# Alternative methods in calculating determinant of matrix

I found literature that calculated a matrix without showing method of doing it. Using conventional approach, my method does not match with literature. In literature, for the following matrix:

$M = \begin{bmatrix} a & b & 0 & d \\ b & c & e & 0 \\ 0 & g & i & h \\ f & 0 & h & j\end{bmatrix}$

The determinant of M is found to be

$( aj - df)(ci-eg) - X$

and X are other terms. I am aware that results above can be calculated conventional way and probably extrapolated to yield the above. Is there another way of doing determinant that yields the above instead of doing conventional way such as:

$M = a\begin{bmatrix} c & e & 0 \\ g & i & h \\ 0 & h & j\end{bmatrix} - b \begin{bmatrix} b & e & 0 \\ 0 & i & h \\ f & h & j\end{bmatrix}- d\begin{bmatrix} b & c & e \\ 0 & g & i \\ f & 0 & h\end{bmatrix}$

In general, there are many ways to calculate determinants. Special matrices allow to use special methods. For an $n$ by $n$ matrix, let us regard element indices modulo $n$. Thus, the main diagonal is the elements whose index difference is zero. Other diagonals have constant index difference other than zero. The determinant of a matrix with just one diagonal is the product of the elements in that diagonal with a plus or minus sign. The matrix $M$ you have is a tridiagonal one. The main diagonal is $(a, c, i, j)$, and next upper and lower diagonals are $(b, e, h, f)$ and $(b, g, h, d)$. The determinant of $M$ is the sum of two products of two $2$ by $2$ submatrices minus the three diagonal products: $$(a j-d f)(c i-e g) + (a c-b b)(i j-h h) - (a c i j) - (b e h f) - (b g h d).$$ This can be generalized but the results may not be worth pursuing.

• Thanks for answering my question. It solves my problem. Do you know any literature on this type of method you used? I like to learn more about this stuff. It is a very interesting method, – Aschoolar May 23 '17 at 14:49
• The en.wikipedia.org/wiki/Tridiagonal_matrix article mentions that the determinant of tridiagonal matrices has a linear recursion. Your matrix is almost tridiagonal except for the last entry in the first row and the first entry in the last row, so the recursion has to be modified. – Somos May 23 '17 at 15:32

In fact, the matrix you have is a block matrix, let $X$ be a block matrix of size $2n$ for example, and $A,B,C,D$ matrices of size $n$:

$X=\begin{bmatrix} A & B\\C & D \end{bmatrix}$

Suppose $D$ is invertible, then:

$$det(X)=det(\begin{bmatrix} A & B\\C & D \end{bmatrix})=det(\begin{bmatrix} A-CD^{-1}B & 0\\C & D \end{bmatrix})=det(A-CD^{-1}B)det(D)$$

If, in addition, $CD=DC$, then you can conclude that:

$det(X)=det(DA-CB)$