Row operation?

Question

How do i do this? use row operation to solve this system of linear equations:
the underscore(_) means subscript
2x_1 - x_2 + x_3 = 6
x_1 + x_2 + x_3 + 2x_4 = 0
2x_1 + x_2 - x_3 + x_4 = -3
x_1 + 2x_2 + 2x_3 +x_4 = -9

any clues on how this works?

Answer 1

Double the second and fourth equations so all have the same coefficient of x_1:

2x1 - x2 + x3 = 6
2x1 + 2x2 + 2x3 + 4x4 = 0
2x1 + x2 - x3 + x4 = -3
2x1 + 4x2 + 4x3 + 2x4 = -18

Now subtract the first equation from each of the others:

2x1 - x2 + x3 = 6
3x2 + x3 + 4x4 = -6
2x2 - 2x3 + x4 = -9
5x2 + 3x3 + 2x4 = -24

Subtract the 2nd and 3rd equations from the fourth:

2x1 - x2 + x3 = 6
3x2 + x3 + 4x4 = -6
2x2 - 2x3 + x4 = -9
4x3 - 3x4 = -9

Multiply the 2nd equation by 2, and the third by 3:

2x1 - x2 + x3 = 6
6x2 + 2x3 + 8x4 = -12
6x2 - 6x3 + 3x4 = -27
4x3 - 3x4 = -9

Subtract the 2nd eq from the 3rd:

2x1 - x2 + x3 = 6
6x2 + 2x3 + 8x4 = -12
-8x3 - 5x4 = -15
4x3 - 3x4 = -9

Double the 4th eq:

2x1 - x2 + x3 = 6
6x2 + 2x3 + 8x4 = -12
-8x3 - 5x4 = -15
8x3 - 6x4 = -18

Add the 3rd eq to the 4th:


2x1 - x2 + x3 = 6
6x2 + 2x3 + 8x4 = -12
-8x3 - 5x4 = -15
- 11x4 = -33

So x4 = 3. Plug that into 3rd eq and get x3 = 0. Plug both values into 2nd eq and get x2 = -6. Finally, plug all 3 values into 1st eq and get x1=0.

Answer 2

firsrt add the 1st equation to 2nd equation and 3rd equation :
2x_1 - x_2 + x_3 =6
2x_1 +2 x_3 + 2x_4 =6
4_x1 + x_4 = 3
x_1 + 2x_2 + 2x_3 + x_4
we will get this : x_4 = 3 - 4x_1
put it into the 2nd equation we will have : x_3 = (5x_1)/2
do it again with the 4th equation we find that : x_2 = -6 - x_1
Then, we will use all of these in the 1st equation to find the value of x_1 : 11x_1/2 = 0 ; so x_1 = 0, x_2 = -6 ; x_3 = 0 ; x_4 = 3