How can I quickly find the determinant of this matrix

Question

$$ \begin{vmatrix} 14 & 2 & 1 & 3\\ 31 & 4 & 5 & 6\\ 26 & 3 & 7 & 4\\ 10 & 1 & 3 & 2\\ \end{vmatrix} = \begin{vmatrix} 5\cdot2+1+3 & 2 & 1 & 3\\ 5\cdot4+5+6 & 4 & 5 & 6\\ 5\cdot3+7+4 & 3 & 7 & 4\\ 5\cdot1+3+2 & 1 & 3 & 2\\ \end{vmatrix}$$$$ = \begin{vmatrix} 5\cdot2 & 2 & 1 & 3\\ 5\cdot4 & 4 & 5 & 6\\ 5\cdot3 & 3 & 7 & 4\\ 5\cdot1 & 1 & 3 & 2\\ \end{vmatrix} + \begin{vmatrix} 1 & 2 & 1 & 3\\ 5 & 4 & 5 & 6\\ 7 & 3 & 7 & 4\\ 3 & 1 & 3 & 2\\ \end{vmatrix} + \begin{vmatrix} 3 & 2 & 1 & 3\\ 6 & 4 & 5 & 6\\ 4 & 3 & 7 & 4\\ 2 & 1 & 3 & 2\\ \end{vmatrix} $$

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?

Answer 1

The two last determinants in your right side are zero as they both have two equal columns, so you're left with:

$$\begin{vmatrix} 14 & 2 & 1 & 3\\ 31 & 4 & 5 & 6\\ 26 & 3 & 7 & 4\\ 10 & 1 & 3 & 2\\ \end{vmatrix}= \begin{vmatrix} 5\cdot2 & 2 & 1 & 3\\ 5\cdot4 & 4 & 5 & 6\\ 5\cdot3 & 3 & 7 & 4\\ 5\cdot1 & 1 & 3 & 2\\ \end{vmatrix}= 5\begin{vmatrix} 2 & 2 & 1 & 3\\ 4 & 4 & 5 & 6\\ 3 & 3 & 7 & 4\\ 1 & 1 & 3 & 2\\ \end{vmatrix}=0 $$

Of course, much easier, short and clear, after you wrote your first equality sign, is to remark that the determinant is zero directly as

$$C_1=5C_2+C_3+C_4\;,\;\;C_i=i-\text{th column}$$

so that the matrix's rank isn't full and etc.

Answer 2

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).