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Use a cofactor expansion across a row or column to find the determinant of the following matrix

$$B=\begin{pmatrix}1 &c&0&0\\0&1&c&0\\0&0&1&c\\c&0&0&1\end{pmatrix}$$

Clearly indicate the steps you take.

I have tried

$$\det B = 1 \det \begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}+(-c) \det \begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}$$ $$ = \det\begin{pmatrix}1 &c\\0&1\end{pmatrix}+c(-c)det\begin{pmatrix}c &0\\1&c\end{pmatrix}$$ $$=1-c^4$$

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    $\begingroup$ What have you tried so far? Where did you stuck? We are not here to do your homework. $\endgroup$ – Mundron Schmidt Aug 13 '17 at 10:28
  • $\begingroup$ I've edited the post, forgot to include what i've attempted $\endgroup$ – Ben Jones Aug 13 '17 at 10:35
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    $\begingroup$ @BenJones: Your solution is correct. $\endgroup$ – levap Aug 13 '17 at 10:38
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    $\begingroup$ If you see the matrix $B$ as a block matrix, you can find it's determinant easily. $\endgroup$ – Amin235 Aug 13 '17 at 10:38
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Expanding $\det B$ on first column we have that

$$\det B=1\det\begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}-c\begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}=1-c^4$$

The determinant of triangular matrix is the product of the diagonal entries

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Viewing $\rm B$ as a block matrix,

$$\det \left[\begin{array}{ccc|c} 1 & c & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & c\\ \hline c & 0 & 0 & 1\end{array}\right] = \det \left( \begin{bmatrix} 1 & c & 0\\ 0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} - \begin{bmatrix} 0\\ 0\\ c\end{bmatrix} \begin{bmatrix} c\\ 0\\ 0\end{bmatrix}^\top \right) = \det \begin{bmatrix} 1 & c & 0\\ 0 & 1 & c\\ -c^2 & 0 & 1\end{bmatrix} = \color{blue}{1 - c^4}$$

Viewing $\rm B$ as a circulant matrix, its eigenvalues are given by

$$\lambda_k = 1 + c \, \exp \left( i \frac{k \pi}{2} \right)$$

for $k \in \{0, 1, 2, 3\}$. Hence, the determinant of $\rm B$ is

$$\det \begin{bmatrix} 1 & c & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & c\\ c & 0 & 0 & 1\end{bmatrix} = (1 + c) (1 + i c) (1 - c) (1 - i c) = (1 - c^2) (1 + c^2) = \color{blue}{1 - c^4}$$

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