Question:
The value of
\[
\begin{vmatrix}
1 & a & b + c \\
1 & b & c + a \\
1 & c & a + b
\end{vmatrix}
\]
is:
The value of
\[
\begin{vmatrix}
1 & a & b + c \\
1 & b & c + a \\
1 & c & a + b
\end{vmatrix}
\]
is:
Show Hint
Summing variables often reveals identical rows or columns in cyclic determinants.
Updated On: Apr 8, 2026
- $a+b+c$
- $(a-b)(b-c)(c-a)$
- 0
- 1
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Use determinant properties to simplify the calculation.
Step 2: Analysis
Apply the column operation $C_{3} \rightarrow C_{3} + C_{2}$. The third column becomes $(a+b+c, a+b+c, a+b+c)$. Taking $(a+b+c)$ common from $C_{3}$, the column becomes $(1, 1, 1)$.
Step 3: Conclusion
Now $C_{1}$ and $C_{3}$ are identical. A determinant with two identical columns is zero.
Final Answer: (C)
Use determinant properties to simplify the calculation.
Step 2: Analysis
Apply the column operation $C_{3} \rightarrow C_{3} + C_{2}$. The third column becomes $(a+b+c, a+b+c, a+b+c)$. Taking $(a+b+c)$ common from $C_{3}$, the column becomes $(1, 1, 1)$.
Step 3: Conclusion
Now $C_{1}$ and $C_{3}$ are identical. A determinant with two identical columns is zero.
Final Answer: (C)
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