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I am trying to solve the following equation using gauss jordan method but unable to solve due to the type of equations.At the end i am getting unwanted zeros in 2nd and 3rd row.Here is my work...

   1  2  2 |-1...... r1

   2 -1 -1 | 3.......r2

   3  1  1 | 3........r3


   1  2  2 |-1...... r1

   0  -5 -5| 5........r2'=r2-r1*2

   0  -5 -5| 9........r3'=r3-r1*3


   1  2   2|-1........r1

   0  1   1|-1.........r2''=r2'/(-5)   

   0 -5  -5| 9.........r3'


   1  2   2|-1.....r1
   1  0   0| 1.....r2'''=r1-r2''*2
   0  0   0|-4......r3''=r3'+r2''*5 

how to solve this so that i can get the following form?

1 0 0|value1
0 1 0|value2
0 0 1|value3
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2 Answers 2

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Because the last row is all zeros except for value 3, you know there is no solution since there is no x,y,z such that 0*x + 0*y + 0*z = -4.

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  • $\begingroup$ what would happen if i had 0*x + 0*y + 0*z = 0 in last row ? $\endgroup$ Oct 20, 2014 at 19:39
  • $\begingroup$ then you may get a solution, since that is true for all x,y,z. $\endgroup$ Oct 22, 2014 at 21:29
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In order to have the identity matrix on the left hand side, the solution of the linear system you are trying to solve must be unique.

Now, for a linear system to have a unique solution the rank of the matrix should be the same as the number of unknowns. Taking a look at the system of equations you have the rank of the matrix is 2, while there are 3 unknowns, therefore, there is no unique solution and you cannot get the identity on the LHS.

You can read more about this in Wikipedia - Rank (linear algebra)

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