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: Write the determinant clearly.
We are given \[ \Delta=\begin{vmatrix} x & y & 1 x+y & y+1 & x+1 1 & x & y \end{vmatrix} \] and the condition \[ x+y=-1 \] We need to evaluate the determinant using this condition.

Step 2: Substitute the condition \( x+y=-1 \) into the second row.
Since \[ x+y=-1 \] the second row becomes \[ (x+y,\ y+1,\ x+1)=(-1,\ y+1,\ x+1) \] So the determinant becomes \[ \Delta=\begin{vmatrix} x & y & 1 -1 & y+1 & x+1 1 & x & y \end{vmatrix} \]

Step 3: Use the relation \( x+y=-1 \) to simplify \( y+1 \) and \( x+1 \).
From \[ x+y=-1 \] we get \[ y+1=-x \] and also \[ x+1=-y \] Therefore the second row becomes \[ (-1,-x,-y) \] Hence, \[ \Delta=\begin{vmatrix} x & y & 1 -1 & -x & -y 1 & x & y \end{vmatrix} \]

Step 4: Observe the relation between rows.
Now compare the second and third rows: \[ R_2=(-1,-x,-y) \] and \[ R_3=(1,x,y) \] Clearly, \[ R_2=-R_3 \] That means the second row is exactly the negative of the third row.

Step 5: Use the property of determinants.
A determinant is zero if any two rows are proportional.
Since \[ R_2=-R_3 \] the rows are proportional. Therefore, \[ \Delta=0 \]

Step 6: Verify the conclusion conceptually.
Whenever one row is a constant multiple of another row, the matrix becomes linearly dependent.
A determinant measures linear independence of rows or columns.
Because the rows are dependent here, the determinant must vanish.

Step 7: Final conclusion.
Hence, the value of the determinant is \[ \boxed{0} \] Therefore, the correct option is \[ \boxed{(4)\ 0} \]