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Calculations
 
  
<math>\left(\begin{array}{cccc}2&3|1&0\\4&5|0&1\end{array}\right)</math> -----&gt;<math>\left(\begin{array}{cccc}  2 & 3 | 1  & 0  \\  0 & -1    |  -2  &  1 \end{array}\right)</math>------&gt;<math>\left(\begin{array}{cccc}  2 & 0 | -5  & 3  \\  0 & -1    |  -2  &  1 \end{array}\right)</math> ----&gt; <math>\left(\begin{array}{cccc}  1 & 0    | -5/2  & 3/2  \\  0 & 1    |  2  & -1 \end{array}\right)</math>  
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== Calculations ==
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<math>\left(\begin{array}{cccc}2&3|1&0\\4&5|0&1\end{array}\right)</math> -----&gt;<math>\left(\begin{array}{cccc}  2 & 3 | 1  & 0  \\  0 & -1    |  -2  &  1 \end{array}\right)</math>------&gt;<math>\left(\begin{array}{cccc}  2 & 0 | -5  & 3  \\  0 & -1    |  -2  &  1 \end{array}\right)</math> ----&gt; <math>\left(\begin{array}{cccc}  1 & 0    |   -5/2  &   3/2  \\  0 & 1    |  2  & -1 \end{array}\right)</math>  
  
 
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<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>
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<span class="texhtml">''A''<sup> − 1</sup> = <math>\left(\begin{array}{cccc}  -5/2 & 3/2  \\  2 & -1    \end{array}\right)</math></span>
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<span class="texhtml" /><br> <br><br><br>
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Note: Calculating a Reuduced Row echelon form for a square matrix which n >5 can get complicated and if you get the Reduced row echelon form by consequence you get the Inverse wrong. In some cases

Revision as of 13:40, 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

A B = I n = B A

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) $

<span class="texhtml" />



Note: Calculating a Reuduced Row echelon form for a square matrix which n >5 can get complicated and if you get the Reduced row echelon form by consequence you get the Inverse wrong. In some cases

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

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