|
Size: 1613
Comment:
|
Size: 3574
Comment: Clean up calculation
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 1: | Line 1: |
| ## page was renamed from LinearAlgebra/InverseMatrices = Inverse Matrices = |
= Matrix Inversion = |
| Line 4: | Line 3: |
| == Introduction == | For some matrices '''''A''''', the '''inverse matrix''' ('''''A'''^-1^'') is a matrix which can be multiplied by the original matrix to produce the [[LinearAlgebra/SpecialMatrices#Identity_Matrix|identity matrix]]. The calculation of an inverse matrix, if it exists, is called '''inversion'''. |
| Line 6: | Line 5: |
| An '''inverse matrix''' is a matrix A^-1^ where multiplying it by matrix A results in the identity matrix. | <<TableOfContents>> |
| Line 8: | Line 7: |
| If A^-1^ exists, then A is '''invertible''' and '''non-singular'''. Not all matrices are invertible. Consider the below problem: {{{ ┌ ┐┌ ┐ ┌ ┐ │ 1 3││ a b│ │ 1 0│ │ 2 7││ c d│ = │ 0 1│ └ ┘└ ┘ └ ┘ }}} |
---- |
| Line 21: | Line 11: |
| == Properties == | == Definition == |
| Line 23: | Line 13: |
| For a permutation matrix, the inverse is also the transpose: P^-1^ = P^T^. | An inverse matrix satisfies the equation '''''AA'''^-1^ = '''I'''''. |
| Line 25: | Line 15: |
| For a square matrix A, the left inverse is the same as the right inverse. AA^-1^ = A^-1^A = I | [[LinearAlgebra/Determinants|Determinants]] are the test for '''invertability'''. if ''|'''A'''| != 0'', then '''''A''''' is invertable and non-singular. Conversely, if ''|'''A'''| = 0'', then '''''A''''' is singular and non-invertable. |
| Line 29: | Line 19: |
| == Gauss-Jordan Calculation == | === Properties === |
| Line 31: | Line 21: |
| The inverse matrix can be calculated through elimination and reverse elimination. | The core principle of inversions is that a matrix '''''A''''' can be canceled out from a larger equation. '''''AA'''^-1^ = '''I''''', so the two terms cancel out. |
| Line 33: | Line 23: |
| First step: | An invertible matrix has only one vector in the [[LinearAlgebra/NullSpaces|null space]]: the zero vector. For [[LinearAlgebra/Orthogonality#Matrices|orthogonal matrices]] (such as [[LinearAlgebra/SpecialMatrices#Permutation_Matrices|permutation matrices]]), the inverse is also the [[LinearAlgebra/MatrixTransposition|transpose]]: '''''Q'''^-1^ = '''Q'''^T^''. For a square matrix '''''A''''', the '''left inverse''' is the same as the '''right inverse'''. '''''AA'''^-1^ = '''A'''^-1^'''A''' = '''I''''' ---- == Calculation == Because '''''AA'''^-1^ = '''I''''', applying [[LinearAlgebra/Elimination|elimination]] and [[LinearAlgebra/Elimination#Reduced_Row_Echelon_Form|backwards elimination]] on '''''A''''' augmented with an [[LinearAlgebra/SpecialMatrices#Identity_Matrix|identity matrix]] ('''''I''''') will create '''''A'''^-1^'' in the augmentation. |
| Line 40: | Line 42: |
2 - 1m = 0 m = 2 2 7 0 1 - 1m - 3m - 1m - 0m ____ ____ ____ ____ 0 1 -2 1 ┌ ┐ │ [1] 3 │ 1 0│ │ 0 1 │ -2 1│ └ ┘ }}} Second step: {{{ |
┌ ┐ │ [1] 3 │ 1 0│ │ 0 [1] │ -2 1│ └ ┘ |
| Line 62: | Line 50: |
3 - 1m = 0 m = 3 1 3 1 0 - 0m - 1m - -2m - 1m ____ ____ _____ ____ 1 0 7 -3 ┌ ┐ │ 1 0 │ 7 -3│ │ 0 [1] │ -2 1│ └ ┘ |
┌ ┐ │ [1] 0 │ 7 -3│ │ 0 [1] │ -2 1│ └ ┘ |
