Is there an easy way to compute the determinant of matrix with 1's on diagonal and a's on anti-diagonal? \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Thanks
 A: We have the matrix
$$A= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$
Elininating the $a$s below the diagonal by adding multiples of the first, second and third line of $A$, we obtain the upper triangular matrix $A'$:
$$A'= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & 0 & 1-a^2 & 0 & a-a^2 \\0 & 0 & 0 & 0 & 1-a^2 & 0 \\0 & 0 & 0 & 0 & 0 & 1-a^2 \\ \end{bmatrix}$$
Since adding a mulitple of a row to another row does not alter the determinant, we can say that $\det(A) = \det(A')$. Furthermore, the determinant of an upper triangular matrix is the product of its diagonal entries. Thus, we have:
$$\det(A) = \det(A') =  (1-a^2)^3$$
Alternatively, eliminate the $a$s above the diagonal by adding multiples of the fourth, fifth and sixth row. The determinant of a lower triangular matrix can be obtained in the same way.
A: Partition the matrix into blocks $\begin{bmatrix}A&B\\C&D\end{bmatrix}$, where 
$A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $B = \begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$, $C = \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}$, $D = \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix}$. 
The block matrix determinant formula states that 
$\det \begin{bmatrix}A&B\\C&D\end{bmatrix} = \det(A)\det(D-CA^{-1}B)$. 
Trivially, $A = I_3$ is the $3 \times 3$ identity matrix, so $\det(A) = 1$. Also, $A^{-1} = I_3$. 
Thus, $D-CA^{-1}B = D-CB$ $= \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix} - \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}\begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$ $= \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix} - \begin{bmatrix}a^2&0&a^2\\0&a^2&0\\0&0&a^2\end{bmatrix}$ $= \begin{bmatrix}1-a^2&0&a-a^2\\0&1-a^2&0\\0&0&1-a^2\end{bmatrix}$. 
Since $D-CA^{-1}B$ is upper-triangular, its determinant is simply the product of the diagonal elements, i.e. $\det(D-CA^{-1}B) = (1-a^2)^3$. 
Therefore, $\det \begin{bmatrix}A&B\\C&D\end{bmatrix} = \det(A)\det(D-CA^{-1}B) = 1 \cdot (1-a^2)^3 = (1-a^2)^3$.
