INVERSION OF MATRIX BY GAUSS ELIMINATION METHOD

1. INVERSION OF MATRIX

2. INVERSSION OF MATRIX :
Let us consider we have a matrix [A] & we have to find the inverse of that matrix
. Let [x] be the inverse of matrix [A].
we know that : A ∗ 𝐴−1 = [𝐼]
now : [A]*[X]=[I] as we considered that : [X]= 𝐴−1
Then we convert the system into an augmented matrix
in the form of [A|I ].
then we have to use elementary row operation on the
augmented matrix to get upper triangular matrix by using
Gauss Elimination Method.

3.  After that using : [A]*[X]=[I]
 the relation we have to find the elements of [X] ; which is our required answer .
 For example we take a matrix and try to find it’s inverse :
 A =
2 −2 4
2 3 2
−1 4 −1
 consider the augmented matrix is :

2 −2 4
2 3 2
−1 4 −1
: 1 0 0
: 0 1 0
: 0 0 1
 =
2 −2 4
0 5 −2
0 3 1
: 1 0 0
: −1 1 0
: (
1
2
) 0 1
[using 𝑅′2 = 𝑅2 −
2
2
∗ 𝑅1; 𝑅′
3 = 𝑅3 − (
1
2
)𝑅1]

4.  =
2 −2 4
0 5 −2
0 0 (
11
5
)
: 1 0 0
: −1 1 0
: (
11
10
) −(
3
5
) 1
[using 𝑅′′3 = 𝑅3 −
3
5
∗ 𝑅2]
 The inverse matrix is given as : X=
𝑥11 𝑥12 𝑥13
𝑥21 𝑥22 𝑥23
𝑥31 𝑥32 𝑥33
 Thus we have an equivalent system of three equation :

2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥11
𝑥21
𝑥31
=
1
−1
(
11
10
)
 which gives :2𝑥 − 2𝑦 + 4𝑧 = 1;
 5𝑦 − 2𝑧 = −1;

11
5
∗ 𝑧 = (
11
10
)
 Solving by back substitution method we get ;
 𝑥 = −
1
2
; 𝑦 = 0; 𝑧 = (
1
2
)

5. 
2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥12
𝑥22
𝑥32
=
0
1
(−
3
5
)
 similarly by solving these we get :
 𝑥12 = (
7
11
) ; 𝑥22= (
1
11
); 𝑥32= (−
3
11
) .


2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥13
𝑥23
𝑥33
=
0
0
1
 similarly by solving these we get :
 𝑥13 = (
−8
11
) ; 𝑥22= (
2
11
); 𝑥32= (
5
11
) .
 The required solution is :
 𝐴−1 =
(−
1
2
) (
7
11
) (−
8
11
)
0 (
1
11
) (
2
11
)
(
1
2
) (−
3
11
) (
5
11
)

6. THANK YOU