`" If " |{:(x^(2)+x,,x+1,,x-2),(2x^(2)+3x-1,,3x,,3x-3),(x^(2)+2x+3,,2x-1,,2x-1):}|=xA +B` then

Options

A. `|{:(1,,1,,1),(-1,,-3,,3),(4,,0,,0):}|`
B. `|{:(0,,1,,2),(1,,-2,,3),(-4,,0,,0):}|`
C. `|{:(1,,1,,-2),(-3,,-2,,3),(4,,0,,1):}|`
D. `|{:(0,,1,,-2),(-1,,-3,,3),(4,,0,,0):}|`

Explanation

To solve the given determinant problem, we need to evaluate the determinant of the matrix and express it in the form \( xA + B \).

Given the determinant:

\[ \Delta = \begin{vmatrix} x^2 & x & x + 1 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} \]

### Step 1: Simplify the Determinant
We will perform row operations to simplify the determinant.

1. **Row Operation**: Replace \( R_2 \) with \( R_2 - 2R_1 \) and \( R_3 \) with \( R_3 - R_1 \).

\[ \Delta = \begin{vmatrix} x^2 & x & x + 1 \\ (2x^2 + 3x - 1) - 2(x^2) & 3x - 2x & (3x - 3) - 2(x + 1) \\ (x^2 + 2x + 3) - (x^2) & (2x - 1) - x & (2x - 1) - (x + 1) \end{vmatrix} \]

Calculating the new rows:

- For \( R_2 \):
- First element: \( 2x^2 + 3x - 1 - 2x^2 = x \)
- Second element: \( 3x - 2x = x \)
- Third element: \( 3x - 3 - 2x - 2 = x - 5 \)

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

Thus, we have:

\[ \Delta = \begin{vmatrix} x^2 & x & x + 1 \\ x & x & x - 5 \\ 2x + 3 & x - 1 & x - 2 \end{vmatrix} \]

### Step 2: Further Simplification
Next, we can perform more row operations.

2. **Row Operation**: Replace \( R_2 \) with \( R_2 - R_1 \).

\[ \Delta = \begin{vmatrix} x^2 & x & x + 1 \\ 0 & 0 & -6 \\ 2x + 3 & x - 1 & x - 2 \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant using the first row:

\[ \Delta = x^2 \begin{vmatrix} 0 & -6 \\ x - 1 & x - 2 \end{vmatrix} - x \begin{vmatrix} 0 & -6 \\ 2x + 3 & x - 2 \end{vmatrix} + (x + 1) \begin{vmatrix} 0 & 0 \\ 2x + 3 & x - 1 \end{vmatrix} \]

Calculating the determinants of the 2x2 matrices:

- The first determinant is \( 0 \) because the first column is all zeros.
- The second determinant is also \( 0 \) for the same reason.
- The third determinant is \( 0 \).

Thus, we find that:

\[ \Delta = 0 \]

### Step 4: Express in the Form \( xA + B \)
Since \( \Delta = 0 \), we can express it as:

\[ 0 = xA + B \]

This implies that \( A = 0 \) and \( B = 0 \).

### Final Answer
Thus, we have:

- \( A = \begin{vmatrix} 1 & 1 & 1 \\ -1 & -3 & 3 \\ 0 & 0 & 0 \end{vmatrix} \)
- \( B = \begin{vmatrix} 0 & 1 & -2 \\ -1 & -3 & 3 \\ 0 & 0 & 0 \end{vmatrix} \)