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 
 A: After adding the first line to the second one, your matrix becomes$$\begin{pmatrix}1 & 1 & 0 & 0 & 0\\  0 & 2 & 1 & 0 & 0\\  0 & -1& 1 & 1 & 0\\  0& 0 & -1 & 1 & 1\\  0 & 0 & 0 & -1 & 1\end{pmatrix}$$and therefore$$\det A=\begin{vmatrix}2 & 1 & 0 & 0\\ -1& 1 & 1 & 0\\ 0 & -1 & 1 & 1\\ 0 & 0 & -1 & 1\end{vmatrix}=2\begin{vmatrix}1 & \frac12 & 0 & 0\\ -1& 1 & 1 & 0\\ 0 & -1 & 1 & 1\\ 0 & 0 & -1 & 1\end{vmatrix}.$$Can you take it from here?
A: This is a tridiagonal Toeplitz matrix with dimension $n=5$:
$$T_n=\begin{pmatrix}
a & b & 0 & 0 & 0 & 0 &\\
c  & a & b & 0 &  0 &  0\\
0 & c & a & b & 0 & 0  \\[-1ex]
& &\ddots & \ddots & \ddots \\[-1ex]
&&&\ddots& a& b \\
0 & 0  & 0 & & c & a
\end{pmatrix}$$
You can try to prove the recurrence formula
$$\det T_n =a\det T_{n-1}-bc\,\det T_{n-2},$$
which here yields $$\det T_n =\det T_{n-1}+\det T_{n-2}.$$
A: To do your method, you'd need to  'divide' your matrix into 4 submatrices (sections of your matrix) $A,B,C,D $. You then need to check and see if $D $ is invertible fir this to work. 
A: 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 $=1\times 2\times \dfrac32\times \dfrac53 \times \dfrac85=8$
