First feature: double vertical lines. (A bunch of numbers wrapped in double vertical lines)
$$ \left| \begin{array}{cccc} 1 & 2 \\ 3 & 4 \\ \end{array} \right| $$
Note that you can't write “()” or “[]”, incorrectly: $$ \left( \begin{array}{cccc} 1 & 2 \\ 3 & 4 \\ \end{array} \right) $$ or $$ \left[ \begin{array}{cccc} 1 & 2 \\ 3 & 4 \\ \end{array} \right] $$
Second feature: number of rows = number of columns
(must be equal; if not, it's not a determinant) Correct way to write:
$$ \left| \begin{array}{cccc} 1 & 2 & \\ 3 & 4 & \\ \end{array} \right| $$
Wrong way to write: $$ \left| \begin{array}{cccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{array} \right| $$ or $$ \left| \begin{array}{cccc} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array} \right| $$
Summarize: The above two features must be satisfied at the same time in order to become a determinant, you can't have one without the other.
A determinant is essentially a number. This means that the determinant is like an expression like (1+2), which is computed and the final result is a number.
Different determinants, with different shapes, are calculated differently. It is divided into:
First of all, to be clear, the concept of the diagonal inside the determinant: determinant from the upper left corner to the lower right corner of the diagonal line, called the positive diagonal. From the upper right corner to the lower left corner of the diagonal line, called the anti-diagonal. This means that every determinant has two diagonals. The line formed by the red numbers 1 and 4 in the following equation is the positive diagonal. The line formed by the green numbers 2 and 3 is called the antidiagonal.
$$ \left| \begin{array}{cccc} \color{red}{1} &\color{green}{2} \\ \color{green}{3} &\color{red}{4} \\ \end{array} \right| $$
Definition of upper triangular determinant: a determinant in which all the numbers on the lower-left side of the diagonal are 0s. In the following determinant, the red numbers 1, 5, and 9 form a diagonal, and the area where the blue number 0 is located is the lower left side of the diagonal. The upper triangle refers to the triangle formed by the numbers 1, 2, 3, 5, 6, and 9, which, as the name suggests, is called the upper triangle.
$$ \left| \begin{array}{cccc} \color{red}{1} &2 &3 \\ \color{blue}{0} &\color{red}{5} &6 \\ \color{blue}{0} &\color{blue}{0} &\color{red}{9} \end{array} \right| $$
More research, computation of upper trigonometric determinants (unfinished, to be updated 2024-04-24 9:14 am)... I'm back. Continue. The upper triangular determinant is calculated by directly multiplying the numbers on the diagonal. Example:
$$ \left| \begin{array}{cccc} \color{red}{1} &2 &3 \\ \color{blue}{0} &\color{red}{5} &6 \\ \color{blue}{0} &\color{blue}{0} &\color{red}{9} \end{array} \right| =1*5*9=45 $$
Definition of lower triangular determinant: A determinant in which all the numbers on the upper right side of the diagonal are 0. In the following determinant, the red numbers 1, 5, and 9 form a diagonal, and the area where the blue number 0 is located is the upper right side of the diagonal. The lower triangle refers to: 1, 4, 5, 7, 8, 9 form a triangle called, as the name suggests, the lower triangle.
$$ \left| \begin{array}{cccc} \color{red}{1} &\color{blue}{0} &\color{blue}{0} \\ 4 &\color{red}{5} &\color{blue}{0} \\ 7 &8 &\color{red}{9} \end{array} \right| $$
The lower triangular determinant is calculated by directly multiplying the numbers on the diagonal (exactly the same way as the upper triangular determinant).
$$ \left| \begin{array}{cccc} \color{red}{1} &\color{blue}{0} &\color{blue}{0} \\ 4 &\color{red}{5} &\color{blue}{0} \\ 7 &8 &\color{red}{9} \end{array} \right|=1*5*9=45 $$
Definition of Diagonal Determinant: A determinant in which all numbers except those on the diagonal are zero. In other words, the diagonal determinant is a combination of the upper and lower triangular determinants. Note: It doesn't matter if there are 0's in the numbers on the diagonal itself, as long as all the numbers other than the diagonal are 0's the determinant is a diagonal determinant. The diagonal determinant is calculated by multiplying the numbers on the diagonal (exactly the same as the upper and lower triangular determinants).
