System of linear equations using matrices

Hi again!

So even though I more or less know how to solve a system of linear equations, I still can't help keep getting lost whenever things get more complicated.(Headbang)

For example, I've got this linear system that I can't seem to solve even though the end result is supposed to be rather simple. So, I've got this linear system:

|2x_{1} + x_{2} - 5x_{3 }+ x_{4 }= 8

|x_{1} - 3x_{2} + 0x_{3 }-6x_{4 }= 9

|0x_{1 }+ 2x_{2 }- x_{3 }+ 2x_{4} = -5

|x_{1 }+ 4x_{2 }-7x_{3 }+ 6x_{4 }= 0

And so, using the coefficients, you get this coefficient matrix:

2 1 -5 1 | 8

1 -3 0 -6 | 9

0 2 -1 2 | -5

1 4 -7 6 | 0

Anyone care to help from here on? I've certainly tried, but my level of maths isn't enough to get an answer out of that.

Re: System of linear equations using matrices

Quote:

Originally Posted by

**JacobE** Hi again!

So even though I more or less know how to solve a system of linear equations, I still can't help keep getting lost whenever things get more complicated.(Headbang)

For example, I've got this linear system that I can't seem to solve even though the end result is supposed to be rather simple. So, I've got this linear system:

|2x_{1} + x_{2} - 5x_{3 }+ x_{4 }= 8

|x_{1} - 3x_{2} + 0x_{3 }-6x_{4 }= 9

|0x_{1 }+ 2x_{2 }- x_{3 }+ 2x_{4} = -5

|x_{1 }+ 4x_{2 }-7x_{3 }+ 6x_{4 }= 0

Look at the inverse.

Re: System of linear equations using matrices

Use Gauss–Jordan elimination.

1) reverse rows 1 and 2;

2) subtract row 1 from row 4, multiply row 1 by 2 and subtract it from row 2;

3) subtract row 2 from row 4;

4) add row 3 to row 1, multiply row 3 by 3 and subtract it from row 2;

5) multiply row 2 by 2 and subtract it from row 3;

6) add row 4 to row 3;

7) multiply row 3 by 2 and add it to row 4;

8) devide row 4 by -27;

9) multiply row 4: by 13 and add it row 3, by -7 and add it row 2, by 4 and add it row 1;

10) multiply row 3 by 2 and add it to row 2, add row 3 to row 1;

11) add row 2 to row 1.

Re: System of linear equations using matrices

I take it we could have stopped at stage 7 and said d=1, c-13d=-14 etc.

Re: System of linear equations using matrices

Yes, once you have a matrix row-reduced to upper (or lower) triangular form, you can solve the set of equations by "back-substitution".

Actually, I find the method a little bit peculiar- but perfectly valid. Personally, I tend to get the "0"s in each column, above, as well as below, the "pivot" number as I am working on the column the first time, not go through all columns getting "0" below the pivot, then back, through getting "0"s above the pivot.