mahesh dasari
Last Activity: 12 Years ago
if given equation is AX = B form then for finding X
we write it like [ A : B] = X
so for finding inverse
AA-1 = I
[A : I ] = A-1
I think u may know about row reduction
Applying the method of Gauss-Jordon
elimination, we get inverse of matrix
?? : ??3 =
1 |
2 |
1 |
1 |
0 |
0 |
2 |
-1 |
-1 |
0 |
1 |
0 |
3 |
1 |
1 |
0 |
0 |
1 |
apply row reduction
R2 - 2R1 , R3 - 3R1 ,
1
|
2
|
1
|
1
|
0
|
0
|
0
|
-5
|
-3
|
-2
|
1
|
0
|
0
|
-5
|
-2
|
-3
|
0
|
1 |
- R2 / 5
1
|
2
|
1
|
1
|
0
|
0
|
0
|
1
|
3/5
|
2/5
|
-1/5
|
0
|
0
|
-5
|
-2
|
-3
|
0
|
1 |
R1 - 2R2
R3 + 5R2
1
|
0
|
-1/5
|
1/5
|
2/5
|
0
|
0
|
1
|
3/5
|
2/5
|
-1/5
|
0
|
0
|
0
|
1
|
-1
|
-1
|
1
|
R1 + R3/5
R2 - 3R3/5
1
|
0
|
0
|
0
|
1/5
|
1/5
|
0
|
1
|
0
|
1
|
2/5
|
-3/5
|
0
|
0
|
1
|
-1
|
-1
|
1
|
This is combination of two 3X3 matrices first matrix is Identity matrix and second matrix is A-1 matrix
for finding A-1 we always reduce the first matrix into indentity matrix and second is A-1
So A-1 =
0 |
1/5 |
1/5 |
1 |
2/5 |
-3/5 |
-1 |
-1 |
1 |
like this you can find A-1
Another process is there to find A-1 is
A-1 = adjoint(A) / det( A)
above process is very easy than this
ok thank u , if u like support my answer