Evaluate the Determinants A?

Question

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

Answer 1

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 $=1\times 2\times \dfrac32\times \dfrac53 \times \dfrac85=8$

Answer 2

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?

Answer 3

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 $$\det T_n =\det T_{n-1}+\det T_{n-2}.$$

Answer 4

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.