Given $\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} = -3$, find $\begin{vmatrix}p&2q&5r+4p\\s&2t&5u+4s\\v&2w&5x+4v\end{vmatrix}$. For the real numbers $p$, $q$, $r$, $s$, $t$, $u$, $v$, $w$, $x$,
$$\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} = -3.$$
Find
$$\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix}.$$

My first thought was to find a particular matrix with determinant $-3$, but that would take too long with no guaranteed results.  But I don't know what else to do.
 A: It's about properties of determinant. Multiples of columns (or rows) can be factored out, and the determinant is linear on each column (or row). Also, a repeated column (or row) causes the determinant to be zero, as the determinant does not change when we add a multiple of a column to another column. So 
\begin{align}
\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix}&=\begin{vmatrix} p & 2q & 5r   \\ s & 2t & 5u   \\ v & 2w & 5x   \end{vmatrix}+\begin{vmatrix} p & 2q &   4p \\ s & 2t &   4s \\ v & 2w &   4v \end{vmatrix}\\ \ \\
&=2\times5\times\begin{vmatrix} p & q & r   \\ s & t & u   \\ v & w & x   \end{vmatrix}+0\\ \ \\
&=2\times5\times(-3)=-30.
\end{align}
A: HINT:
$$\begin{vmatrix} p & 2q & 5r+4p \\ s & 2t & 5u+4s \\ v & 2w & 5x+4v \end{vmatrix} $$
$$\begin{vmatrix} p & 2q & 5r \\ s & 2t & 5u \\ v & 2w & 5x \end{vmatrix}+\begin{vmatrix} p & 2q & 4p \\ s & 2t & 4s \\ v & 2w & 4v \end{vmatrix}  $$
$$2*5*\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix}+2*4*\begin{vmatrix} p & q & p \\ s & t & s \\ v & w & v \end{vmatrix}  $$
Multiples of rows and columns can be factored out. And in second determinant, $C_1$ and $C_3$ are same, Thus that one becomes 0
$$2*5*(-3)+0$$
$$-30$$
A: $\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix} = \begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} \begin{vmatrix}1& 0 &4\\0&2&0\\0&0&5\end{vmatrix}$
And $\det(AB) = \det(A)\det(B)$
A: The important thing to understand is that the determinant is linear in each column (or row) and since a determinant with two equal columns (or rows) is zero, you can add and subtract multiples of a row from another row. Algebraically, if $c_i$ are the columns, and $a,b$ are real numbers we have
$$ \begin{vmatrix} (ac_1 + bc'_1) & c_2 & c_3 \end{vmatrix} = a\begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix}+b\begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix} $$
and
$$ \begin{vmatrix} c_1 & c_2+ac_1 & c_3 \end{vmatrix} = \begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix} + a\begin{vmatrix} c_1 & c_1 & c_3 \end{vmatrix} = \begin{vmatrix} c_1 & c_2 & c_3 \end{vmatrix}. $$
In this case, subtract $4$ of the first column from the third column, and then pull the multiple of $2$ out of the second column and the multiple of $5$ out of the new third column.
