Deteminant of a matrix If $P= \left(\begin{array}{ll}
    A & B \\
    C & D \\
  \end{array}\right)$,  where A, B, C, D are tridiagonal  matrices. Then how to define its determinant ?

See Determinants of Block Matrices.  
A block matrix is a matrix that is defined using smaller matrices, called blocks. For example,
$P= \left(\begin{array}{ll}
    A & B \\
    C & D \\
  \end{array}\right)$
where $A$, $B$, $C$, and $D$ are themselves matrices, is a block matrix. In the specific example
$A= \left(\begin{array}{ll}
    a_{(1,1)} & a_{(1,2)} \\
    a_{(2,1)} & a_{(2,2)} \\
  \end{array}\right)$, $B= \left(\begin{array}{ll}
    b_{(1,1)} & b_{(1,2)}& b_{(1,3)} \\
    b_{(2,1)} & b_{(2,2)}& b_{(2,3)} \\
  \end{array}\right)$, $C= \left(\begin{array}{ll}
    c_{(1,1)} & c_{(1,2)} \\
    c_{(2,1)} & c_{(2,2)} \\
    c_{(3,1)} & c_{(3,2)} \\ 
  \end{array}\right)$, $D= \left(\begin{array}{ll}
    d_{(1,1)} & d_{(1,2)}& d_{(1,3)} \\
    d_{(2,1)} & d_{(2,2)}& d_{(2,3)} \\
    d_{(3,1)} & d_{(3,2)}& d_{(3,3)} \\ 
  \end{array}\right)$
Therefore, it is the matrix $P$
$$P= \left(\begin{array}{ll}
 p_{(1,1)} & p_{(1,2)} & \dots & p_{(1,n)} \\
p_{(2,1)} & p_{(2,2)} & \dots & p_{(2,n)} \\
\vdots &  \vdots & \ddots & \vdots \\
p_{(n,1)} & p_{(n,2)} & \dots & p_{(n,n)} 
  \end{array}\right)=
 \left(\begin{array}{ll}
    a_{(1,1)}  & a_{(1,2)} & b_{(1,1)} & b_{(1,2)}& b_{(1,3)}\\
    a_{(2,1)} & a_{(2,2)} &    b_{(2,1)} & b_{(2,2)}& b_{(2,3)} \\
    c_{(1,1)} & c_{(1,2)} &  d_{(1,1)} & d_{(1,2)}& d_{(1,3)} \\
    c_{(2,1)} & c_{(2,2)} &  d_{(2,1)} & d_{(2,2)}& d_{(2,3)} \\
    c_{(3,1)} & c_{(3,2)} &  d_{(3,1)} & d_{(3,2)}& d_{(3,3)} \\ 
  \end{array}\right)$$
See Blok Matrix
Where the determinant of a matrix $P$ of arbitrary size $n \times n$ can be defined by the a Leibniz formul or the Laplace formula.
