Determinants and the following question I understand  determinants but I cannot understand the following question, can someone explain it to me ?
Suppose that a $4 x 4$ matrix  with rows $v_1,v_2,v_3$ and $v_4$ has determinant det A = -1. Find the following determinants: 
$$det \begin{bmatrix}v_1\\6v_2\\v_3\\v_4 \end{bmatrix}=$$
$$det \begin{bmatrix}v_2\\v_1\\v_4\\v_3 \end{bmatrix}=$$
$$det \begin{bmatrix}v_1\\v_2+7v_1\\v_3\\v_4 \end{bmatrix}=$$
I know how to find the determinate of a matrix but do not understand how this works.
 A: Split the determinant across multiplication and notice that:
$\begin{bmatrix} v_1 \\ 6v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} 1 \ 0\ 0\ 0 \\ 0\ 6\ 0\ 0 \\ 0\ 0\ 1\ 0\\ 0\ 0\ 0\ 1 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$
$\begin{bmatrix} v_2 \\ v_1 \\ v_4 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 1\ 0\ 0 \\ 1\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\\ 0\ 0\ 1\ 0 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$
$\begin{bmatrix} v_1 \\ v_2 +7V_1 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} 1 \ 0\ 0\ 0 \\ 7\ 1\ 0\ 0 \\ 0\ 0\ 1\ 0\\ 0\ 0\ 0\ 1 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$
A: Hints:


*

*the determinant is multilinear on the rows.

*the sign changes when you swap two rows.

*the determinant is zero when two rows are the same


With this, you can compute all three determinants in your question
Note: multilinear means $\det(ar_1+br_1',r_2,r_3,r_4)=a\det(r_1,r_2,r_3,r_4)+b\det(r_1',r_2,r_3,r_4)$ in the first. 
And likewise for other entries.
A: Hint: For the first one, use the property that $$\det(A\cdot B)=\det(A)\cdot\det(B).$$See if you can change the identity matrix just right to get the 'new' matrix, where the second row is multiplied by $6$.
For the second one, use a similar idea, but this time the rows of the identity matrix get jumbled up a little bit. Determine which row has a $1$ in the first column, which row has a $1$ in the second column, etc. Now calculate this determinant and multiply it by the determinant of your original matrix.
For the third one, this builds on the previous exercise. Now you can begin with the identity matrix and change one of the zero entries so that when you multiply it to your original matrix, you get $v_2+7v_1$.
