How to find the adjoint matrix A of the 3rd order matrix 15

Use the algebraic coundroller or the adjoint matrix of the equation a =|a|*a^-1a^*=

Let's start with the concept of "algebraic courishin":

Let d be an nth-order determinant and aij (i, j are the lower corners) be the element on row i, column j in d. In d.

After the i-th and j-columns of AIJ are crossed out, the remaining n-1 determinant is called the "coincidental" of the element AIJ, which is denoted as mij. Put aij = (-1) (i+j) *

The mij is called the "algebraic coundant" of the element aij. symbol for power operation).

First, find the covariant formulas for each algebra.

a11 = (-1)^2 * a22 * a33 - a23 * a32) = a22 * a33 - a23 * a32

a12 = (-1)^3 * a21 * a33 - a23 * a31) = -a21 * a33 + a23 * a31

a13 = (-1)^4 * a21 * a32 - a22 * a31) = a21 * a32 - a22 * a31

a21 = (-1)^3 * a12 * a33 - a13 * a32) = -a12 * a33 + a13 * a32

a33 = (-1)^6 * a11 * a22 - a12 * a21) = a11 * a22 - a12 * a21

And then the adjoining matrix is.

a11 a21 a31

a12 a22 a32

a13 a23 a33 adjoint matrix=

Find the first cross-out first row and the first column 2*3-4*1 2 the second 2*3-1*3 3 and the third 2

So the first column.

Then find the second line... Get the second column.

Third column. So the accompanying array is.

The invertible matrix has the formula a*=laia -1=2a -1, and brings into the original formula i-3 2*a -1l = (-3 2) 3*la -1l.

A1(B2·C3-B3·C2) -A2(B1C3-B3·C1) +A3(B1·C2-B2·C1) uses the determinant operation: that is, the determinant is equal to each number in its first row multiplied by its coponder, or equal to each number in the first column multiplied by its cognant, and then the sum is calculated after adding a sign to each item according to the law of +.

Any line or column – algebraic remainder:

The covariant of an element of the determinant: the determinant crosses out the elements of the row and column where the element is located, and the remaining elements are arranged as they are, resulting in a new determinant.

Algebraic coundifier of an element of a determinant: The product of the covariant of an element of the determinant and the positive and negative signs corresponding to that element.

That is, the determinant can be the sum of the product of the elements and their corresponding algebraic remainders in a row or column.

The inverse of the matrix is equal to the adjoint matrix divided by the determinant of the matrix, so now only the determinant of the original matrix is required.

a^*=a^(-1)|a|, both sides take the determinant at the same time.

a^*|a|2 (because it is a third-order matrix).

Again|a^*|4,|a|>0, so |a|=2

So a (-1) = a (*2, which is the adjoint matrix divided by 2.

Special Methods: 1) When the matrix is greater than or equal to the second order:

The main diagonal element is to remove the row and column of the element in the original matrix and then find the determinant, and the non-main diagonal element is the element of the conjugate position of the element in the original matrix to remove the column and find the determinant multiplied by <>

x,y is the ordinal number of the row and column of the element at the conjugate position of the element, starting from 1. The primary diagonal element is actually a special case of a non-primary diagonal element, and since x=y, it <>

It's always a positive number, and there's no need to think about the notation of the main diagonal element.

2) When the order of the matrix orange is equal to the first order, the adjoint matrix is a square matrix of first-order units.

3) The formula for finding the second-order matrix: the main diagonal elements are swapped, and the secondary diagonal elements are added with a negative sign.

Method 1: Using the definition of adjoint matrix, first find the corresponding algebraic remainder of each element, and then transpose Method 2: use the adjoint matrix (only in the case of invertible matrices), the relationship with the determinant and the inverse matrix

Find the determinant |a|

Then use the elementary row transformation to find the inverse matrix.

According to the formula.

Special Finding.

1) When the matrix is greater than or equal to the second order:

The main diagonal element is to remove the row and column of the element in the original matrix and then find the determinant, and the non-main diagonal element is the element of the conjugate position of the element in the original matrix to remove the column and find the determinant multiplied by <>

The ordinal number of the row and column of the element at the conjugate position of the element, starting with 1. The main diagonal element is actually a special case of a non-primary diagonal element.

Because so. It's always a positive number, and there's no need to think about the notation of the main diagonal element.

2) When the order of the matrix is equal to the first order, the adjoint matrix is a square matrix of first-order units.

3) The formula for finding the second-order matrix: the main diagonal elements are interchanged, and the secondary diagonal elements are changed.

Use the algebraic coundroller or the adjoint matrix of the equation a =|a|*a -1a *=1 -2 70 1 -20 0 1 First of all, the concept of "algebraic coundations" is introduced

Let d be an nth-order determinant and aij (i, j are the lower corners) be the element on row i, column j in d. In d.

After the i-th and j-columns of AIJ are crossed out, the remaining n-1 determinant is called the "coincidental" of the element AIJ, which is denoted as mij. Put aij = (-1) (i+j) *

The mij is called the "algebraic coundant" of the element aij. The symbol denotes the power operation) First find the covariant of each algebra a11 = (-1) 2 * a22 * a33 - a23 * a32) = a22 * a33 - a23 * a32 a12 = (-1) 3 * a21 * a33 - a23 * a31) = -a21 * a33 + a23 * a31 a13

-1)^4 * a21 * a32 - a22 * a31) = a21 * a32 - a22 * a31 a21 = (-1)^3 * a12 * a33 - a13 * a32)

a12 * a33 + a13 * a32 ……a33 = (-1) 6 * a11 * a22 - a12 * a21) = a11 * a22 - a12 * a21 Then the adjoint matrix is a11 a21 a31 a12 a22 a32 a13 a23 a33

Adjoint matrix = 1 -2 -10 1 20 0 1

Finally, the matrix formed by the algebraic remainder formula needs to be transposed.

3a)[(3a)^(1)-2a*]=e-6|a|e=e-3e=-2e

3a)^(1)-2a*|=2e|/|3a|=-8 (27 (1 2))=16 27,2,2,Woo Woo Woo T