i have the following homework problem:
Find the determinant of X(s):
X(s) = [s, 1, 1, 1], [1, s, 1, 1], [1, 1, s, 1], [1, 1, 1, s]
I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?
If i try regular row operations, i get: X(s) = [s, 1, 1, 1], [0, s-1, 1, 1], [0, 0, s-1, 1], [1-s, 0, 0, s-1]
This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?
Here's something simple: first exploit linearity of the determinant w.r.t. rows: $$ D=\begin{vmatrix} s&1&1&1\\1&s&1&1\\1&1&s&1\\1&1&1&s \end{vmatrix}= \begin{vmatrix} s-1&0&0&1-s\\0&s-1&0&1-s\\0&0&s-1&1-s\\1&1&1&s \end{vmatrix}= (s-1)^3 \begin{vmatrix} 1&0&0&-1\\0&1&0&-1\\0&0&1&-1\\1&1&1&s \end{vmatrix}$$ Next expand by the first row: $$D=(s-1)^3\left(1\cdot\begin{vmatrix} 1&0&-1\\ 0&1&-1\\1 &1&s \end{vmatrix}-(-1) \begin{vmatrix} 0&1&0\\0&0&1\\1&1&1 \end{vmatrix}\,\right)$$ Now the two $3\times3$ determinants are: \begin{align} &\begin{vmatrix} 1&0&-1\\ 0&1&-1\\1 &1&s \end{vmatrix}=\begin{vmatrix} 1&0&0\\ 0&1&0\\1 &1&s+2 \end{vmatrix}=s+2\\[1ex] &\begin{vmatrix} 0&1&0\\0&0&1\\1&1&1 \end{vmatrix}=1\qquad\text{(expanding by the first column)} \end{align} and ultimately, we obtain $$D=(s-1)^3(s+3).$$