Question:
If \(
\begin{vmatrix}
x+y+2z & x & y
z & y+z+2x & y
z & x & z+x+2y
\end{vmatrix}
= k(x+y+z)^3 \), then the value of \( k \) is
If \(
\begin{vmatrix}
x+y+2z & x & y
z & y+z+2x & y
z & x & z+x+2y
\end{vmatrix}
= k(x+y+z)^3 \), then the value of \( k \) is
Show Hint
For symmetric determinants, try adding rows/columns to extract common factors like \( (x+y+z) \).
Updated On: Apr 23, 2026
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The Correct Option is B
Solution and Explanation
Concept:
Use row/column operations and symmetry in determinants.
Step 1: Apply row operations. Add all rows: \[ R_1 + R_2 + R_3 \] Each column becomes: \[ (x+y+2z) + z + z = x + y + 4z \] Similarly simplifying leads to factorization: \[ \text{Determinant} = 2(x+y+z)^3 \]
Step 2: Compare with given form: \[ k(x+y+z)^3 \] Final Answer: \[ k = 2 \]
Step 1: Apply row operations. Add all rows: \[ R_1 + R_2 + R_3 \] Each column becomes: \[ (x+y+2z) + z + z = x + y + 4z \] Similarly simplifying leads to factorization: \[ \text{Determinant} = 2(x+y+z)^3 \]
Step 2: Compare with given form: \[ k(x+y+z)^3 \] Final Answer: \[ k = 2 \]
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