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Given that matrix A and B satisfy the relation AB = a + 2B, where a = 423, find B110 - 1223
Because AB = a + 2B
So (a-2e) B = a
(A-2E,A) =
4 2 3 1 0 0
1 1 0 0 1 0
-1 2 3 0 0 1
r1-4r2,r3+r2
0 -2 3 1 -4 0
1 1 0 0 1 0
0 3 3 0 1 1
r3*(1/3),r1+2r3,r2-r3
0 0 5 1 -10/3 2/3
1 0 -1 0 2/3 -1/3
0 1 1 0 1/3 1/3
r1*(1/5),r2+r1,r3-r1
0 0 1 1/5 -2/3 2/15
1 0 0 1/5 0 -1/5
0 1 0 -1/5 1 1/5
Exchange bank
1 0 0 1/5 0 -1/5
0 1 0 -1/5 1 1/5
0 0 1 1/5 -2/3 2/15
X =
1/5 0 -1/5
-1/5 1 1/5
1/5 -2/3 2/15
Given the matrix AB = a + 2B, find B
Matrix A=
0,3,3
1,1,0
-1,2,3
From ab = 2B + A, (a-2e) B = a
(A-2E,A) =
-2 3 3 0 3 3
1 -1 0 1 1 0
-1 2 1 -1 2 3
r1+2r2,r3+r2
0 1 3 2 5 3
1 -1 0 1 1 0
0 1 1 0 3 3
r2+r1,r3-r1
0 1 3 2 5 3
1 0 3 3 6 3
0 0 -2 -2 -2 0
r3*(-1/2),r1-3r3,r2-3r3
0 1 0 -1 2 3
1 0 0 0 3 3
0 0 1 1 1 0
Exchange bank
1 0 0 0 3 3
0 1 0 -1 2 3
0 0 1 1 1 0
B =
0 3 3
-1 2 3
1 1 0
Let a = {301 0014} and satisfy AB = a + 2B, find the matrix of B
Because AB = a + 2B, (a-2e) B = a (A-E, a) = 1013011 - 101001204r2-r111013010 - 1 - 1 - 21 - 101204r3 + R2, R2 * (- 1) 101301012 - 11001 - 223r1-r3, r2-r31005 - 2 - 20104 - 3 - 200
The proof of elementary transformation of matrix!
It is proved that the standard forms of matrices A and B are the same if and only if they are equivalent
Necessity: if a and B are equivalent, let the canonical form d of a be obtained by elementary transformation, then a and D are equivalent. According to the transitivity of equivalence, B and D are also equivalent, so D is also the canonical form of matrix B, that is, their canonical forms are the same
Sufficiency: if the standard forms of matrix A and B are the same, they are all D. then a and D are equivalent. Similarly, B and D are also equivalent. According to the transitivity of equivalence, a and B are also equivalent
Elementary transformation exercises of matrix
1 2 2
2 1 -2
2 -2 1
① R2-2 * R1, r3-2 * R1: 1220 - 3 - 60 - 6 - 3, ② & # 8531; R2, &# 8531; R3: 1220 - 1 - 20 - 2 - 1, ③ r3-2 * R2: 1220 - 1 - 2003, ④ R2 + R1: 12201003, ⑤ & # 8531; R3, r1-2 * R2: 10201001, ⑥ r1-2 * R3: 1001
I want to ask how to use the elementary transformation to find the inverse matrix,
for instance
The matrix of 1 - 1 1 is transformed into 1 - 1 1 | 100
1 1 3 1 1 3 |0 1 0
2 -3 2 2 -3 2| 0 0 1
Then we tried to make the front into an identity matrix. That's what our teacher said. However, he spent a long time on the platform and didn't understand it. Which elder brother helped me to make it more detailed. What I don't understand is what he said. If the first line doesn't change and the third line multiplies a number to add to the first line, can the first line multiply a number to add to other lines, What's the rule of transformation?
