If A is a square matrix of order n such that its elements are polynomila in x and its r-rows become identical for x = k, then

Options

A. `(x -k)^(r)` is a factor of `|A|`
B. `(x -k)^(r) -1` is a factor of `|A|`
C. `(x -k)^(r) +1` is a factor `|A|`
D. `(x +k)^(r)` is a factor of `|A|`

Answer

To solve the problem, we need to analyze the implications of having a square matrix \( A \) of order \( n \) with polynomial elements in \( x \) and \( r \) rows becoming identical when \( x = k \).

### Step 1: Understand the implications of identical rows
When \( r \) rows of the matrix \( A \) become identical for \( x = k \), it means that substituting \( x = k \) into the elements of these rows results in the same values for all elements in those rows.

**Hint:** Identical rows in a matrix lead to a determinant of zero.

### Step 2: Factor out the common term
Since the \( r \) rows become identical when \( x = k \), we can factor out \( (x - k) \) from each of these \( r \) rows. This is because the difference between the elements of these rows will involve \( (x - k) \).

**Hint:** When rows are identical, the determinant can be expressed with a common factor.

### Step 3: Formulate the determinant
By factoring out \( (x - k) \) from each of the \( r \) identical rows, we can express the determinant as:
\[ \text{det}(A) = (x - k)^r \cdot \text{det}(B) \]
where \( B \) is a new matrix formed by the remaining rows and the modified rows after factoring out \( (x - k) \).

**Hint:** The power of the factor corresponds to the number of identical rows.

### Step 4: Conclude the result
Thus, we can conclude that \( (x - k)^r \) is a factor of the determinant of \( A \). This leads us to the final answer.

### Final Answer:
Therefore, the correct option is:
\[ x - k \text{ whole to the power } r \text{ is a factor of } \text{det}(A). \]