The determinant of a square matrix is a single number.
The determinant is zero when the matrix has no inverse (singular).
sign as \( \det(A) \) or \( |A| \).(A is a square matrix)
\( det(A^{-1})=\frac{1}{det(A)} \)

\( P: \) permutation matrix
\( det(PA)=\pm det(A) \)
\( \Rightarrow \det(P)\det(A)=\det(L)\det(U) \)
\( \Rightarrow \det(A)=\pm\det(U) \)

example:
\( det \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 1 \), \( det \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = -1 \)

\( \det((PA)^T)=\det((LU)^T) \)
\( \rightarrow \det(A^T)\det(P^T) = \det(U^T)\det(L^T) \)
hint:
\( P \): permutation matrix
\( P^TP=I \)
\( P^{-1}=P^T \)
*** \( \det(A+B) \neq \det(A)+\det(B) \)
\( \det(AB)=\det(A)\cdot\det(B) \)
\( \det(A^T)=\det(A) \)
\( \begin{vmatrix} ta & tb \\ c & d \end{vmatrix} = t \begin{vmatrix} a & b \\ c & d \end{vmatrix} \)
example:
\( \begin{vmatrix} 4 & 8 & 8 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4 \begin{vmatrix} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)

*** \( \det(2A)\neq 2\det(A) \)
\( t\in R, A\in F^{m\times n}, m=n \)
\( \det(tA)=t^m\det(A) \)
\( \begin{vmatrix} a+a' & b+b' \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a' & b' \\ c & d \end{vmatrix} \)
example:
一次提一行
\( \begin{vmatrix} 4 & 8 & 8 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 8 & 8 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} + 8 \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4 \)

一次提一列
\( \begin{vmatrix} 4 & 8 & 8 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} = \begin{vmatrix} 4 & 8 & 8 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix} + \begin{vmatrix} 0 & 8 & 8 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} \Rightarrow \text{混合提純量} = 4 \begin{vmatrix} 1 & 8 & 8 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix} + 8 \begin{vmatrix} 0 & 1 & 1 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} = 28 \)

也可以分解一行或一列的向量
example:
\( \begin{vmatrix} 4 & 8 & 8 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} \)
第一列 \( \begin{bmatrix} 4 & 8 & 8 \end{bmatrix} = \begin{bmatrix} 3 & 6 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 5 \end{bmatrix} \)
所以可以分成
\( \begin{vmatrix} 4 & 8 & 8 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} = \begin{vmatrix} 3 & 6 & 3 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} + \begin{vmatrix} 1 & 2 & 5 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} \)

或者分解第2行
\( \begin{bmatrix} 8 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ 1 \\ \end{bmatrix} + \begin{bmatrix} 3 \\ 1 \\ 0 \\ \end{bmatrix} \)
則
\( \begin{vmatrix} 4 & 8 & 8 \\ 4 & 1 & 1 \\ 12 & 1 & 2 \end{vmatrix} = \begin{vmatrix} 4 & 5 & 8 \\ 4 & 0 & 1 \\ 12 & 1 & 2 \end{vmatrix} + \begin{vmatrix} 4 & 3 & 8 \\ 4 & 1 & 1 \\ 12 & 0 & 2 \end{vmatrix} \)

equal rows && zero rows

\( \det(A)=0 \)
example:
\( \begin{vmatrix} a & b \\ a & b \end{vmatrix} =0 \) , \( \begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix} =0 \) ,
有相同行的的行列式也等於0
\( \begin{vmatrix} f & f \\ g & g \end{vmatrix} =0 \)

without permutation, gauss elimination

\( \det(A) \) unchanged
example:
\( \ell \in R \)
\( \begin{vmatrix} a & b \\ c-a\ell & d-b\ell \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} +(-\ell) \begin{vmatrix} a & b \\ a & b \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} \)

triangular matrix

if \( A \) is triangular \( \rightarrow \) product of diagonal entries.
\( \begin{vmatrix} a & b \\ 0 & d \end{vmatrix} = ad \), \( \begin{vmatrix} a & 0 \\ c & d \end{vmatrix} = ad \)
\( \begin{vmatrix} a & b \\ c & d \end{vmatrix} =0 \leftrightarrow \text{singular} \)
\( |AB|=|A||B| \)

When the \( B \) is \( A^{-1} \), this rule says that the determinant of \( A^{-1} \) is \( \frac{1}{\det A} \)
\( \det (AA^{-1})= (\det A)(\det A^{-1}) =\det I =1 \)

determinant method