Question #6404

The following system of equations has a unique solution.
x−5y−9z=−21
−2x+5y−2z=−33
3x−4y+5z=42
You will solve this system using the method of Gauss-Jordan elimination.
Note, you CANNOT interchange rows of the matrix at any step.
Please follow exactly the instructions provided

Expert's answer

Here is the matrix of the given system:

& 1 -5 -9 | -21

-2& 5 -2 | -33

& 3 -4& 5 |& 42

Let's perform Gauss-Jordan elimination:

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 11 32 |& 105

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 0 -12 |& 60

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 0 -1& | -5

& 1 -5& 0& |& 24

& 0 -5& 0& |& 25

& 0& 0 -1& | -5

& 1 -5& 0& |& 24

& 0 -1& 0& |& 5

& 0& 0& 1& |& 5

& 1& 0& 0& | -1

& 0 -1& 0& |& 5

& 0& 0& 1& |& 5

& 1& 0& 0& | -1

& 0& 1& 0& | -5

& 0& 0& 1& |& 5

So, solution is (-1,-5,5).

& 1 -5 -9 | -21

-2& 5 -2 | -33

& 3 -4& 5 |& 42

Let's perform Gauss-Jordan elimination:

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 11 32 |& 105

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 0 -12 |& 60

& 1 -5 -9& | -21

& 0 -5 -20 | -75

& 0& 0 -1& | -5

& 1 -5& 0& |& 24

& 0 -5& 0& |& 25

& 0& 0 -1& | -5

& 1 -5& 0& |& 24

& 0 -1& 0& |& 5

& 0& 0& 1& |& 5

& 1& 0& 0& | -1

& 0 -1& 0& |& 5

& 0& 0& 1& |& 5

& 1& 0& 0& | -1

& 0& 1& 0& | -5

& 0& 0& 1& |& 5

So, solution is (-1,-5,5).

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