Use Gaussian elimination to solve the system
$2 x - 3 y = - 5$
$3 x + y = 9$
Find the value of $x$ .

Question

Use Gaussian elimination to solve the system  2  x  −  3  y  =  −  5  3  x  +  y  =  9Find the value of   x.

Syllingbs · Accepted Answer

Since we want to find the value of   x, we need to eliminate y from both the equations i.e. from the equations                    2        x        −        3        y                            =        −        5                    Equation         1                            3        x        +        y                            =        9                    Equation         2            Though you can do it directly, without resorting to matrices, to enable you to understand how to work with matrices, I am going to do this in the matrix form.The two equations can be written in a matrix form as shown below.                              [                                                    2                                            −                3                                                                    3                                            1                                              ]                          [                                                    x                                                                    y                                              ]                =                  [                                                    −                5                                                                    9                                              ]                    Typically, Gauss elimination (  L  U factorization) involves making the entries below the main diagonal zero using row-operations i.e.                              [                                                    2                                            −                3                                                                    3                                            1                                              ]                →                  [                                                    ×                                            ×                                                                    0                                            ×                                              ]                    However, you could also do Gauss elimination by making the entries above the main diagonal zero using row-operations i.e.                              [                                                    2                                            −                3                                                                    3                                            1                                              ]                →                  [                                                    ×                                            0                                                                    ×                                            ×                                              ]                    Since you are interested in   x, it makes sense to proceed along the second line and convert                              [                                                    2                                            −                3                                                                    3                                            1                                              ]                    to a lower triangular matrix.This can be done as follows. If we denote the first equation as       R    1   and the second equation as       R    2  , then the operation       R    1    =      R    1    +  3      R    2   converts                              [                                                    2                                            −                3                                                                    3                                            1                                              ]                          [                                                    x                                                                    y                                              ]                =                  [                                                    −                5                                                                    9                                              ]                    to                              [                                                    2                +                3                ×                3                                            −                3                +                3                ×                1                                                                    3                                            1                                              ]                          [                                                    x                                                                    y                                              ]                =                  [                                                    −                5                +                3                ×                9                                                                    9                                              ]                 i.e.                   [                                                    11                                            0                                                                    3                                            1                                              ]                          [                                                    x                                                                    y                                              ]                =                  [                                                    22                                                                    9                                              ]                    The first row/equation can now be used to read of the value for   x.  11  x  =  22    ⟹    x  =  2If you are interested in y as well, plug in the value for   x and obtain   y from the second equation, i.e.  3  ×  2  +  y  =  9    ⟹    y  =  3

Hugo Brady · Accepted Answer

I had the same problem, thanks!

Use Gaussian elimination to solve the system 2

Answered question

Answer & Explanation

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