Answer:
option C)
(19,2,5)
Explanation:
| 4 | x
| 12 | y
| 5 | z
After reducing the matrix in row-echelon form we are left with three equation as following
Equation 1
0x + 0y + 1z = 5
Equation 2
0x + y1 + 2z = 12
Equation 3
1x + 0y -3z = 4
Solution
As in first equation
0 + 0 + z = 5
z = 5
Put this value of z in equation 2 and 3
y + 2(5) = 12
y + 10 =12
y = 12 - 10
y = 2
x + 0 -3(5) = 4
x - 15 = 4
x = 4 + 15
x = 19
So the solution of the system is (x,y,z) = (19,2,5)