Determinant and its properties

Now let's move on to the second part of the course —— determinant , This section will focus on the determinant of the equation .

One 、 The nature of determinants

A determinant is a number corresponding to a matrix , It reflects the properties of the square matrix .

• nature 1： Unit matrix $I$,$|I|=1$
• nature 2： After swapping two lines , The value sign of determinant is opposite

for instance ：

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=1\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=-1$$
For one $2\times 2$ Matrix , The way to calculate the determinant is to multiply the main diagonal elements and then subtract them ：
$$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$$

• nature 3 The common multiple of a row can be outside the determinant
  $$\left|\begin{array}{cc}ta& tb\\ c& d\end{array}\right|=t\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$$

• nature 4 Determinants can be added to each other ( The determinant is linear between lines )
  $$\left|\begin{array}{cc}a+a^{\prime }& b+b^{\prime }\\ c& d\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+\left|\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c& d\end{array}\right|$$

The following properties can be determined from the above 4 Personality quality export ：

• nature 5 The same row will make the determinant 0
  Utilization property 2 This property can be obtained . Because swapping peers will change their symbols , But the determinant is still the original determinant , The determinant should be the same , The only way to exchange symbols without changing them is 0 了 .

The same row causes the determinant to be 0, The same row makes the matrix irreversible .

• nature 6 Subtract multiples of one row from other rows , The value of the determinant does not change
  This property tells us that the elimination of a matrix does not change the value of its determinant .
  By nature 5 We can draw ：
  $$\left|\begin{array}{cc}a& b\\ c-la& d-lb\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+\left|\begin{array}{cc}a& b\\ -la& -lb\end{array}\right|$$
  Again by nature 4 There can be ：
  $$\left|\begin{array}{cc}a& b\\ c-la& d-lb\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|+\left|\begin{array}{cc}a& b\\ -la& -lb\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|-l\left|\begin{array}{cc}a& b\\ a& b\end{array}\right|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$$

• nature 7 Zero rows will cause the determinant to be 0
  This nature 3 Easy to come by , Put forward a 0 The coefficient is OK

• nature 8 The determinant of the upper triangular matrix is equal to the product of the diagonal
  $$det(U)=\left|\begin{array}{ccccc}{d}_1& \ast & \cdots & \ast & \ast \\ 0& d_2& \ast & \cdots & \ast \\ 0& 0& 0& \cdots & \ast \\ 0& 0& 0& 0& d_n\end{array}\right|=d_1{d}_2\cdots d_n$$
  This property is also very easy to understand , Utilization property 3 Extract the numbers from each line , Reuse properties 6 Put these asterisks “ get some soy sauce ”, And make use of the nature 1 You can get ：
  $$d_1{d}_2\cdots d_n\left|\begin{array}{ccccc}1& \ast & \cdots & \ast & \ast \\ 0& 1& \ast & \cdots & \ast \\ 0& 0& 0& \cdots & \ast \\ 0& 0& 0& 0& 1\end{array}\right|=d_1{d}_2\cdots d_n$$

• nature 9 If and only if $A$ It's a singular matrix （ Irreversible , There are zero rows ）,$|A|$ zero ;$|A|\ne 0$ Matrices are nonsingular （ reversible , There are no zero lines ）

The first nine properties are all about determinants , Next, we will add two important properties about matrices ：

• nature 10 $|AB|=|A||B|$
  The determinant of the product of two square matrices is equal to the determinant of the two square matrices and then the product .
  inference 10.1: The determinant of an invertible matrix and the determinant of its inverse matrix are reciprocal
  $$|A^{-1}|=\frac{1}{|A|}$$
  inference 10.2:
  $$|A^n|=|A{|}^n$$

• nature 11 The determinant of a matrix is equal to the corresponding determinant of the matrix after its transpose
  prove ： Let this matrix be $A$, Make it $LU$ decompose
  $$|A^T|=|(LU{)}^T|=|U^T{L}^T|=|U^T||L^T|$$
  Because the transpose of the upper and lower triangular matrix does not affect the value of its determinant , therefore ：
  $$|A^T|=|U||L|$$
  That is to say ：
  $$|A|=|A^T|$$
  inference ： If a determinant has a column of zero vectors , So this determinant is 0. This will be used in the next section .