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

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.
