If `|(x^2+2x,2x+1,1),(2x+1,x+2,1),(3,3,1)|=(x-1)^(k)` then k =

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the determinant equation \( |(x^2 + 2x, 2x + 1, 1), (2x + 1, x + 2, 1), (3, 3, 1)| = (x - 1)^k \), we will simplify the determinant step by step.

### Step 1: Write the determinant
We start with the determinant:
\[ D = \begin{vmatrix} x^2 + 2x & 2x + 1 & 1 \\ 2x + 1 & x + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} \]

### Step 2: Perform row operations
We can simplify the determinant using row operations. Let's perform \( R_2 - R_1 \) and \( R_3 - R_1 \):
\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1 \]
This gives us:
\[ D = \begin{vmatrix} x^2 + 2x & 2x + 1 & 1 \\ (2x + 1) - (x^2 + 2x) & (x + 2) - (2x + 1) & 0 \\ 3 - (x^2 + 2x) & 3 - (2x + 1) & 0 \end{vmatrix} \]

Calculating the new rows:
- For \( R_2 \):
- First element: \( 2x + 1 - (x^2 + 2x) = 1 - x^2 \)
- Second element: \( x + 2 - (2x + 1) = 1 - x \)

- For \( R_3 \):
- First element: \( 3 - (x^2 + 2x) = 3 - x^2 - 2x \)
- Second element: \( 3 - (2x + 1) = 2 - 2x \)

So we have:
\[ D = \begin{vmatrix} x^2 + 2x & 2x + 1 & 1 \\ 1 - x^2 & 1 - x & 0 \\ 3 - x^2 - 2x & 2 - 2x & 0 \end{vmatrix} \]

### Step 3: Factor out common terms
Notice that the last column has zeros in the second and third rows. We can factor out \( (x - 1) \) from the second row and from the third row:
\[ D = (x - 1) \cdot \begin{vmatrix} x^2 + 2x & 2x + 1 \\ 1 - x^2 & 1 - x \\ \end{vmatrix} \]

### Step 4: Calculate the remaining determinant
Now we calculate the remaining \( 2 \times 2 \) determinant:
\[ D' = \begin{vmatrix} x^2 + 2x & 2x + 1 \\ 1 - x^2 & 1 - x \end{vmatrix} \]
Calculating this determinant:
\[ D' = (x^2 + 2x)(1 - x) - (2x + 1)(1 - x^2) \]

Expanding both terms:
1. First term: \( x^2 + 2x - x^3 - 2x^2 = -x^3 - x^2 + 2x \)
2. Second term: \( 2x + 1 - 2x^3 - 1 = -2x^3 + 2x \)

Combining:
\[ D' = -x^3 - x^2 + 2x + 2x^3 - 2x = x^3 - x^2 \]

### Step 5: Final determinant expression
Thus, we have:
\[ D = (x - 1)(x^3 - x^2) = (x - 1)^2 \cdot x(x + 1) \]

### Step 6: Set equal to the original expression
Setting this equal to \( (x - 1)^k \):
\[ (x - 1)^2 \cdot x(x + 1) = (x - 1)^k \]

### Step 7: Determine the value of k
For the equality to hold, we need \( k = 2 \) since the \( x(x + 1) \) term does not contribute to the power of \( (x - 1) \).

Thus, the value of \( k \) is:
\[ \boxed{2} \]