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The procedure for finding an inverse of a square matrix, A, is as follows:
- Form the augmented matrix that has A on the left and the identity matrix, I, on the right.
- Perform Gauss-Jordan elimination on the augmented matrix. The reduced row echelon form will either be an identity matrix on the left, or will have a row of zeros on the bottom of the left half of the augmented matrix.
- If the procedure produces a row of zeros, then the matrix A is not invertible. If it produces an itendity matrix, then the matrix on the right of the augmented matrix is the inverse, A-1.
Example:
Find the inverse of the matrix
A = é
ê
ë3
1
0-1
1
20
2
4ù
ú
û
Solution:
é
ê
ë3
1
0-1
1
20
2
4ï
ï
ï1
0
00
1
00
0
1ù
ú
ûwrite the augmented matrix [ A | I ]
é
ê
ë1
2
01
-1
22
0
4ï
ï
ï0
1
01
0
00
0
1ù
ú
ûSwap rows 1 and 2
(to bring a 1 to the upper left)
é
ê
ë1
0
01
-4
22
-6
4ï
ï
ï0
1
01
-3
00
0
1ù
ú
ûAdd -3 times row 1 to row 2
(to get a 0 in the first column)
é
ê
ë1
0
01
2
-42
4
-6ï
ï
ï0
0
11
0
-30
1
0ù
ú
ûSwap second and third rows
(this will make the computation easier)
é
ê
ë1
0
01
2
02
4
2ï
ï
ï0
0
11
0
-30
1
2ù
ú
ûAdd 2 times row 2 to row 3
(to get a zero in the second column)
é
ê
ë1
0
01
1
02
2
1ï
ï
ï0
0
1/21
0
-3/20
1/2
1ù
ú
ûDivide rows 2 and 3 by 2
(to get leading ones)
é
ê
ë1
0
00
1
00
2
1ï
ï
ï0
0
1/21
0
-3/2-1/2
1/2
1ù
ú
ûAdd -1 times row 2 to row 1
(to get zeros above the leading 1's)
é
ê
ë1
0
00
1
00
0
1ï
ï
ï0
-1
1/21
3
-3/2-1/2
-3/2
1ù
ú
ûAdd -2 times row 3 to row 2
(to get a zero above the leading 1)This has reduced the left-hand side to the identity matrix, so the right-hand side is the inverse of A.
A-1 = é
ê
ë0
-1
1/21
3
-3/2-1/2
-3/2
1ù
ú
ûTo check the answer, we compute the product A-1A and see if it equals I. If it does, then we have the inverse.
A-1A =
é
ê
ë0
-1
1/21
3
-3/2-1/2
-3/2
1ù
ú
ûé
ê
ë0
-1
1/21
3
-3/2-1/2
-3/2
1ù
ú
û=
é
ê
ë0(3)+1(1)+(-1/2)(0)
-1(3)+3(1)+(-3/2)(0)
(1/2)(3)+(-3/2)(1)+1(0)0(-1)+1(1)+(-1/2)(2)
-1(-1)+3(1)+(-3/2)(2)
(1/2)(-1)+(-3/2)(1)+1(2)0(0)+1(2)+(-1/2)(4)
-1(0)+3(2)+(-3/2)(4)
(1/2)(0)+(-3/2)(2)+1(4)ù
ú
û=
é
ê
ë0+1+0
-3+3+0
3/2-3/2+00+1-1
1+3-3
-(1/2)-(3/2)+10+2-2
0+6-6
0-3+4ù
ú
û=
é
ê
ë1
0
00
1
00
0
1ù
ú
û= I Since
A-1A = I , we know we have the correct inverse.
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