Difference between revisions of "Matrix inversion"
| Line 13: | Line 13: | ||
# [[Adjugate Formula]] | # [[Adjugate Formula]] | ||
| − | Transformation matrices have a special structure, that is described in the [[Transformations|transformations]] chapter. For this special matrix structure an easier method to invert the matrix exists. This method is presented in [[Inverse | + | Transformation matrices have a special structure, that is described in the [[Transformations|transformations]] chapter. For this special matrix structure an easier method to invert the matrix exists. This method is presented in |
| + | # [[Inverse transformation]]. | ||
<br/> | <br/> | ||
Revision as of 14:43, 12 June 2014
| ← Back: Minors and cofactors | Overview: Matrix inversion | Next: Gauß-Jordan-Algorithm → |
The inverse of an n-by-n square matrix 
is denoted as 
and defined such that
where 
is the n-by-n identity matrix.
Prerequesite for the inversion is, that is an n-by-n square matrix and that is regular. Regular means that the row and column vectors are linearly independent and so the determinant is nonzero:
Otherwise the matrix is called singular.
Before determining the inverse of a matrix it is always useful to compute the determinant and check whether the matrix is regular or singular. If it is singular it is not possible to determine the inverse because there is no inverse. The following two subarticles describe two of the common procedures to determine the inverse of a matrix.
Transformation matrices have a special structure, that is described in the transformations chapter. For this special matrix structure an easier method to invert the matrix exists. This method is presented in
 Example: inverse of a 2-by-2 matrix
This is a simple example for the inverse of a 2-by-2 matrix: |
![\mathbf{A}_2 =
\left[\begin{array}{cc}
2 & 3\\
1 & 2
\end{array}\right]
,\quad
{\mathbf{A}_2}^{-1} =
\left[\begin{array}{cc}
2 & -3\\
-1 & 2
\end{array}\right]](/wiki/robotics/images/math/7/b/7/7b75cee085886960d31619e58b18aea7.png)
![\begin{align}
{\mathbf{A}_2}{\mathbf{A}_2}^{-1} &=
\left[\begin{array}{cc}
2 & 3\\
1 & 2
\end{array}\right]\cdot
\left[\begin{array}{cc}
2 & -3\\
-1 & 2
\end{array}\right]\\&=
\left[\begin{array}{cc}
2\cdot2+3\cdot(-1) & 2\cdot(-3)+3\cdot2\\
1\cdot2+2\cdot(-1) & 1\cdot(-3)+2\cdot2
\end{array}\right]\\&=
\left[\begin{array}{cc}
4-3 & -6+6\\
2-2 & -3+4
\end{array}\right]\\&=
\left[\begin{array}{cc}
{\color{Green}\mathbf{1}} & 0\\
0 & {\color{Green}\mathbf{1}}
\end{array}\right]=
\mathbf{I}_2
\end{align}](/wiki/robotics/images/math/b/c/b/bcba94166ac9c6e7dbfc0b224f3c9288.png)
| Example: inverse of a 3-by-3 matrix
This is an example for the inverse of a 3-by-3 matrix: |
![\mathbf{A}_3 =
\left[\begin{array}{ccc}
1&0&1\\
3&1&0\\
1&0&2
\end{array}\right]
