If vmatrix 2 & 3+i & -1 3-i & 0 & i-1 -1 & -1-i & 1 vmatrix = A +...

Given the determinant:

$\begin{vmatrix} 2 \text{ and } 3+i \text{ and } -1 \\ 3-i \text{ and } 0 \text{ and } i-1 \\ -1 \text{ and } -1-i \text{ and } 1 \end{vmatrix} = A + iB$

Expanding the determinant along the first row, we get:

$D = 2 \begin{vmatrix} 0 \text{ and } i-1 \\ -1-i \text{ and } 1 \end{vmatrix} - (3+i) \begin{vmatrix} 3-i \text{ and } i-1 \\ -1 \text{ and } 1 \end{vmatrix} - 1 \begin{vmatrix} 3-i \text{ and } 0 \\ -1 \text{ and } -1-i \end{vmatrix}$

Evaluating each $2 \times 2$ minor:

First minor:
$\begin{vmatrix} 0 \text{ and } i-1 \\ -1-i \text{ and } 1 \end{vmatrix} = (0)(1) - (i-1)(-1-i) = 0 - (-(i^2 - 1)) = 0 - (-(-1 - 1)) = -2$

Second minor:
$\begin{vmatrix} 3-i \text{ and } i-1 \\ -1 \text{ and } 1 \end{vmatrix} = (3-i)(1) - (i-1)(-1) = 3 - i + i - 1 = 2$

Third minor:
$\begin{vmatrix} 3-i \text{ and } 0 \\ -1 \text{ and } -1-i \end{vmatrix} = (3-i)(-1-i) - (0)(-1) = -3 - 3i + i + i^2 = -3 - 2i - 1 = -4 - 2i$

Substituting these values back into the expansion:

$D = 2(-2) - (3+i)(2) - 1(-4 - 2i)$

$D = -4 - (6 + 2i) + (4 + 2i)$

$D = -4 - 6 - 2i + 4 + 2i$

$D = -6$

Comparing this with $A + iB$, we get:
$A = -6$ and $B = 0$

Therefore, $A + B = -6 + 0 = -6$.

Answer: $-6$