Difference between revisions of "Adjugate Formula"
| Line 19: | Line 19: | ||
|Title=inverse of a 4-by-4 matrix using the adjugate formula | |Title=inverse of a 4-by-4 matrix using the adjugate formula | ||
|Contents= | |Contents= | ||
| − | <br/><math> | + | <br/> |
| + | <math> | ||
\mathbf{A}_e = | \mathbf{A}_e = | ||
\left[\begin{array}{cccc} | \left[\begin{array}{cccc} | ||
| Line 28: | Line 29: | ||
\end{array}\right]</math> | \end{array}\right]</math> | ||
<br/><br/> | <br/><br/> | ||
| − | The calculation of two of the cofactors of <math>\mathbf{A}_e</math> has already been described in the example for [[Minors_and_cofactors|minors and cofactors]] | + | The calculation of two of the cofactors of <math>\mathbf{A}_e</math> has already been described in the example for [[Minors_and_cofactors|minors and cofactors]]. The resulting cofactor matrix of matrix <math>\mathbf{A}_e</math> is<br/><br/> |
| − | <math> | + | <math> |
| − | \mathbf{C}(\mathbf{A}_e) | + | \mathbf{C}(\mathbf{A}_e)= |
\left[\begin{array}{cccc} | \left[\begin{array}{cccc} | ||
1 & 0 & 3 & -6\\ | 1 & 0 & 3 & -6\\ | ||
| Line 36: | Line 37: | ||
-2 & 1 & -6 & 12\\ | -2 & 1 & -6 & 12\\ | ||
0 & 0 & 1 & -1 | 0 & 0 & 1 & -1 | ||
| − | \end{array}\right] | + | \end{array}\right]</math> |
| − | \mathbf{C}(\mathbf{A}_e)^T | + | <br/><br/> |
| + | The transposed of the cofactor matrix then appears as | ||
| + | <br/><br/><math>\mathbf{C}(\mathbf{A}_e)^T= | ||
\left[\begin{array}{cccc} | \left[\begin{array}{cccc} | ||
1 & 0 & -2 & 0\\ | 1 & 0 & -2 & 0\\ | ||
| Line 43: | Line 46: | ||
3 & -1 & -6 & 1\\ | 3 & -1 & -6 & 1\\ | ||
-6 & 2 & 12 & -1 | -6 & 2 & 12 & -1 | ||
| − | \end{array}\right]=\text{adj}(\mathbf{A}_e) | + | \end{array}\right]=\text{adj}(\mathbf{A}_e)</math><br/><br/> |
| − | \mathbf{A}_e^{-1} | + | The determinant is computed in the example for a [[Determinant_of_a_4-by-4_matrix|determinant of a 4-by-4 matrix]] and equals 1. |
| + | <math> | ||
| + | \mathbf{A}_e^{-1}=\frac{1}{\det(\mathbf{A}_e)}\text{adj}(\mathbf{A}_e) | ||
=\frac{1}{1} | =\frac{1}{1} | ||
\left[\begin{array}{cccc} | \left[\begin{array}{cccc} | ||
| Line 59: | Line 64: | ||
-6 & 2 & 12 & -1 | -6 & 2 & 12 & -1 | ||
\end{array}\right] | \end{array}\right] | ||
| − | + | </math> | |
| + | |||
}} | }} | ||
Revision as of 16:18, 9 May 2014
The adjugate formula defines the inverse of an n-by-n square matrix 
as
where 
is the so called adjugate matrix of . The adjugate matrix is the transposed of the cofactor matrix:
And the cofactor matrix 
is just a matrix where each cell corresponds to the related cofactor:
![\mathbf{C}(\mathbf{A})=\left[\begin{array}{cccc}
C_{1,1}(\mathbf{A}) & C_{1,2}(\mathbf{A}) & \cdots & C_{1,n}(\mathbf{A})\\
C_{2,1}(\mathbf{A}) & C_{2,2}(\mathbf{A}) & & C_{2,n}(\mathbf{A})\\
\vdots & & \ddots & \vdots\\
C_{n,1}(\mathbf{A}) & C_{n,2}(\mathbf{A}) & \cdots & C_{n,n}(\mathbf{A})
\end{array}\right]](/wiki/robotics/images/math/c/b/7/cb7505ec4a107c95fd31601fac888472.png)
So to determine the inverse of an n-by-n square matrix you have to compute the n square cofactors, then transpose the resulting cofactor matrix and divide all the values by the determinant.
 Example: inverse of a 4-by-4 matrix using the adjugate formula
|
![\mathbf{A}_e =
\left[\begin{array}{cccc}
1 & 2 & 0 & 0\\
3 & 0 & 1 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 2 & 1
\end{array}\right]](/wiki/robotics/images/math/a/3/8/a386cb4eac26aca26ac89fd336061a0e.png)
has already been described in the example for ![\mathbf{C}(\mathbf{A}_e)=
\left[\begin{array}{cccc}
1 & 0 & 3 & -6\\
0 & 0 & -1 & 2\\
-2 & 1 & -6 & 12\\
0 & 0 & 1 & -1
\end{array}\right]](/wiki/robotics/images/math/2/c/1/2c171afbf8b3bb79274d1b6d160574ad.png)
![\mathbf{C}(\mathbf{A}_e)^T=
\left[\begin{array}{cccc}
1 & 0 & -2 & 0\\
0 & 0 & 1 & 0\\
3 & -1 & -6 & 1\\
-6 & 2 & 12 & -1
\end{array}\right]=\text{adj}(\mathbf{A}_e)](/wiki/robotics/images/math/4/d/6/4d60f99b73fcd16a1e7f14e381e93148.png)
![\mathbf{A}_e^{-1}=\frac{1}{\det(\mathbf{A}_e)}\text{adj}(\mathbf{A}_e)
=\frac{1}{1}
\left[\begin{array}{cccc}
1 & 0 & -2 & 0\\
0 & 0 & 1 & 0\\
3 & -1 & -6 & 1\\
-6 & 2 & 12 & -1
\end{array}\right]
=
\left[\begin{array}{cccc}
1 & 0 & -2 & 0\\
0 & 0 & 1 & 0\\
3 & -1 & -6 & 1\\
-6 & 2 & 12 & -1
\end{array}\right]](/wiki/robotics/images/math/1/f/3/1f3b7de9ef6df9ca9270bf2b7d321cb6.png)
