Answer
$\det (B)=12$
Work Step by Step
Let $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=-6$
Multiplying the second row of A by $-2$, we obtain:
$A_1=\begin{bmatrix}
a & b & c\\
-2d & -2e &-2f\\
g & h & i
\end{bmatrix}$
$\det (A_1)=(-2)\det (A)=-2*(-6)=12$
Multiplying the first row of $A_1$ by $-1$, we obtain:
$A_2=\begin{bmatrix}
-a & -b & -c\\
-2d & -2e &-2f\\
g & h & i
\end{bmatrix}$
$\det (A_2)=(-1)\det (A_1)=(-1)(12)=-12$
Interchanging the first and third rows of $A_2$ by $-1$, we obtain:
$B=\begin{bmatrix}
g & h & i\\
-2d & -2e &-2f\\
-a & -b & -c
\end{bmatrix}$
$\det (B)=(-1)\det (A_2)=(-1)(-12)=12$
