Answer
$\det (B)=-2$
Work Step by Step
Let $A=\begin{bmatrix}
a & b\\
c&d
\end{bmatrix}$ and $\det (A)=1$
$\rightarrow A^T=\begin{bmatrix}
a & c\\
b &d
\end{bmatrix}$
We obtain the matrix $A_1=\begin{bmatrix}
a & c\\
3a+b & 3c+d
\end{bmatrix}$ by adding the first row of $A^T$ multiplied by 3 to its second row.
And then by multiplying the first row of $A_1$ by $-2$ we obtain the matrix B$=\begin{bmatrix}
-2a & -2c\\
3a+b & 3c+d
\end{bmatrix}$
The determinant of B is:
$\det (B)=(-2)\det (A_1)=(-2).1=-2$