1 -1 1 1 0 0
1 1 3 0 1 0
2 -3 2 0 0 1
R2-r1 (the first line multiplied by - 1 and added to the second line, or the first line subtracted by one time from the second line, the same below), r3-2r1
1 -1 1 1 0 0
0 2 2 -1 1 0
0 -1 0 -2 0 1
R2R3 (exchange lines 2 and 3)
1 -1 1 1 0 0
0 -1 0 -2 0 1
0 2 2 -1 1 0
r1-r2, r3+2r2
1 0 1 3 0 -1
0 -1 0 -2 0 1
0 0 2 -5 1 2
r2*(-1), r3*(1/2)
1 0 1 3 0 -1
0 1 0 2 0 -1
0 0 1 -5/2 1/2 1
r1-r3
1 0 0 11/2 -1/2 -2
0 1 0 2 0 -1
0 0 1 -5/2 1/2 1
The transformation rules are used in three kinds of line transformation (in the textbook)
The purpose of the transformation is to transform the left side into the identity matrix, and the right side is the inverse of A
If you don't understand, ask or inform me
Find the inverse of the following matrix by elementary row transformation
Matrix: 1 - 3 2
-3 0 1
1 1 -1
Matrix formula: 1-3 2
-3 0 1
1 1 -1
When finding the inverse of a matrix with elementary row change,
The matrix (a, e) is transformed into (E, b) by row transformation, then B is equal to the inverse of A
ad locum
(A,E)=
1 -3 2 1 0 0
-3 0 1 0 1 0
1 1 - 1 001 line 2 plus line 3 × 3, line 3 minus line 1
1 -3 2 1 0 0
0 3 -2 0 1 3
0 4 - 3 - 1 01 add line 2 to line 1 and subtract line 2 from line 3
1 0 0 1 1 3
0 3 -2 0 1 3
0 1 - 1 - 1 - 1 - 2 Line 2 minus line 3 × 2
1 0 0 1 1 3
0 1 0 2 3 7
0 1 - 1 - 1 - 1 - 2 line 3 minus line 2, line 3 * (- 1)
1 0 0 1 1 3
0 1 0 2 3 7
0 0 1 3 4 9
In this way, (a, e) ～ (E, a ^ - 1) has been transformed by elementary row transformation,
So the inverse matrix of the original matrix is obtained
1 1 3
2 3 7
3 4 9
Finding inverse matrix by elementary transformation of matrix
1234 2312 111-1 10-2-6
For the specific transformation process, if you can take a picture better,
Solution: (a, e)=
1 2 3 4 1 0 0 0
2 3 1 2 0 1 0 0
1 1 1 -1 0 0 1 0
1 0 -2 -6 0 0 0 1
r1-r3,r2-2r3,r4-r3
0 1 2 5 1 0 -1 0
0 1 -1 4 0 1 -2 0
1 1 1 -1 0 0 1 0
0 -1 -3 -5 0 0 -1 1
ri-r4,i=1,2,3
0 0 -1 0 1 0 -2 1
0 0 -4 -1 0 1 -3 1
1 0 -2 -6 0 0 0 1
0 -1 -3 -5 0 0 -1 1
r1*(-1),r2+4r1,r3+2r1,r4+3r1
0 0 1 0 -1 0 2 -1
0 0 0 -1 -4 1 5 -3
1 0 0 -6 -2 0 4 -1
0 -1 0 -5 -3 0 5 -2
r2*(-1),r3+6r2,r4+5r2
0 0 1 0 -1 0 2 -1
0 0 0 1 4 -1 -5 3
1 0 0 0 22 -6 -26 17
0 -1 0 0 17 -5 -20 13
R4 * (- 1)
1 0 0 0 22 -6 -26 17
0 1 0 0 -17 5 20 -13
0 0 1 0 -1 0 2 -1
0 0 0 1 4 -1 -5 3
So a ^ - 1=
22 -6 -26 17
-17 5 20 -13
-1 0 2 -1
4 -1 -5 3
Finding the inverse of a matrix by elementary row transformation
A=
1 2 2
2 1 -2
2 -2 1
Find a ^ - 1
A. E-e.a ^ - 1 is difficult to simplify
1 2 2 1 0 0
2 1 -2 0 1 0
2-21001 to
1 0 0 X X X
0 1 0 X X X
0 0 1 X X X
(A,E)=1 2 2 1 0 02 1 -2 0 1 02 -2 1 0 0 1r2-2r1,r3-2r11 2 2 1 0 00 -3 -6 -2 1 00 -6 -3 -2 0 1r3-r21 2 2 1 0 00 -3 -6 -2 1 00 0 9 2 -2 1r2*(-1/3),r3*(1/9)1 2 2 1 0 00 1 2 2/3 -1/3 00 0 1 2/9 -2/9 1/9r2...
Finding the inverse of a matrix by elementary transformation
1 2 3
2 1 2
1 3 4
The inverse matrix is - 2 1
-6 1 4
5 -1 -3
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