Q.The value of the determinant $\begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix}$ is
(A) $9x^2(x + y)$
(B) $9y^2(x + y)$
(C) $3y^2(x + y)$
(D) $7x^2(x + y)$

Concept: Determinant Properties Cyclic — Use row operations to exploit the cyclic pattern and simplify.

Step 1: Let the determinant be $\Delta$. Apply $R_1 \to R_1 + R_2 + R_3$:

Each entry in $R_1$ becomes $(x)+(x+2y)+(x+y) = 3x+3y = 3(x+y)$. So

Step 2: Factor $3(x+y)$ from $R_1$:

Step 3: Use $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$\Delta = 3(x+y) \begin{vmatrix} 1 & 0 & 0 \\ x+2y & -2y & -y \\ x+y & y & -y \end{vmatrix}.$ …

The key insight here is that the matrix has a cyclic structure — each row is a shift of the previous one. When you see a pattern like $x$, $x+y$, $x+2y$ cycling around, the determinant often simplifies dramatically by adding all rows together or using column operations. This isn't a brute-force expansion problem; it's about spotting the invariance.

Let’s work through it step by step.

1. Write the determinant clearly

2. Add all rows to the first row — a classic trick for cyclic matrices. $R_1 \to R_1 + R_2 + R_3$ The new first row becomes:

So the determinant becomes:

3. Factor out the common factor from the first row $3(x+y)$ is common to all three entries in row 1:

4. Use column operations to create zeros $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$: The first row becomes $(1, 0, 0)$. The second row:
   • Column 2: $x - (x+2y) = -2y$
   • Column 3: $(x+y) - (x+2y) = -y$ The third row:
   • Column 2: $(x+2y) - (x+y) = y$
   • Column 3: $x - (x+y) = -y$ So we have:

5. Expand along the first row Only the $1$ in the top-left contributes (since the other entries in row 1 are zero). The determinant reduces to the $2 \times 2$ minor: …

Method: Reducing a Cyclic-Pattern Determinant by Summing All Rows

When every row of a determinant is the same set of terms in a different order (a "cyclic" arrangement, e.g. $x,\,x+y,\,x+2y$ shifting position each row), summing all the rows into one is far faster than direct expansion, because the shifted terms collapse into an identical total in every column.

Steps

Step 1: Add every row into one row to expose a common factor

Apply $R_1\to R_1+R_2+R_3$ (this doesn't change the determinant's value). Because the same three terms appear once in each column across the three rows, every entry in the new row 1 becomes the same sum — here $3x+3y=3(x+y)$ in all three columns. Recognising that a cyclic pattern always produces this "same total in every column" behaviour is the key insight of the method.

Step 2: Factor the common value out of that row

Pull the shared factor (here $3(x+y)$) out in front of the determinant, leaving a row of all 1's in its place:

Step 3: Clear the row of 1's with column operations, then expand …

Common Mistakes

Mistake 1: Mis-signing the $2\times2$ minor after the row/column reduction

Why it's wrong: after $C_2\to C_2-C_1$, $C_3\to C_3-C_1$, the reduced $2\times2$ block often has entries that are negatives of each other (like $-2y$ and $y$), and it's easy to drop a sign while computing $(-2y)(-y)-(-y)(y)$. A sign slip here turns $3y^2$ into $y^2$ or $-y^2$, and the option chosen changes accordingly. Correct approach: expand the $2\times2$ minor one multiplication at a time, keeping every negative sign explicit before combining.

Mistake 2: Forgetting to reattach the factor pulled out in the row-sum step …