divisible by and Explanation: To find the value of the determinant and check its divisibility, we apply elementary row operations to simplify the matrix. 1. Apply Row Operations: We want to create zeros in the first column to make expansion easier. Let's perform the following operations: Applying these to the determinant : 2. Simplify the Determinant: 3. Expand along the First Column: Since the first column now has two zeros, expanding along is straightforward: 4. Analyze Divisibility: The final expression for the determinant is . This expression is a product of and . Therefore, is clearly divisible by both and . Conclusion: The determinant is divisible by and . Correct Option: C) divisible by and

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NIMCET 2022 — Mathematics PYQ

NIMCET | Mathematics | 2022

If $D = 111 am p; 1 am p; 2 + x am p; 1 am p; 1 am p; 1 am p; 2 + y, x \neq = 0, y \neq = 0$ , then $D$ is

Choose the correct answer:

Explanation

To find the value of the determinant $D$ and check its divisibility, we apply elementary row operations to simplify the matrix.

1. Apply Row Operations:

We want to create zeros in the first column to make expansion easier. Let's perform the following operations:

$R_2 \to R_2 - R_1$

$R_3 \to R_3 - R_1$

Applying these to the determinant $D$ :

$D = \begin{vmatrix} 1 & 1 & 1 \\ 1-1 & (2+x)-1 & 1-1 \\ 1-1 & 1-1 & (2+y)-1 \end{vmatrix}$

2. Simplify the Determinant:

$D = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 + x & 0 \\ 0 & 0 & 1 + y \end{vmatrix}$

3. Expand along the First Column:

Since the first column now has two zeros, expanding along $C_1$ is straightforward:

$D = 1 \cdot \begin{vmatrix} 1 + x & 0 \\ 0 & 1 + y \end{vmatrix}$

$D = 1 \cdot [(1 + x)(1 + y) - (0 \cdot 0)]$

$D = (1 + x)(1 + y)$

4. Analyze Divisibility:

The final expression for the determinant is $D = (x + 1)(y + 1)$ .

This expression is a product of $(x + 1)$ and $(y + 1)$ .

Therefore, $D$ is clearly divisible by both $(x + 1)$ and $(y + 1)$ .

Conclusion:

The determinant is divisible by $(x + 1)$ and $(y + 1)$ .

Correct Option:

C) divisible by $(x + 1)$ and $(y + 1)$