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Using the elementary transformation method to find the inverse matrix of the matrix, the first line is 12-1, the second line is 34-2, and the third line is 5-41
Answer: first line - 157 - 1 second line - 6.53 - 0.5 third line - 167 - 1
The first line is 223, the second line is 1 - 10, the third line is - 121
Using the elementary row transformation method to find the inverse matrix of the matrix, the first row is 223, the second row is 1-10, and the third row is-121
Solution: (a, e)=
2 2 3 1 0 0
1 -1 0 0 1 0
-1 2 1 0 0 1
r1-2r2,r3+r2
0 4 3 1 -2 0
1 -1 0 0 1 0
0 1 1 0 1 1
r1-4r3,r2+r3
0 0 -1 1 -6 -4
1 0 1 0 2 1
0 1 1 0 1 1
r2+r1,r3+r1,r1*(-1)
0 0 1 -1 6 4
1 0 0 1 -4 -3
0 1 0 1 -5 -3
Exchange will do
1 0 0 1 -4 -3
0 1 0 1 -5 -3
0 0 1 -1 6 4
So a ^ - 1=
1 -4 -3
1 -5 -3
-1 6 4
The inverse matrix of the following matrices is calculated by elementary transformation of matrix. The first row is 21 - 1, the second row is 0 21, and the third row is 52 - 3
(A,E) =
2 1 -1 1 0 0
0 2 1 0 1 0
5 2 -3 0 0 1
r3-2r1
2 1 -1 1 0 0
0 2 1 0 1 0
1 0 -1 -2 0 1
r1-2r3
0 1 1 5 0 -2
0 2 1 0 1 0
1 0 -1 -2 0 1
r2-2r1
0 1 1 5 0 -2
0 0 -1 -10 1 4
1 0 -1 -2 0 1
r1+r2,r3-r2
0 1 0 -5 1 2
0 0 -1 -10 1 4
1 0 0 8 -1 -3
R2 * (- 1), exchange line
1 0 0 8 -1 -3
0 1 0 -5 1 2
0 0 1 10 -1 -4
So a ^ - 1=
8 -1 -3
-5 1 2
10 -1 -4
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