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}.
Is this a problem like $2x=8$, so $4x=16$? I can't figure out how to solve this.
On
Here's an additional hint:
The determinant is also linear: that is, if all but one of the rows/columns are fixed and the remaining row/column is of the form $a \mathbf{b} + c \mathbf{d}$, then the determinant function splits up linearly -- that is, it will be a sum of two determinants where one of them will contain, in the same position as the concerned row/column of the initial matrix, the row/column $a \mathbf{b}$ and the other will contain the other term. Now you can easily bring out the scalar from these determinants and use hints given in the previous answer.
I am unable you write equations to at the moment to explain this answer more clearly, but yes you're right the answer is -30.
Hints:
Multiplying a column by a scalar $c$ causes the determinant to also be multiplied by $c$.
Adding [a scalar multiple of] a column to a different column does not change the determinant.