The following determinant, the red numbers form a diagonal line, the blue number 0 where the region are 0, as for the diagonal line there is no 0 does not matter, so the following two are diagonal determinant
$$ \left| \begin{array}{cccc} \color{red}{1} &\color{blue}{0} &\color{blue}{0} \\ \color{blue}0 &\color{red}{5} &\color{blue}{0} \\ \color{blue}0 &\color{blue}{0} &\color{red}{9} \end{array} \right| $$ 、 $$ \left| \begin{array}{cccc} \color{red}{1} &\color{blue}{0} &\color{blue}{0} \\ \color{blue}0 &\color{red}{5} &\color{blue}{0} \\ \color{blue}0 &\color{blue}{0} &\color{red}{0} \end{array} \right| $$
The diagonal determinant is calculated by multiplying the numbers on the diagonal directly (exactly the same way as the upper triangular determinant and the lower triangular determinant).
$$ \left| \begin{array}{cccc} \color{red}{1} &\color{blue}{0} &\color{blue}{0} \\ \color{blue}0 &\color{red}{5} &\color{blue}{0} \\ \color{blue}0 &\color{blue}{0} &\color{red}{0} \end{array} \right|=1*5*0=0 $$
Definition of antidiagonal determinant: a determinant in which all numbers except those on the antidiagonal are 0 (it doesn't matter if there are 0's in the antidiagonal). The following determinant, the red numbers 3, 5 and 7 form the antidiagonal, and the blue number 0 is in the region of 0, which meets the definition, so it is an antidiagonal determinant.
$$\left| \begin{array}{cccc} \color{blue}{0} &\color{blue}{0} &\color{red}{3} \\ \color{blue}{0} &\color{red}{5} &\color{blue}{0} \\ \color{red}{7} &\color{blue}{0} &\color{blue}{0} \end{array} \right| $$
But the calculation of the antiangular determinant is not the same as the previous calculation. I was told to go to work and then continue to write as I learned after work 。。。。 I'm back. The calculation of the antiangular determinant is as follows.
$$\left| \begin{array}{cccc} \color{blue}{0} &\color{blue}{0} &\color{red}{3} \\ \color{blue}{0} &\color{red}{5} &\color{blue}{0} \\ \color{red}{7} &\color{blue}{0} &\color{blue}{0} \end{array} \right| $$ $$ = (-1)^{n*(n-1)/2}*3*5*7 $$ $$ = (-1)^{3*(3-1)/2}*3*5*7 $$ $$ = -105 $$
Are you smart enough to spot the calculation patterns? Look carefully and look more. Look further down. Where n is the number of rows or columns in the determinant, followed by 3, 5, and 7 which are the numbers on the opposite diagonal. Why do you calculate this way? Let's just wonder, and then we'll move on, and one day we'll figure it out.
There's a paragraph here that needs explaining, so leave it blank and move on.