,\quad
{\mathbf{A}_3}^{-1} =
\left[\begin{array}{ccc}
2&0&-1\\
-6&1&3\\
-1&0&1
\end{array}\right]](/wiki/robotics/images/math/6/5/a/65a81fe76e04f3ae0dd31374cdeaa36f.png)
![\begin{align}
{\mathbf{A}_3}{\mathbf{A}_3}^{-1} &=
\left[\begin{array}{ccc}
1&0&1\\
3&1&0\\
1&0&2
\end{array}\right]\cdot
\left[\begin{array}{ccc}
2&0&-1\\
-6&1&3\\
-1&0&1
\end{array}\right]\\&=
\left[\begin{array}{ccc}
1\cdot2+0\cdot(-6)+1\cdot(-1) & 1\cdot0+0\cdot1+1\cdot0 & 1\cdot(-1)+0\cdot3+1\cdot1\\
3\cdot2+1\cdot(-6)+0\cdot(-1) & 3\cdot0+1\cdot1+0\cdot0 & 3\cdot(-1)+1\cdot3+0\cdot1\\
1\cdot2+0\cdot(-6)+2\cdot(-1) & 1\cdot0+0\cdot1+2\cdot0 & 1\cdot(-1)+0\cdot3+2\cdot1
\end{array}\right]\\&=
\left[\begin{array}{ccc}
2+0-1 & 0+0+0 & -1+0+1\\
6-6+0 & 0+1+0 & -3+3+0\\
2+0-2 & 0+0+0 & -1+0+2
\end{array}\right]\\&=
\left[\begin{array}{ccc}
{\color{Green}\mathbf{1}} & 0 & 0\\
0 & {\color{Green}\mathbf{1}} & 0\\
0 & 0 & {\color{Green}\mathbf{1}}
\end{array}\right]=
\mathbf{I}_3
\end{align}](/wiki/robotics/images/math/4/a/2/4a241ecb528bafdfeeb28da3ae7c356d.png)
| Example: inverse of a 4-by-4 matrix
This example is a proof of equation 3.40 in the robotics script (see page 3-61): |
![^R\mathbf{T}_N =
\left[\begin{array}{cccc}
0 & 1 & 0 & 2a\\
0 & 0 & -1 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right]
,\quad
{^R\mathbf{T}_N}^{-1} =
\left[\begin{array}{cccc}
0 & 0 & -1 & 0\\
1 & 0 & 0 & -2a\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right]](/wiki/robotics/images/math/2/6/9/26972fd789b662c263fb7ac640f9d040.png)
![\begin{align}
{^R\mathbf{T}_N}{^R\mathbf{T}_N}^{-1} &=
\left[\begin{array}{cccc}
0 & 1 & 0 & 2a\\
0 & 0 & -1 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right]\cdot
\left[\begin{array}{cccc}
0 & 0 & -1 & 0\\
1 & 0 & 0 & -2a\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right]\\&=
\left[\begin{array}{cccc}
0\cdot0+1\cdot1+0\cdot0+2a\cdot0 & 0\cdot0+1\cdot0+0\cdot(-1)+2a\cdot0 & 0\cdot(-1)+1\cdot0+0\cdot0+2a\cdot0 & 0\cdot0+1\cdot(-2a)+0\cdot0+2a\cdot1\\
0\cdot0+0\cdot1+(-1)\cdot0+0\cdot0 & 0\cdot0+0\cdot0+(-1)\cdot(-1)+0\cdot0 & 0\cdot(-1)+0\cdot0+(-1)\cdot0+0\cdot0 & 0\cdot0+0\cdot(-2a)+(-1)\cdot0+0\cdot1\\
(-1)\cdot0+0\cdot1+0\cdot0+0\cdot0 & (-1)\cdot0+0\cdot0+0\cdot(-1)+0\cdot0 & (-1)\cdot(-1)+0\cdot0+0\cdot0+0\cdot0 & (-1)\cdot0+0\cdot(-2a)+0\cdot0+0\cdot1\\
0\cdot0+0\cdot1+0\cdot0+1\cdot0 & 0\cdot0+0\cdot0+0\cdot(-1)+1\cdot0 & 0\cdot(-1)+0\cdot0+0\cdot0+1\cdot0 & 0\cdot0+0\cdot(-2a)+0\cdot0+1\cdot1\\
\end{array}\right]\\&=
\left[\begin{array}{cccc}
{\color{Green}\mathbf{1}} & 0 & 0 & 0\\
0 & {\color{Green}\mathbf{1}} & 0 & 0\\
0 & 0 & {\color{Green}\mathbf{1}} & 0\\
0 & 0 & 0 & {\color{Green}\mathbf{1}}
\end{array}\right]=
\mathbf{I}_4
\end{align}](/wiki/robotics/images/math/b/3/1/b31a1fac2dc30fbc82ade8e45cf7ea1e.png)
