Answer
$\det (B)=8$
Work Step by Step
We have $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=4$
Adding $-1$ times of the second row to the first row, we obtain:
$A_1=\begin{bmatrix}
a-d & b-e & c-f\\
d & e &f\\
g & h & i
\end{bmatrix}$
$\det (A_1)= \det (A)=4$
Multiplying the second row of $A_1$ by $-1$, we obtain:
$A_2=\begin{bmatrix}
a-d & b-e & c-f\\
-d & -e &-f\\
g & h & i
\end{bmatrix}$
$\det (A_2)=-\det (A_1)=-4$
Multiplying the third row of $A_2$ by $2$, we obtain:
$A_3=\begin{bmatrix}
a-d & b-e & c-f\\
-d & -e &-f\\
2g & 2h & 2i
\end{bmatrix}$
$\det (A_3)=2\det (A_2)=2.(-4)=-8$
Permutting the second and third rows of $A_3$ we obtain:
$B=\begin{bmatrix}
a-d & b-e & c-f\\
2g & 2h & 2i\\
-d & -e &-f
\end{bmatrix}$
$\det (B)=-\det (A_3)=8$
