Why does this $6\times 6$ matrix has a null determinant? about this matrix:
$$
  \left(
   \begin{array}{ c c c c c c}
      1& 0& 0& \mathit{1}& 0& 0\\
    0.5& 0.5& 0& 0.5& \mathit{0.5}& 0 \\ 
    0.5& 0& 0.5& 0.5& 0& \mathit{0.5} \\
    \mathit{0.25}& 0.5& 0.25& 0& 1& 0 \\ 
    0 & \mathit{0.5}& 0.5& 0& 0.5& 0.5 \\ 
    0& 0& \mathit{1}& 0& 0& 1\\
   \end{array}
   \right)
$$
I have calculated its determinant by two techniques and they all give 0.
I do not understand why, because:


*

*There is no entire row equal to zero

*Two rows or columns are not equal

*No row either column is a constant multiple of another row or column (or I cannot see)

*There is diagonal line (noted in italic) without any 0


Does someone has a idea why?
thanks.

Looking at the first row, I notice that there are exactly two columns with an $1$ there, column 1 and column 4. Therefore my first thought was to take the difference of those:
$$\begin{pmatrix} 1\\0.5\\0.5\\0.25\\0\\0 \end{pmatrix}
- \begin{pmatrix} 1\\0.5\\0.5\\0\\0\\0 \end{pmatrix}
= \begin{pmatrix} 0\\0\\0\\0.25\\0\\0 \end{pmatrix}$$
Only a single entry remains. That's a good start! So let's see if we can get the same result from the other columns as well. Columns 2 and 5 indeed work:
$$\begin{pmatrix} 0\\0.5\\0\\0.5\\0.5\\0 \end{pmatrix}
- \begin{pmatrix} 0\\0.5\\0\\1\\0.5\\0 \end{pmatrix}
= \begin{pmatrix} 0\\0\\0\\-0.5\\0\\0 \end{pmatrix}$$
So now it's obvious that 2 times the first difference plus the second difference gives zero, therefore the determinant vanishes.