| Line 77: | Line 56: |
| The inverse matrix of A is: | '''''A'''^-1^'' is: |
| Line 86: | Line 65: |
| ---- == Determinant and Cofactor Matrix == Given the [[LinearAlgebra/Determinants|determinant]] of '''''A''''', it can also be simple to compute '''''A'''^-1^'' as ''(1/|'''A'''|)'''C'''^T^''. '''''C''''' is the cofactor matrix, where ''c,,i j,,'' is the cofactor of ''a,,i j,,''. For example, given a 2 x 2 '''''A''''' like: {{{ ┌ ┐ │ a b│ │ c d│ └ ┘ }}} The cofactor matrix '''''C''''' is: {{{ ┌ ┐ │ d -c│ │ -b a│ └ ┘ }}} But this must be transposed to '''''C'''^T^'': {{{ ┌ ┐ │ d -b│ │ -c a│ └ ┘ }}} And then '''''A'''^-1'' is: {{{ ┌ ┐ │ (1/det A) * d (1/det A) * -b│ │ (1/det A) * -c (1/det A) * a│ └ ┘ }}} The above example fits into this formula. The [[LinearAlgebra/Elimination|elimination]] and [[LinearAlgebra/Elimination#Reduced_Row_Echelon_Form|backwards elimination]] prove that the determinant of that '''''A''''' is 1. The more fundamental formula ''ad - bc'' expands to ''1 * 7 - 2 * 3'' which also reveals a determinant of 1. As such, ''(1/|'''A'''|)'' is trivially 1. So simply plug the given (''a'', ''b'', ''c'', ''d'') into the transposed cofactor matrix to find the inverse. |
Matrix Inversion
For some matrices A, the inverse matrix (A-1) is a matrix which can be multiplied by the original matrix to produce the identity matrix. The calculation of an inverse matrix, if it exists, is called inversion.
Definition
An inverse matrix satisfies the equation AA-1 = I.
Determinants are the test for invertability. if |A| != 0, then A is invertable and non-singular. Conversely, if |A| = 0, then A is singular and non-invertable.
Properties
The core principle of inversions is that a matrix A can be canceled out from a larger equation. AA-1 = I, so the two terms cancel out.
An invertible matrix has only one vector in the null space: the zero vector.
For orthogonal matrices (such as permutation matrices), the inverse is also the transpose: Q-1 = QT.
For a square matrix A, the left inverse is the same as the right inverse. AA-1 = A-1A = I
Calculation
Because AA-1 = I, applying elimination and backwards elimination on A augmented with an identity matrix (I) will create A-1 in the augmentation.
┌ ┐ │ [1] 3 │ 1 0│ │ 2 7 │ 0 1│ └ ┘ ┌ ┐ │ [1] 3 │ 1 0│ │ 0 [1] │ -2 1│ └ ┘ ┌ ┐ │ 1 3 │ 1 0│ │ 0 [1] │ -2 1│ └ ┘ ┌ ┐ │ [1] 0 │ 7 -3│ │ 0 [1] │ -2 1│ └ ┘
A-1 is:
┌ ┐ │ 7 -3│ │ -2 1│ └ ┘
Determinant and Cofactor Matrix
Given the determinant of A, it can also be simple to compute A-1 as (1/|A|)CT. C is the cofactor matrix, where ci j is the cofactor of ai j.
For example, given a 2 x 2 A like:
┌ ┐ │ a b│ │ c d│ └ ┘
The cofactor matrix C is:
┌ ┐ │ d -c│ │ -b a│ └ ┘
But this must be transposed to CT:
┌ ┐ │ d -b│ │ -c a│ └ ┘
And then A^-1 is:
┌ ┐ │ (1/det A) * d (1/det A) * -b│ │ (1/det A) * -c (1/det A) * a│ └ ┘
The above example fits into this formula. The elimination and backwards elimination prove that the determinant of that A is 1. The more fundamental formula ad - bc expands to 1 * 7 - 2 * 3 which also reveals a determinant of 1. As such, (1/|A|) is trivially 1. So simply plug the given (a, b, c, d) into the transposed cofactor matrix to find the inverse.