<s>[INST] Determinants using elementary row operations Let matrix $A$ be defined as \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}
And $det(A)=d$
Let $B=$
\begin{pmatrix}
2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\
2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\
\end{pmatrix} $$+$$
\begin{pmatrix}
3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}
Find $det(B)$ using elementary row operations. I think the answer is $3^{(n-1)}*5*d$, but I'm not sure.
 [/INST] $$\begin{pmatrix}
2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\
2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\
\end{pmatrix} +
\begin{pmatrix}
3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}=\begin{pmatrix}
5a_{11} & 5a_{12} & \cdots & 5a_{1n} \\
3a_{21} & 3a_{22} & \cdots & 3a_{2n} \\
\vdots  & \vdots  & \vdots  & \vdots  \\
3a_{n1} & 3a_{n2} & \cdots & 3a_{nn}\\
\end{pmatrix}$$
Since the determinant is a $n$-linear function (considering that each line is an element of $\mathbb R^n$), it follows that, since the first line of $B$ is the first line of $A$ multiplied by $5$ and since all the other $n-1$ lines of $B$ are the same $n-1$ lines of $A$ multiplied by $3$, it follows that
$\det(B)=5.3^{n-1}.det(A)=5.3^{n-1}.d$
Your answer is correct.
</s>