Answer:
x = 3, y = - 1, z = 1
Explanation:
listing the coefficients
1 - 2 1 6 ← R1
3 1 - 1 7 ← R2
4 - 1 2 15 ← R3
We require the first entry in R2 to be 0 while retaining R1 and the first 2 entries of R3 to be 0, thus
R2 - 3R1 and R3 - 4R1
1 - 2 1 6 ← R1
0 7 - 4 - 11 ← R2
0 7 - 2 - 9 ← R3
Now R3 - R2
1 - 2 1 6 ← R1
0 7 - 4 - 11 ← R2
0 0 2 2 ← R3
From R3
2z = 2 ⇒ z = 1
Substitute z = 1 into R2
7y - 4(1) = - 11
7y - 4 = - 11 ( add 4 to both sides )
7y = - 7 ⇒ y = - 1
Substitute y = - 1, x = 1 into R1
x - 2(- 1) + 1 = 6
x + 2 + 1 = 6
x + 3 = 6 ( subtract 3 from both sides )
x = 3
Solution is (3, - 1, 1 )