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.
On
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
$$2*5*(-3)+0$$ $$-30$$
On
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.
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}