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| − | + | [[Category:MA265Fall2011Walther]] | |
| + | [[Category:bonus point project]] | ||
| + | [[Category:MA265]] | ||
| + | [[Category:linear algebra]] | ||
| + | [[Category:math]] | ||
| − | = | + | <center><font size= 4> |
| + | '''The Inverse of a Matrix''' | ||
| + | </font size> | ||
| + | Student project for [[MA265]] | ||
| + | </center> | ||
| + | ---- | ||
---- | ---- | ||
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Steps: | Steps: | ||
| − | 1) Get each of the 9 cofactor matrices by eliminating the row and the column of each element | + | 1) Get each of the 9 cofactor matrices by eliminating the row and the column of each element. |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| + | 2) Get each determinat from the 2x2 matrices | ||
| + | 3) Place the correct sign for each determinant. (If the sum of i and j = even its positive, odd negative). | ||
| + | <br> | ||
| + | <br> | ||
<math>A= \left(\begin{array}{cccc}1&0&-1\\1&1&1\\2&1&2\end{array}\right)</math> | <math>A= \left(\begin{array}{cccc}1&0&-1\\1&1&1\\2&1&2\end{array}\right)</math> | ||
| Line 96: | Line 105: | ||
<math>M31=\left(\begin{array}{cccc}0&-1\\1&1\end{array}\right)</math> <math>M32=\left(\begin{array}{cccc}1&-1\\1&1\end{array}\right)</math> <math>M33=\left(\begin{array}{cccc}1&0\\1&1\end{array}\right)</math> | <math>M31=\left(\begin{array}{cccc}0&-1\\1&1\end{array}\right)</math> <math>M32=\left(\begin{array}{cccc}1&-1\\1&1\end{array}\right)</math> <math>M33=\left(\begin{array}{cccc}1&0\\1&1\end{array}\right)</math> | ||
| + | <br> | ||
| + | |||
| + | A11=(+)1 A12=(-)0 A13=(+)-1 | ||
| + | |||
| + | A21=(-) 1 A22=(+) 4 A32=(-) 1 | ||
| + | |||
| + | A31=(+) 1 A32=(-)2 A33=(+) 1 | ||
| + | |||
| + | <br> | ||
| + | |||
| + | You get the matrix | ||
| + | |||
| + | <math>C= \left(\begin{array}{cccc}1&0&-1\\-1&4&-1\\1&-2&1\end{array}\right)</math><br> | ||
| + | |||
| + | If you multiply A by the adjacent matrix in this case "C" you get the Identity matrix multiplied by the determinant. | ||
| + | |||
| + | Lets prove it:<br> | ||
| + | |||
| + | <math>\left(\begin{array}{cccc}1&0&-1\\1&1&1\\2&1&2\end{array}\right)</math> * <math>\left(\begin{array}{cccc}1&0&-1\\-1&4&-1\\1&-2&1\end{array}\right)</math> = <math>\left(\begin{array}{cccc}2&0&0\\0&2&0\\0&0&2\end{array}\right)</math> | ||
| + | |||
| + | <br> | ||
| + | With this you get the formula | ||
| − | + | <span class="texhtml">''A''<sup> − 1</sup> = 1 / ''d''''e''''t''(''A'') * ''a''''d''''j''(''A'')</span><br> | |
| − | + | With this formula you can calculate the inverse of a matrix with a different approach, and if needed you can also calculate a single entry of the inverse. | |
| − | + | [[Category:MA265Fall2011Walther]] | |
Latest revision as of 08:07, 11 April 2013
The Inverse of a Matrix
Student project for MA265
Definition: Let A be a square matrix of order n x n(square matrix). If there exists a matrix B such that
Then B is called the inverse matrix of A.
Conditions
A n x n is invertible (non-singular) if:
- Ax=0 has a unique solution
- There is a B matrix such that A B = In
- Ax=b has a unique solution for any b---x=A − 1
Properties
- (AB) − 1 = B − 1A − 1
- (A1 A2.....Ar) − 1=Ar − 1A'''r − 1 − 1...A1 − 1
- (A − 1) − 1 = A
- (A − 1)T = (AT) − 1
Calculations
- Steps:
1) Put the original matrix and in the left side the corresponding identity matrix (In) in the right side.
