Question

use elementary row operations to transform each augmented coefficient matrix to echelon form. Then solve the system by back substitution.

2x_1+5x_2+12x_3=6
3x_1+x_2+5x_3=12
5x_1+8x_2+21x_3=17

Answer 1

2 5 12 6
3 1 5 12
5 8 21 17

that's how I put it in the matrix.

Then I did -3/2(R1+R2)
and got..

2 5 12 6
0 -13/2 -13 3
5 8 21 17

then I did.. -5/2(R1+R3)
and got..

2 5 12 6
0 -13/2 -13 3
0 -9/2 -9 2

Answer 2

@UnkleRhaukus can you help me? :) please

Answer 3

why did you use those row operations?

Answer 4

PS i dont think this system has a solution

Answer 5

I'm not familiar with this topic.. so im just trying things out.. I'm not sure how to choose the 'right' row operations. how do you know if it doesnt have a solution?

Answer 6

i have a matrix solving program that is helpful sometimes.

.

Answer 7

***
lets try some different row operations

(LESS mistakes this time)

2 5 12 6
3 1 5 12
5 8 21 17


2 5 12 6
1 -4* -7 6 R_2 becomes R_2-R_1
1 -2 -3 5 R_3 becomes R_3-2R_1

Answer 8

now

0 9 18 -4 R_1 becomes R_1-2R_3
0 -6* -10* 1 R_2 becomes R_2-R_3
1 -2 -3 5

Answer 9

1 -2 -3 5 R_1 becomes R_3
0 1 5/3 -1/6 R_2 becomes R_2/-6
0 9 18 -4 R_3 becomes R_1


1 -2 -3 5
0 1 5/3 -1/6
0 1 2 -4/9 R_3 becomes R_3/9

1 -2 -3 5
0 1 5/3 -1/6
0 0 1/3 -5/18 R_3 becomes R_3-R_2

1 -2 -3 5
0 1 5/3 -1/6
0 0 1 -5/6 R_3 becomes 3*R_3

Answer 10

1 0 1/3 14/3 R_1 becomes R_1+2R_2
0 1 0 11/9 R_2 becomes R_2-5/3 R_3
0 0 1 -5/6 R_3 becomes 3*R_3


1 0 0 89/18 R_1 becomes R_1-1/3 R_3
0 1 0 11/9
0 0 1 -5/6

lets see if this works
___
2x_1+5x_2+12x_3=6

2(89/18)+5(11/9)+12(-5/6)= 6 \(\checkmark\)
___
3x_1+x_2+5x_3=12

3(89/9)+11/9+5(-5/6) ~ 11.888... ≠12 (not right)
___
5x_1+8x_2+21x_3=17

5(89/9)+8(11/9)+21(-5/6) ~ 41.72... ≠ 17 (not right)

so i have made mistakes in the working somewhere

Answer 11

i dont want to do this any more

Answer 12

I figured it out a different way. Thank you for your help though.

Answer 13

can you tell me what you got as an answer ?