# Is there an easy way to compute this determinant

\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

-

## 2 Answers

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.

-

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, $\det \begin{bmatrix}A&B\\C&D\end{bmatrix} = \det(A)\det(D-CA^{-1}B) = 1 \cdot (1-a^2)^3 = (1-a^2)^3$.

-