How to find the determinant of the matrix? 
if $\det \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} = 1$
then $\det \begin{bmatrix}a - 6g&7b - 42h&c - 6i\\d&7e&f\\g&7h&i\end{bmatrix} = ?$

Yeah, I have no idea how to solve this nor do I know what is going on in the first row??
Here is how the problem is presented to me, I tried my best to copy it.



 A: $\det \begin{bmatrix}a - 6g&7b - 42h&c - 6i\\d&7e&f\\g&7h&i\end{bmatrix}=\det \begin{bmatrix}a &7b&c\\d&7e&f\\g&7h&i\end{bmatrix}+\det \begin{bmatrix}- 6g& - 42h& - 6i\\d&7e&f\\g&7h&i\end{bmatrix}$ 
by the multilinear property, now the second determinate is $0$ because the first and the third row are linearity dependent; so
$\det \begin{bmatrix}a &7b &c \\d&7e&f\\g&7h&i\end{bmatrix}=7 \det \begin{bmatrix}a &b&c\\d&e&f\\g&h&i\end{bmatrix}=7$
A: I think they mean $\begin{pmatrix}a-6g & 7b -42h & c-6i\end{pmatrix}$ for the first row.
Hint: The determinant is a multilinear function of its columns.
For instance: $$\begin{align} \det\begin{pmatrix}a-6g&7b-42h&c-6i\\
d&7e&f\\
g&7h&i
\end{pmatrix} &= 7\det\begin{pmatrix}a-6g&b-6h&c-6i\\
d&e&f\\
g&h&i
\end{pmatrix} \\ &= 
7\left(\det\begin{pmatrix}a&b-6h&c-6i\\
d&e&f\\
g&h&i
\end{pmatrix}
+
\det\begin{pmatrix}-6g&b-6h&c-6i\\
0&e&f\\
0&h&i
\end{pmatrix}
\right)\\ &= \cdots
\end{align} $$
Edit: It's actually smarter to use a row operation! See Gianluca's answer.
A: The matrix is supposed to be
$$\begin{pmatrix}a-6g&7b-42h&c-6i\\
d&7e&f\\
g&7h&i
\end{pmatrix}$$
Hint: This matrix can be written in terms of the original matrix by manipulating the rows and columns.
