Question

Find the value of the determinant \( \begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix} \).

Hint:

When evaluating determinants, always look for an opportunity to apply the operation \(R_1 \to R_1+R_2+R_3\) or \(C_1 \to C_1+C_2+C_3\). If this creates a common factor in a row or column, it will simplify the determinant significantly.

Answer 1

Step 1: Understanding the Concept:
The problem is to evaluate a 3x3 determinant. The key is to look for symmetries and apply determinant properties to simplify before expanding. A common technique is to perform row or column operations to create zeros or factor out common terms.
Step 2: Key Formula or Approach:
1. Use the property that adding a multiple of one column (or row) to another does not change the determinant. Specifically, apply \(C_1 \to C_1 + C_2 + C_3\).
2. Factor out common terms from the first column.
3. Apply row operations to create zeros in the first column, simplifying expansion.
4. Expand the determinant along the first column.
Step 3: Detailed Explanation:
Let \(\Delta\) be the determinant: \[ \Delta = \begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix} \] Apply \(C_1 \to C_1 + C_2 + C_3\): \[ \Delta = \begin{vmatrix} 2x+2y+2z & x & y \\ 2x+2y+2z & y+z+2x & y \\ 2x+2y+2z & x & z+x+2y \end{vmatrix} \] Factor out \(2(x+y+z)\) from the first column: \[ \Delta = 2(x+y+z) \begin{vmatrix} 1 & x & y \\ 1 & y+z+2x & y \\ 1 & x & z+x+2y \end{vmatrix} \] Apply row operations: \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\): \[ \Delta = 2(x+y+z) \begin{vmatrix} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{vmatrix} \] Expand along the first column: \[ \Delta = 2(x+y+z) \cdot \begin{vmatrix} x+y+z & 0 \\ 0 & x+y+z \end{vmatrix} = 2(x+y+z)(x+y+z)^2 \] \[ \Delta = 2(x+y+z)^3 \] Step 4: Final Answer:
The value of the determinant is \(\boxed{2(x+y+z)^3}\).