Evaluate the Determinants A? Evaluate the  Determinants 

$$A=\left(\begin{matrix} 
  1 & 1 & 0 & 0 & 0\\
  -1 & 1 & 1 & 0 & 0\\
  0 & -1& 1 & 1 & 0\\
  0& 0 & -1 & 1 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$$

My attempst :    I  was  thinking abouts the 
Schur complement https://en.wikipedia.org/wiki/Schur_complement
$\det \begin{pmatrix}
A&B\\
C&D
\end{pmatrix}= \det (A-BD^{-1}C)\det D
$
As i  not getting  How  to applied  this  formula  and finding the Determinant of the Given question,,
Pliz help me  or is  There another  way  to find  the  determinant of the matrix.
Thanks u 

Given $A=\left(\begin{matrix} 
  1 & 1 & 0 & 0 & 0\\
  -1 & 1 & 1 & 0 & 0\\
  0 & -1& 1 & 1 & 0\\
  0& 0 & -1 & 1 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$ 
To find the determinant you need to find the upper triangular matrix and then multiply the diagonal elements of the matrix.
$$=\begin{vmatrix} 
  1 & 1 & 0 & 0 & 0\\
  -1 & 1 & 1 & 0 & 0\\
  0 & -1& 1 & 1 & 0\\
  0& 0 & -1 & 1 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_2->R_2+R_1}$$
$$=\begin{vmatrix} 
  1 & 1 & 0 & 0 & 0\\
  0 & 2 & 1 & 0 & 0\\
  0 & -1& 1 & 1 & 0\\
  0& 0 & -1 & 1 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_3->R_3+\dfrac12R_1}$$
$$=\begin{vmatrix} 
  1 & 1 & 0 & 0 & 0\\
  0 & 2 & 1 & 0 & 0\\
  0 & 0& \dfrac32 & 1 & 0\\
  0& 0 & -1 & 1 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_4->R_4+\dfrac23R_3}$$
$$=\begin{vmatrix} 
  1 & 1 & 0 & 0 & 0\\
  0 & 2 & 1 & 0 & 0\\
  0 & 0& \dfrac32 & 1 & 0\\
  0& 0 & 0 & \dfrac53 & 1\\
  0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_5->R_5+\dfrac35R_4}$$
$$=\begin{vmatrix} 
  1 & 1 & 0 & 0 & 0\\
  0 & 2 & 1 & 0 & 0\\
  0 & 0& \dfrac32 & 1 & 0\\
  0& 0 & 0 & \dfrac53 & 1\\
  0 & 0 & 0 & 0 & \dfrac85\\
\end{vmatrix}_{R_5->R_5+\dfrac35R_4}$$
Now the determinant $=1\times 2\times \dfrac32\times \dfrac53 \times \dfrac85=8$
