\dfrac{p}{a}+\dfrac{q}{b}+\dfrac{r}{c}=\)
If \(\left|\begin{matrix}p& q-b& r-cp-a& q& r-cp-a& q-b& r\end{matrix}\right|=0\), then the value of \(\dfrac{p}{a}+\dfrac{q}{b}+\dfrac{r}{c}=\)
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In determinant problems, use row and column operations to reduce the determinant into a simple algebraic condition.
Updated On: Jul 3, 2026
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The Correct Option is B
Solution and Explanation
Concept:
We are given a determinant condition. Such determinants are simplified by applying row or column operations and then comparing the resulting condition.
Step 1: Write the determinant condition. \[ \left|\begin{matrix} p& q-b& r-c\\ p-a& q& r-c\\ p-a& q-b& r \end{matrix}\right|=0 \] The determinant being zero means the rows or columns are linearly dependent.
Step 2: Use the known result for this determinant form. For this standard determinant form, after simplification, the condition becomes \[ \frac{p}{a}+\frac{q}{b}+\frac{r}{c}=2 \]
Step 3: Final answer. \[ \boxed{2} \]
We are given a determinant condition. Such determinants are simplified by applying row or column operations and then comparing the resulting condition.
Step 1: Write the determinant condition. \[ \left|\begin{matrix} p& q-b& r-c\\ p-a& q& r-c\\ p-a& q-b& r \end{matrix}\right|=0 \] The determinant being zero means the rows or columns are linearly dependent.
Step 2: Use the known result for this determinant form. For this standard determinant form, after simplification, the condition becomes \[ \frac{p}{a}+\frac{q}{b}+\frac{r}{c}=2 \]
Step 3: Final answer. \[ \boxed{2} \]
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