TITLE: Justify, without evaluating, that the determinant of the following matrix is zero
QUESTION [18 upvotes]: I am currently stuck at this question and have no idea how to solve.
I just started out learning linear and I'm really weak in this field.

Justify, without evaluating, that the determinant of the following matrix is zero

Here's the matrix A:
$$\begin{pmatrix}
     1 &  0  &  2 &  4\\
    -2 & 3 &   8 &  6\\
    -1 &  3 &  10 & 10\\
     6 &  6 &  -3 &  7\\
\end{pmatrix}$$ 
I searched online but couldn't find something similar.
What I found though was that if it was skew-symmetric ($A^t= -A$)
then the determinant could directly be said to be equal to zero.
But in this case it didn't work with me.
Thank you.

REPLY [2 votes]: The easiest solution is by far the one from @A.Goodier.
Still, if you're stuck, you could try to solve
$$
\left\{ 
\begin{array}{c}
a &+ 0b &+ 2c &+ 4d&=0 \\ 
-2a &+ 3b&+8c&+6d&=0 \\
-a&+ 3b&+10c&+10d&=0 \\
6a&+ 6b&-3c&+7d&=0
\end{array}
\right. 
$$
which is the equivalent of trying to find the vectors $V$ for which $ M * V = 0$.
If you find a solution $V$ which isn't the zero vector, it means the columns are linearly dependent and the determinant of the matrix is $0$.
Solving the equation, you find that:
$$
\left\{ 
\begin{array}{c}
b &= \frac{a}{33} \\
c &= \frac{15 a}{22} \\
d &= -\frac{13 a}{22}\\ 
\end{array}
\right. 
$$
For example, with $a = 66$, you find
$$
V=
    \begin{bmatrix}
    66 \\
    2 \\
    45 \\
    -39
    \end{bmatrix}
$$