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 $\det (A)=4$
Adding $-5$ times second row of $A_2$ to the first row, we obtain
$A_1=\begin{bmatrix}
a-5d & b-5e & c-5f\\
d & e &f\\
g & h & i
\end{bmatrix}$
$\det (A_1)=4$
Multiplying the third row of $A_1$ by $3$, we obtain:
$A_2=\begin{bmatrix}
a-5d & b-5e & c-5f\\
d & e &f\\
3g & 3h & 3i
\end{bmatrix}$
$\det (A_2)=3\det (A_1)=12$
Multiplying the third second row of $A_2$ by $-1$, we obtain:
$A_3=\begin{bmatrix}
a-5d & b-5e & c-5f\\
-d & -e & -f\\
3g & 3h & 3i
\end{bmatrix}$
$\det (A_3)=-\det (A_2)=-12$
Permutting the second and third rows of $A_3$ we obtain:
$A_4=\begin{bmatrix}
a-5d & b-5e & c-5f\\
3g & 3h & 3i\\
-d & -e & -f
\end{bmatrix}$
$\det (A_4)=-\det (A_3)=12$
Adding the second row of $A_4$ to the third row, we obtain
$B=\begin{bmatrix}
a-5d & b-5e & c-5f\\
3g & 3h & 3i\\
-d+3g & -e+3h & -f+3i
\end{bmatrix}$
$\det (B)=\det (A_4)=12$