The determinant with two rows and two columns is also known as the second-order determinant and can be calculated directly using the formula. The formula is:
$$ \left| \begin{array}{cccc} a & b \\ c & d \\ \end{array} \right|=ad-bc $$
Examples:$$ \left| \begin{array}{cccc} 1 & 2 \\ 3 & 4 \\ \end{array} \right|=1*4-2*3=-2 $$
A three-row, three-column determinant is also known as a third-order determinant, and similar to a second-order determinant, it can be computed directly using the formula for:
$$ \left| \begin{array}{cccc} a & b & c\\ d & e & f \\ g & h & i \\ \end{array} \right| $$ =aei+bfg+cdh-ceg-afh-bdi
How's that for a formula that's a little hard to remember? I'll tell you the law: there are six terms on the right side of the equation, the first three are positive, the last three are negative. Let's start with the first three terms. The first term, aei, is the multiplication of the numbers on the positive diagonal. The second bfg is to remove the aei only six numbers, and then from these six numbers to find three different rows and columns of numbers multiplied by the formation of bfg. third cdh is based on the previous, remove the aei, remove bfg, the remaining three numbers multiplied by the formation of cdh. then say the last three, the first ceg is the anti-diagonal multiplication of the numbers. The second term, afh, is the result of removing ceg, leaving only six numbers, and then multiplying those six numbers by three numbers in different rows and columns to form afh. The third term, bdi, is the result of removing ceg, removing afh, and multiplying the remaining three numbers to form bdi. Note that the last three terms have a minus sign in front of them. Please be sure to practice. (To understand this formula, please practice more on paper, later I will add animation to help illustrate).
example:
$$
\left|
\begin{array}{cccc}
1 & 2 & 3\\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{array}
\right|
$$
=aei+bfg+cdh-ceg-afh-bdi
=1*5*9+2*6*7+3*4*8-3*5*7-1*6*8-2*4*9
=0
The second-order determinant and third-order determinant learned earlier can be calculated by the formula, but there is no formula for determinants greater than the third order. Then how to calculate it, the idea is to make the complex into simple, the unknown into known, the high price into low order, the fourth order into the third or second order. Calculation method is: descending order method (also called expanding by ranks). Note: the fourth order or more than four orders can only use the descending order method, while the third order can use both the formula method, you can also use the descending order method, the third order down to the second order.
Example: using the step-down method to reduce a third order to a second order. I'm a bit lazy and won't demonstrate the fourth order. Step 1: Select any row or column from the determinant. At this point there are six options: you can choose the first line 1,0,-2; you can choose the second line 1,1,3; you can choose the third line -2,3,1; you can choose the first column 1,1,-2; you can choose the second column 0,1,3; you can choose the first column -2,3,1. From the six kinds of selection of any choice can be selected, because no matter what kind of selection, the final answer is the same. For example, we'll choose the first row, i.e. 1,0,2. Step 2: Follow the descending set, which is Coefficients * (-1)^Exponents * Determinants. Since it's 3 rows and 3 columns, there are 3 coefficients and 3 determinants, so the result is First coefficient*(-1)^first exponent *first determinant + Second coefficient*(-1)^second exponent *second determinant + Third coefficient*(-1)^third exponent *third determinant. Where (-1) is the fixed part of the set, this is just remembered, (-1) precedes Since 1 is in the first row and first column, the first coefficient is 1, and the first exponent is 1+1. Since 0 is in the first row and second column, the second coefficient is 0, and the second exponent is 1+2. Since -2 is in the third column of the first row, the third coefficient is -2, and the exponent is 1+3. Then: 1. 3, 3, 1; 1, 3, -2, 1; 1,1,-2,3; are the determinants left after removing the rows and columns where 1, 0, and -2 are located, respectively. (Here please read a few more times, if it does not make you understand, wait for me to demonstrate later with animation.)
$$ \left| \begin{array}{cccc} 1 & 0 & -2\\ 1 & 1 & 3 \\ -2 & 3 & 1 \\ \end{array} \right| $$ $$ = 1\times \left ( -1 \right ) ^{1+1} \times \begin{bmatrix} 1 & 3 \\ 3 & 1\end{bmatrix}$$ $$+ 0\times \left ( -1 \right )^{1+2} \times \begin{bmatrix} 1 & 3\\ -2 & 1 \end{bmatrix} $$ $$+ \left ( -2 \right )\times \left ( -1 \right ) ^{1+3} \times \begin{bmatrix} 1 & 1\\ -2 & 3 \end{bmatrix} $$ $$ =-8+0-10 $$ $$ =-18 $$
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