How to find the determinant of the matrix?

Question

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.

Answer 1

$\det \begin{bmatrix}a - 6g&7b - 42h&c - 6i\\d&7e&f\\g&7h&i\end{bmatrix}=\det \begin{bmatrix}a &7b&c\\d&7e&f\\g&7h&i\end{bmatrix}+\det \begin{bmatrix}- 6g& - 42h& - 6i\\d&7e&f\\g&7h&i\end{bmatrix}$

by the multilinear property, now the second determinate is $0$ because the first and the third row are linearity dependent; so

$\det \begin{bmatrix}a &7b &c \\d&7e&f\\g&7h&i\end{bmatrix}=7 \det \begin{bmatrix}a &b&c\\d&e&f\\g&h&i\end{bmatrix}=7$

Answer 2

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.

Answer 3

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.