Answer
$\det (B)=-18$
Work Step by Step
Let $A=\begin{bmatrix}
a & b & c\\
d & e &f\\
g & h & i
\end{bmatrix}$ and $\det (A)=-6$
Permuting the first and second rows of $A$:
$A_1=\begin{bmatrix}
d & e &f\\
a & b & c\\
g & h & i
\end{bmatrix}$
$\det (A_1)=-\det (A_1)=6$
Multiplying the second row of $A_1$ by $-3$, we obtain:
$A_2=\begin{bmatrix}
d & e &f\\
-3a & -3b & -3c\\
g & h & i
\end{bmatrix}$
$\det (A_2)=(-3)\det (A_1)=-3.(6)=-18$
Adding 4 times first row of $A_2$ to the third row, we obtain:
$B=\begin{bmatrix}
d & e &f\\
-3a & -3b & -3c\\
g-4d & h-4e & f-4f
\end{bmatrix}$
$\det (B)=\det (A_2)=-18$
