I received a matrix for which I need to calculate its determinant. $$ A = \begin{pmatrix} 0 & 1 & 1 & \cdot & \cdot & \cdot & 1 \\ 1 & 0 & 1 & 1 & \cdot & \cdot & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 1 & 1 & \cdot & \cdot & 1 & 0 & 1 \\ 1 & 1 & \cdot & \cdot & \cdot & 1 & 0 \\ \end{pmatrix} \in \operatorname{Mat}_n (\mathbb{R}) $$
My calculation procedure is as follows: $$\begin{align} A \cong & \prod^{n-1}_{i=0} E_{i,n,-1} \cdot \begin{pmatrix} 0 & 1 & 1 & \cdot & \cdot & \cdot & 1 \\ 1 & 0 & 1 & 1 & \cdot & \cdot & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 1 & 1 & \cdot & \cdot & 1 & 0 & 1 \\ 1 & 1 & \cdot & \cdot & \cdot & 1 & 0 \\ \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 & \cdot & \cdot & 0 & 1 \\ 0 & -1 & 0 & 0 & \cdot & 0 & 1 \\ 0 & 0 & -1 & 0 & 0 & 0 & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot & \cdot & 0 & -1 & 1 \\ 1 & 1 & \cdot & \cdot & \cdot & 1 & 0 \\ \end{pmatrix} = A_1 \\ \\ \\ A_1 \cong & \prod^{n-1}_{i=0} E_{n,i,1} \cdot A_1 = \begin{pmatrix} -1 & 0 & 0 & \cdot & \cdot & 0 & 1 \\ 0 & -1 & 0 & 0 & \cdot & 0 & 1 \\ 0 & 0 & -1 & 0 & 0 & 0 & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot & \cdot & 0 & -1 & 1 \\ 0 & 0 & \cdot & \cdot & \cdot & 0 & n - 1 \\ \end{pmatrix} = A_2 \end{align}$$
Explanation of the above:
- $E$ is an elementary matrix.
The sign $E_{i,j,\lambda}$ means "Add to row number #i the row number #j multiplied by $\lambda$". - $n$ is the matrix size.
The matrix $A_2$ is equivalent ("row-equivalent") to $A$, using only the "Row addition" operation , therefore their determinants are equal.
We can see that matrix $A_2$ is an upper triangular matrix, therefore its determinant is equal to the multiplication of the items in the main diagonal: $(-1)^{n-1} \cdot (n-1)$.
Please confirm my solution. Also, do you see any easier way for solving this?
You could try to diagonalize it. Let $J = \begin{pmatrix} 1 & \cdots & 1 \\ 1 & ... &1 & \\ 1 & \cdots & 1 \\ \end{pmatrix} $
Then $A=J-Id$
$rgo(J)=1$ so $0$ is eigenvalue with multiplicity $n-1$
$J\cdot(1,...,1)^t=(n,...,n)^t$ so $n$ is eigenvalue too.
Then $J \simeq \begin{pmatrix} 0 & \cdots &0& 0 \\ 0 & ... & 0 &0 & \\ 0 & \cdots & 0 & n \\ \end{pmatrix} $ so $ A = J-Id \simeq \begin{pmatrix} -1 & \cdots &0& 0 \\ 0 & ... & -1 &0 & \\ 0 & \cdots & 0 & n-1 \\ \end{pmatrix} $
so $\det(A)=(-1)^{n-1}(n-1)$