Answer
$\det (B)=-12$
Work Step by Step
Let $A=\begin{bmatrix}
a & b\\
c&d
\end{bmatrix}$ and $\det (A)=1$
$\rightarrow A_1=\begin{bmatrix}
a & c\\
b &d
\end{bmatrix}$
We obtain the matrix $A_2=\begin{bmatrix}
3c & 3d\\
a & b
\end{bmatrix}$ by multiplying the first row of $A_1$ by 3.
And then by multiplying the second row of $A_2$ by $4$ we obtain the matrix B$=\begin{bmatrix}
3c & 3d\\
4a & 4b
\end{bmatrix}$
The determinant of B is:
$\det (B)=4\det (A_2)=-12$
