Adjugate Formula

From Robotics
Jump to: navigation, search
← Back: Gauß-Jordan-Algorithm Overview: Matrix inversion Next: Transformations

The adjugate formula defines the inverse of an n-by-n square matrix \mathbf{A} as

\mathbf{A}^{-1}=\frac{1}{\det(\mathbf{A})}\text{adj}(\mathbf{A})

where \text{adj}(\mathbf{A}) is the so called adjugate matrix of \mathbf{A}. The adjugate matrix is the transposed of the cofactor matrix:


\text{adj}(\mathbf{A})=\mathbf{C}(\mathbf{A})^T

And the cofactor matrix \mathbf{C}(\mathbf{A}) is just a matrix where each cell corresponds to the related cofactor:


\mathbf{C}(\mathbf{A})=\left[\begin{array}{cccc}
C_{11}(\mathbf{A}) & C_{12}(\mathbf{A}) & \cdots & C_{1n}(\mathbf{A})\\
C_{21}(\mathbf{A}) & C_{22}(\mathbf{A}) &  & C_{2n}(\mathbf{A})\\
\vdots &  & \ddots & \vdots\\
C_{n1}(\mathbf{A}) & C_{n2}(\mathbf{A}) & \cdots & C_{nn}(\mathbf{A})
\end{array}\right]

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 2-by-2 matrix using the adjugate formula

The following matrix is already used in examples for the minors and cofactors and for the determinant.


\mathbf{A}_2  = 
\left[\begin{array}{cc}
2&3\\
1&2
\end{array}\right]

The computation of the cofactors is easy in this case because the minors correspond to the determinants of 1-by-1 matrices, that are the value itself. So the cofactor matrix of matrix \mathbf{A}_3 is


\mathbf{C}(\mathbf{A}_2)=
\left[\begin{array}{cc}
(-1)^{1+1}\det
\left[\begin{array}{cc}
\Box&\Box\\
\Box&2
\end{array}\right] 
& (-1)^{1+2}\det
\left[\begin{array}{cc}
\Box&\Box\\
1&\Box
\end{array}\right] 
\\ \\
(-1)^{2+1}\det
\left[\begin{array}{cc}
\Box&3\\
\Box&\Box
\end{array}\right]  
& (-1)^{2+2}\det
\left[\begin{array}{cc}
2&\Box\\
\Box&\Box
\end{array}\right] 
\end{array}\right]=
\left[\begin{array}{cc}
(-1)^{1+1}\det[2] & (-1)^{1+2}\det[1]\\
(-1)^{2+1}\det[3] & (-1)^{2+2}\det[2]
\end{array}\right]=
\left[\begin{array}{cc}
2 & -1\\
-3 & 2
\end{array}\right]

The transposed of the cofactor matrix which corresponds to the adjugate matrix then appears as

\mathbf{C}(\mathbf{A}_2)^T=
\left[\begin{array}{cc}
2&-3\\
-1&2
\end{array}\right]=\text{adj}(\mathbf{A}_2)

The determinant is computed in the example for a determinant of a matrix and equals 1. So the inverse of \mathbf{A}_2 is determined as


\mathbf{A}_2^{-1}=\frac{1}{\det(\mathbf{A}_2)}\text{adj}(\mathbf{A}_2)
=\frac{1}{1}
\left[\begin{array}{cc}
2&-3\\
-1&2
\end{array}\right]
=
\left[\begin{array}{cc}
2&-3\\
-1&2
\end{array}\right]

For a proof please have a look at the example in the main article of matrix inversion.

Example: inverse of a 3-by-3 matrix using the adjugate formula

The following matrix is already used in examples for the minors and cofactors and for the determinant.


\mathbf{A}_3  = 
\left[\begin{array}{ccc}
1&0&1\\
3&1&0\\
1&0&2
\end{array}\right]

The calculation of two of the cofactors of \mathbf{A}_3 has already been described in the example for minors and cofactors. The resulting cofactor matrix of matrix \mathbf{A}_3 is


\mathbf{C}(\mathbf{A}_3)=
\left[\begin{array}{ccc}
2&-6&-1\\
0&1&0\\
-1&3&1
\end{array}\right]

The transposed of the cofactor matrix which corresponds to the adjugate matrix then appears as

\mathbf{C}(\mathbf{A}_3)^T=
\left[\begin{array}{ccc}
2&0&-1\\
-6&1&3\\
-1&0&1
\end{array}\right]=\text{adj}(\mathbf{A}_3)

The determinant is computed in the example for a determinant of a matrix and equals 1. So the inverse of \mathbf{A}_3 is determined as


\mathbf{A}_3^{-1}=\frac{1}{\det(\mathbf{A}_3)}\text{adj}(\mathbf{A}_3)
=\frac{1}{1}
\left[\begin{array}{ccc}
2&0&-1\\
-6&1&3\\
-1&0&1
\end{array}\right]
=
\left[\begin{array}{ccc}
2&0&-1\\
-6&1&3\\
-1&0&1
\end{array}\right]

For a proof please have a look at the example in the main article of matrix inversion.

Example: inverse of a 4-by-4 matrix using the adjugate formula

In this example again the transformation matrix ^R\mathbf{T}_N that is introduced in the robotics script in chapter 3 on page 3-37 is used:


^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]

The calculation of two of the cofactors of ^R\mathbf{T}_N has already been described in the example for minors and cofactors. The resulting cofactor matrix of matrix ^R\mathbf{T}_N is


\mathbf{C}(^R\mathbf{T}_N)=
\left[\begin{array}{cccc}
0&1&0&0\\
0&0&-1&0\\
-1&0&0&0\\
0&-2a&0&1
\end{array}\right]

The transposed of the cofactor matrix which corresponds to the adjugate matrix then appears as

\mathbf{C}(^R\mathbf{T}_N)^T=
\left[\begin{array}{cccc}
0&0&-1&0\\
1&0&0&-2a\\
0&-1&0&0\\
0&0&0&1
\end{array}\right]=\text{adj}(^R\mathbf{T}_N)

The determinant is computed in the example for a determinant of a matrix and equals 1. So the inverse of ^R\mathbf{T}_N is determined as


^R\mathbf{T}_N^{-1}=\frac{1}{\det(^R\mathbf{T}_N)}\text{adj}(^R\mathbf{T}_N)
=\frac{1}{1}
\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&0&-1&0\\
1&0&0&-2a\\
0&-1&0&0\\
0&0&0&1
\end{array}\right]

For a proof please have a look at the example in the main article of matrix inversion.