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*<span class="texhtml">(AB)<sup> − 1</sup></span> = <span class="texhtml">''B''<sup> − 1</sup>''A''<sup> − 1</sup></span> | *<span class="texhtml">(AB)<sup> − 1</sup></span> = <span class="texhtml">''B''<sup> − 1</sup>''A''<sup> − 1</sup></span> | ||
| − | *(A1 A2.....Ar)<span class="texhtml"><sup> − 1</sup></span>=<span class="texhtml">''A''r'''''<b><sup> − 1</sup></b>'''''A''''''''r'' − 1<sup> − 1</sup>...''A''1<sup> − 1</sup></span> | + | *(A1 A2.....Ar)<span class="texhtml"><sup> − 1</sup></span>=<span class="texhtml">''A''r'''''<b><sup> − 1</sup></b>'''''A''''''''r'' − 1<sup> − 1</sup>...''A''1<sup> − 1</sup>'''</span> |
*<span class="texhtml">(''A''<sup> − 1</sup>)<sup> − 1</sup> = ''A''</span><span class="texhtml"><sup></sup></span> | *<span class="texhtml">(''A''<sup> − 1</sup>)<sup> − 1</sup> = ''A''</span><span class="texhtml"><sup></sup></span> | ||
*<span class="texhtml">(''A''<sup> − 1</sup>)<sup>''T''</sup> = (''A''<sup>''T''</sup>)<sup> − 1</sup></span> | *<span class="texhtml">(''A''<sup> − 1</sup>)<sup>''T''</sup> = (''A''<sup>''T''</sup>)<sup> − 1</sup></span> | ||
| + | <br> | ||
| + | Calculations | ||
| − | + | <math>\left(\begin{array}{cccc}2&3|1&0\\4&5|0&1\end{array}\right)</math> -----><math>\left(\begin{array}{cccc} 2 & 3 | 1 & 0 \\ 0 & -1 | -2 & 1 \end{array}\right)</math>------><math>\left(\begin{array}{cccc} 2 & 0 | -5 & 3 \\ 0 & -1 | -2 & 1 \end{array}\right)</math> ----> <math>\left(\begin{array}{cccc} 1 & 0 | -5/2 & 3/2 \\ 0 & 1 | 2 & -1 \end{array}\right)</math> | |
| + | |||
| + | <br> | ||
| + | |||
| + | <span class="texhtml">''A''<sup> − 1</sup> = <math>\left(\begin{array}{cccc} -5/2 & 3/2 \\ 2 & -1 \end{array}\right)</math></span><br> <br><br><br> | ||
Revision as of 14:33, 13 December 2011
Inverse of a Matrix
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
$ \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) $
