TITLE: How can I quickly find the determinant of this matrix
QUESTION [6 upvotes]: $$
        \begin{vmatrix}
        14 & 2 & 1 & 3\\
        31 & 4 & 5 & 6\\
        26 & 3 & 7 & 4\\
        10 & 1 & 3 & 2\\
        \end{vmatrix}
       =
        \begin{vmatrix}
        5\cdot2+1+3 & 2 & 1 & 3\\
        5\cdot4+5+6 & 4 & 5 & 6\\
        5\cdot3+7+4 & 3 & 7 & 4\\
        5\cdot1+3+2 & 1 & 3 & 2\\
        \end{vmatrix}$$$$
       =
        \begin{vmatrix}
        5\cdot2 & 2 & 1 & 3\\
        5\cdot4 & 4 & 5 & 6\\
        5\cdot3 & 3 & 7 & 4\\
        5\cdot1 & 1 & 3 & 2\\
        \end{vmatrix}
        +
        \begin{vmatrix}
        1 & 2 & 1 & 3\\
        5 & 4 & 5 & 6\\
        7 & 3 & 7 & 4\\
        3 & 1 & 3 & 2\\
        \end{vmatrix}
        +
        \begin{vmatrix}
        3 & 2 & 1 & 3\\
        6 & 4 & 5 & 6\\
        4 & 3 & 7 & 4\\
        2 & 1 & 3 & 2\\
        \end{vmatrix}
$$
However I am not able to proceed beyond. The answer given is zero. Is there any simple determinant property that I am not able to guess?

REPLY [3 votes]: You have found—and expressed it in the first equality—that the first column is a linear combination of other columns: $$Col_1 = 5\cdot Col_2+Col_3+Col_4$$ And the simple determinant property you can't guess is: determinant with linearly dependent columns is $0$ (and vice versa, if it is zero, it has linearly dependent columns—and rows, too).