Line 32: Line 32:
 
<br>  
 
<br>  
  
 +
<br>
  
== Calculations ==
+
== 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>  
 
<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>  
Line 40: Line 40:
 
<br>  
 
<br>  
  
<span class="texhtml">''A''<sup> 1</sup> = <math>\left(\begin{array}{cccc}  -5/2 & 3/2  \\  2 & -1    \end{array}\right)</math></span>
+
<span class="texhtml"><math>A^{-1}=\left(\begin{array}{cccc}  -5/2 & 3/2  \\  2 & -1    \end{array}\right)</math></span>  
  
<span class="texhtml" /><br> <br><br><br>
+
&lt;span class="texhtml" /&gt;<br> <br><br><br>  
  
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
+
Note: Calculating a Reuduced Row echelon form for a square matrix which n &gt;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 05:58, 16 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

Questions/answers with a recent ECE grad

Ryne Rayburn