Let
`Delta(x,y)=|(1,x,y),(1,x+y,y),(1,x,x+y)|`
Then `Delta(-3,2)` equals

Options

A. 13
B. `-6`
C. `12`
D. `-3`

Answer

To find the value of \( \Delta(-3, 2) \), we start by substituting \( x = -3 \) and \( y = 2 \) into the determinant:

\[ \Delta(x, y) = \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix} \]

Substituting \( x = -3 \) and \( y = 2 \):

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

Calculating \( -3 + 2 = -1 \) and \( -3 + 2 = -1 \):

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

Next, we calculate the determinant using the cofactor expansion along the first row:

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

Now we calculate each of the 2x2 determinants:

1. For \( \begin{vmatrix} -1 & 2 \\ -3 & -1 \end{vmatrix} \):
\[ = (-1)(-1) - (2)(-3) = 1 + 6 = 7 \]

2. For \( \begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix} \):
\[ = (1)(-1) - (2)(1) = -1 - 2 = -3 \]

3. For \( \begin{vmatrix} 1 & -1 \\ 1 & -3 \end{vmatrix} \):
\[ = (1)(-3) - (-1)(1) = -3 + 1 = -2 \]

Now substituting back into the determinant expression:

\[ \Delta(-3, 2) = 1 \cdot 7 - (-3)(-3) + 2(-2) \]
\[ = 7 - 9 - 4 \]
\[ = 7 - 13 = -6 \]

Thus, the final value of \( \Delta(-3, 2) \) is:

\[ \boxed{-6} \]