tutorial 02 matrix inversion - ihaltas.com · tutorial 2 - 1 the university of new brunswick...
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Tutorial 2 - 1
The University of New BrunswickDepartment of Electrical and Computer Engineering
Fredericton, NB, E3B 5A3 Canada
Tutorial 2 - 2
Matrix Inversion Matrix Inversion
Tutorial 2 - 3
MATRIX INVERSION
Using minors
( ) ji-1ij
AA =
detAAji is the algebraic complement of the matrix A.
The definition of the algebraic complement Aji:
1. Cross out jth row and ith column
2. Calculate det of resulting matrix (minor)
3. Multiply the result by (-1)i+j
Tutorial 2 - 4
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Obtain the inverse of the matrix A
Tutorial 2 - 5
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Step 1. Calculate determinant:
1 0 01 3
det(A)= 2 1 3 1 20 2
0 0 2= × =
Tutorial 2 - 6
1 0 0det(A)= 2 1 3 B1-B2=2-0=2
0 0 2=
Step 1. Calculate determinant:
Continued...
( ) ( ) ( )B1= 1×1×2 + 0×3×0 + 0×2×0 2 0 0 2= + + =⎡ ⎤⎣ ⎦
( ) ( ) ( )B2= 0×1×0 + 2×0×2 + 1×0×3 0 0 0 0= + + =⎡ ⎤⎣ ⎦
Tutorial 2 - 7
1 0 0det(A) = 2 1 3
0 0 2
Step 1. Calculate determinant:Continued...
1 02 10 0
1 02 10 0
Tutorial 2 - 8
Step 1. Calculate determinant:Continued...
( ) ( ) ( )B1= 1×1×2 + 0×3×0 + 0×2×0 2 0 0 2= + + =⎡ ⎤⎣ ⎦
1 0 0 1 0det(A)= 2 1 3 2 1
0 0 2 0 0
Tutorial 2 - 9
Step 1. Calculate determinant:Continued...
( ) ( ) ( )B2= 0×1×0 + 2×0×2 + 1×0×3 0 0 0 0= + + =⎡ ⎤⎣ ⎦
1 0 0 1 0det(A)= 2 1 3 2 1
0 0 2 0 0
Tutorial 2 - 10
1 0 0det(A)= 2 1 3 B1-B2=2-0=2
0 0 2=
Step 1. Calculate determinant:Continued...
( ) ( ) ( )B1= 1×1×2 + 0×3×0 + 0×2×0 2 0 0 2= + + =⎡ ⎤⎣ ⎦
( ) ( ) ( )B2= 0×1×0 + 2×0×2 + 1×0×3 0 0 0 0= + + =⎡ ⎤⎣ ⎦
Tutorial 2 - 11
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Step 2. Calculate minors and algebraic components:
( )1+111
1 3A = -1 2
0 2× = ( )2+1
210 0
A = -1 00 2
× = ( )3+131
0 0A = -1 0
1 3× =
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Tutorial 2 - 12
( )1+212
2 3A = -1 4
0 2× = − ( )2+2
221 0
A = -1 20 2
× = ( )3+232
1 0A = -1 3
2 3× = −
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Step 2. Calculate minors and algebraic components:
Tutorial 2 - 13
( )1+313
2 1A = -1 0
0 0× = ( )2+3
231 0
A = -1 00 0
× = ( )3+333
1 0A = -1 1
2 1× =
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 0A= 2 1 3
0 0 2
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Step 2. Calculate minors and algebraic components:
Tutorial 2 - 14
( )1+111
1 3A = -1 2
0 2× =
( )1+212
2 3A = -1 4
0 2× = −
( )1+313
2 1A = -1 0
0 0× =
( )2+121
0 0A = -1 0
0 2× =
( )2+222
1 0A = -1 2
0 2× =
( )2+323
1 0A = -1 0
0 0× =
( )3+131
0 0A = -1 0
1 3× =
( )3+232
1 0A = -1 3
0 3× = −
( )3+333
1 0A = -1 1
2 1× =
Step 3. Calculate A-1
11 21 31-1
12 22 32
13 23 33
1A =det(A)
A A AA A AA A A
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
-12 0 0
1A = 4 2 32
0 0 1
⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠
Step 2. Calculate minors and algebraic components:
Tutorial 2 - 15
-12 0 0
1A = 4 2 32
0 0 1
⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠
-12 / 2 0 / 2 0 / 2
A = 4 / 2 2 / 2 3/ 20 / 2 0 / 2 1/ 2
⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠
-11 0 0
A = 2 1 3/ 20 0 1/ 2
⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎝ ⎠
Step 3. Calculate A-1
Tutorial 2 - 16
1 0 0 1 0 0 1 0 02 1 3 2 1 3/ 2 0 1 00 0 2 0 0 1/ 2 0 0 1
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
-1AA =I
Step 4. Check the result, if you wish
Obtain the inverse of the matrix A
Tutorial 2 - 17
8 6A=
5 7⎛ ⎞⎜ ⎟⎝ ⎠
Obtain the inverse of the following (2x2 )matrix
A Spacial Casa
Tutorial 2 - 18
( ) ( )8 6= 8 7 5 6 56 30 26
5 7∆ = × − × = − =
Step 1: Determinant
8 6det( ) =
5 7A A= = ∆
Tutorial 2 - 19
Step 2: Minors
1 1 211
8 6A = ( 1) (7) ( 1) (7) 7
5 7+= − = − =
2 1 321
8 6A = ( 1) (6) ( 1) (6) 6
5 7+= − = − = −
1 2 312
8 6A = ( 1) (5) ( 1) (5) 5
5 7+= − = − = −
2 2 422
8 6A = ( 1) (8) ( 1) (8) 8
5 7+= − = − =
11 21ji
12 22A =
A AA A
⎛ ⎞⎜ ⎟⎝ ⎠
ji7 6
A =5 8
−⎛ ⎞⎜ ⎟−⎝ ⎠
Tutorial 2 - 20
Step 3: Inverse
( ) ji-1ij
AA =
detA
-1 7 61A =5 826
−⎛ ⎞⎜ ⎟−⎝ ⎠
-1
7 626 26A =
5 826 26
−⎛ ⎞⎜ ⎟⎜ ⎟
−⎜ ⎟⎜ ⎟⎝ ⎠
8 6A=
5 7⎛ ⎞⎜ ⎟⎝ ⎠
-1 7 61A =5 826
−⎛ ⎞⎜ ⎟−⎝ ⎠
Check the patern between theoriginal matrix and its minorsfor a (2x2) matrix.