Question

Let \( \Delta = \begin{vmatrix} x & y & 1 x+y & y+1 & x+1 1 & x & y \end{vmatrix} \). If \( x+y=-1 \), then the value of \( \Delta \) is equal to

• A. \( 3 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( 0 \)
• E. \( -3 \)

Hint:

In determinant problems, always check whether the given condition can make two rows or two columns proportional. That is often the fastest route to the answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves evaluating a determinant using its properties. The given condition \(x+y=-1\) is a key hint that certain row or column operations might simplify the determinant significantly.
Step 2: Key Formula or Approach:
A fundamental property of determinants is that if a row or column is a linear combination of other rows or columns, the determinant is zero. Another property is that adding a multiple of one column (or row) to another column (or row) does not change the value of the determinant. We will use this to create a column of zeros.
Step 3: Detailed Explanation:
Let's apply the column operation \(C_1 \to C_1 + C_2 + C_3\). This means we replace the first column with the sum of all three columns. The value of the determinant remains unchanged.
\[ \Delta = \begin{vmatrix} x+y+1 & y & 1
(x+y)+(y+1)+(x+1) & y+1 & x+1
1+x+y & x & y \end{vmatrix} \] Let's simplify the elements of the new first column:
Row 1, Column 1: \(x+y+1\)
Row 2, Column 1: \(2x+2y+2 = 2(x+y+1)\)
Row 3, Column 1: \(1+x+y\)
Now, substitute the given condition \(x+y=-1\):
Row 1, Column 1: \((-1)+1 = 0\)
Row 2, Column 1: \(2((-1)+1) = 2(0) = 0\)
Row 3, Column 1: \(1+(-1) = 0\)
The determinant becomes:
\[ \Delta = \begin{vmatrix} 0 & y & 1
0 & y+1 & x+1
0 & x & y \end{vmatrix} \] A determinant with an entire column of zeros is always equal to 0. This can be seen by expanding the determinant along the first column:
\[ \Delta = 0 \cdot \begin{vmatrix} y+1 & x+1
x & y \end{vmatrix} - 0 \cdot \begin{vmatrix} y & 1
x & y \end{vmatrix} + 0 \cdot \begin{vmatrix} y & 1
y+1 & x+1 \end{vmatrix} = 0 \] Step 4: Final Answer:
The value of the determinant is 0. Therefore, option (D) is the correct answer.