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.
 A: Notice that $$\textrm{Row}_1 + 2\textrm{Row}_5 = 2\textrm{Row}_2 + \textrm{Row}_6$$
Thus the rows do not form a linearly independent set.
A: An $n\times n$ matrix is singular (and hence has a zero determinant) if and only if its rows (or columns) form a linearly dependent set. If $n$ is larger than $2$, linear independence of the rows is a stronger condition than just saying that no row is a scalar multiple of another row. For instance, the matrix
$$ \begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix} $$
is singular because $(7,8,9)=2\cdot (4,5,6)-(1,2,3)$, but no row is a scalar multiple of another row.
A: What you are are going to want to do to find this determinant (without going about it with cofactor expansion which though possible, is very time consuming) is to put the matrix in Reduced Echelon Form (Not Row-Reduced Echelon form [difference is explained here]) and then multiply the elements down the diagonal. In so doing you will find that $Det(M)=0$.
So finding the  determinant may look like this:
(I used this calculator to put this in REF. Instructions are printed how to do it with manual calculations once values are inputted) 
$$
  \left(
   \begin{array}{ c c c c c c}
      1& 0& 0& \mathit{1}& 0& 0\\
    0.5& 0.5& 0& 0.5& \mathit{0.5}& 0 \\ 
    0.5& 0& 0.5& 0.5& 0& \mathit{0.5} \\
    \mathit{0.25}& 0.5& 0.25& 0& 1& 0 \\ 
    0 & \mathit{0.5}& 0.5& 0& 0.5& 0.5 \\ 
    0& 0& \mathit{1}& 0& 0& 1\\
   \end{array}
   \right)
\longrightarrow REF \longrightarrow \left(\begin{array} \\ 
1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1\\
0 & 0 & 0 & 1 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{array}\right) $$
Now, just simply multiply the diagonal and you will see that $Det(M)=0$. 
Hope this helped.
A: There is a faster way to see that the matrix is singular if you are familiar with determinants of block matrices. Partition your matrix into four sub-blocks of equal sizes. Then it is in the form of $M=\pmatrix{A&A\\ B&C}$. Since the top two sub-blocks commute, we get $\det M=\det(CA-BA)=\det(C-B)\det(A)$. Yet $C-B$ clearly has two zero rows. Therefore its determinant and in turn the determinant of your matrix are zero.
Put it another way, if you subtract the first three columns of your matrix from the last three columns, you get a block-lower triangular matrix whose lower diagonal block has two zero rows. Therefore lower diagonal block is singular and your matrix is singular too.
A: 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.