2) You compute rref in the left side, keep in mind that the operations also have an effect on the right side.
3) After you have a reduced row echelon form in the left side, the matrix that is left on the right side is the Inverse of the original matrix.
$ \left(\begin{array}{cccc}2&3|1&0\\4&5|0&1\end{array}\right) $ ----->$ \left(\begin{array}{cccc} 2 & 3 | 1 & 0 \\ 0 & -1 | -2 & 1 \end{array}\right) $------>$ \left(\begin{array}{cccc} 2 & 0 | -5 & 3 \\ 0 & -1 | -2 & 1 \end{array}\right) $ ----> $ \left(\begin{array}{cccc} 1 & 0 || -5/2 & 3/2 \\0&1 || 2 & -1 \end{array}\right) $
$ A^{-1}=\left(\begin{array}{cccc} -5/2 & 3/2 \\ 2 & -1 \end{array}\right) $
Note: Calculating the Reuduced Row echelon form for a square matrix with n >5 can get complicated and if you get the Reduced row echelon form wrong by consequence you get the Inverse wrong. In some cases it is better to use the adjacent matrix as I will show on the next section.
Adjacent Matrix
A is a n x n matrix.
Def: the (i,j) minor of A is the submatrix of A obtained by erasing row i and column j of A. there are n2 submissions.
To each Mij we associate the number
Aij = ( − 1)i + j * d'e't(M'i'j)
Finally let C be the "cofactor matrix" with entries Cij = Aij.
Note: The cofactor or adjacent matrix does not have the same order of entries as A.
Example 1:
Normal Matrix Adjacent Matrix
$ \left(\begin{array}{cccc}A11&A12&A13\\A21&A22&A23\\A31&A32&A33\end{array}\right) $ $ \left(\begin{array}{cccc}A11&A21&A31\\A12&A22&A32\\A13&A23&A33\end{array}\right) $
Example:
Steps:
1) Get each of the 9 cofactor matrices by eliminating the row and the column of each element.
2) Get each determinat from the 2x2 matrices
3) Place the correct sign for each determinant. (If the sum of i and j = even its positive, odd negative).
$ A= \left(\begin{array}{cccc}1&0&-1\\1&1&1\\2&1&2\end{array}\right) $
$ M11=\left(\begin{array}{cccc}1&1\\1&2\end{array}\right) $ $ M12=\left(\begin{array}{cccc}1&1\\2&2\end{array}\right) $ $ M13=\left(\begin{array}{cccc}1&1\\2&1\end{array}\right) $
$ M21=\left(\begin{array}{cccc}0&-1\\1&2\end{array}\right) $ $ M22=\left(\begin{array}{cccc}1&-1\\2&2\end{array}\right) $ $ M23=\left(\begin{array}{cccc}1&0\\2&1\end{array}\right) $
$ M31=\left(\begin{array}{cccc}0&-1\\1&1\end{array}\right) $ $ M32=\left(\begin{array}{cccc}1&-1\\1&1\end{array}\right) $ $ M33=\left(\begin{array}{cccc}1&0\\1&1\end{array}\right) $
A11=(+)1 A12=(-)0 A13=(+)-1
A21=(-) 1 A22=(+) 4 A32=(-) 1
A31=(+) 1 A32=(-)2 A33=(+) 1
You get the matrix
$ C= \left(\begin{array}{cccc}1&0&-1\\-1&4&-1\\1&-2&1\end{array}\right) $
If you multiply A by the adjacent matrix in this case "C" you get the Identity matrix multiplied by the determinant.
Lets prove it:
$ \left(\begin{array}{cccc}1&0&-1\\1&1&1\\2&1&2\end{array}\right) $ * $ \left(\begin{array}{cccc}1&0&-1\\-1&4&-1\\1&-2&1\end{array}\right) $ = $ \left(\begin{array}{cccc}2&0&0\\0&2&0\\0&0&2\end{array}\right) $
With this you get the formula
A − 1 = 1 / d'e't(A) * a'd'j(A)
With this formula you can calculate the inverse of a matrix with a different approach, and if needed you can also calculate a single entry of the inverse.
