Solving systems of linear equations

Answer: a = 2, b = 3, c = -1,

See below for the detailed solution.

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************ Solving Linear Equations ************
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2a + 3b - 2c = 15
a + b = 5
2a - b + 5c = -4

Number of Unknows are: 3 ( a, b, c,  )

Writing in Matrix form:  [A][X] = [B]

Where [A] = 
      [   2   3  -2   |
      |   1   1   0   |
      |   2  -1   5   ]

[X] = 
      [  a  |
      |  b  |
      |  c  ]

[B] = 
      [  15  |
      |  5   |
      |  -4  ]

Combining [A] & [B] for the purpose of gaussian elimination, we get

[   2   3  -2   15   |
 |   1   1   0   5    |
 |   2  -1   5  -4    ]

Converting the combined matrix to a upper triangular matrix by doing rows manipulations

Performing: 
R2 = R2 - (1/2) R1
R3 = R3 - (2/2) R1

[   2   3    -2   15    |
     |   0  -0.5   1  -2.5   |
     |   0  -4     7  -19    ]

Performing: 
R3 = R3 - (-4/-0.5) R2

[   2   3    -2   15    |
     |   0  -0.5   1  -2.5   |
     |   0   0    -1   1     ]

Re-writing back to the equations, we get

2a + 3b - 2c = 15			Equation 1
    -0.5b + c = -2.5			Equation 2
    -c = 1			Equation 3
  From Equation 3, we get
  c = -1
  Substituting c,  in Equation 2(As required), we get
  b = 3
  Substituting c, b,  in Equation 1(As required), we get
  a = 2

Solving Linear Equations Using Gaussian Elimination Method

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