Answer
$\det (B)=-12$
Work Step by Step
We have $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=4$
$\rightarrow A^T=\begin{bmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{bmatrix}$ and $\det (A^T)=4$
Multiplying the second row of $A^T$ by $-1$, we obtain:
$A_1=\begin{bmatrix}
-a & -d & -g\\
b & e & h\\
c & f & i
\end{bmatrix}$
$\det (A_1)=- \det (A^T)=-4$
Multiplying the second row of $A_1$ by $3$, we obtain:
$A_2=\begin{bmatrix}
-a & -d & -g\\
3b & 3e & 3h\\
c & f & i
\end{bmatrix}$
$\det (A_2)=-\det (A_1)=-4$
Adding $2$ times the first row to the the third row, we obtain:
$A_3=\begin{bmatrix}
-a & -d & -g\\
3b & 3e & 3h\\
c-2a & f-2a & i-2g
\end{bmatrix}$
$\det (A_3)=\det (A_2)=-12$
Permutting the first and second rows of $A_3$ we obtain:
$A_4=\begin{bmatrix}
3b & 3e & 3h\\
-a & -d & -g\\
c-2a & f-2a & i-2g
\end{bmatrix}$
$\det (A_4)=-\det (A_3)=12$
Permutting the second and third rows of $A_4$ we obtain:
$B=\begin{bmatrix}
3b & 3e & 3h\\
c-2a & f-2a & i-2g\\
-a & -d & -g
\end{bmatrix}$
$\det (B)=-\det (A_4)=-12$
