How do you solve the system #3x-y+z=5#, #x+3y+3z=-6#, and #x+4y-2z=12#?
1 Answer
Explanation:
Given:
#"(i)":color(white)(1/1)3x-y+z=5#
#"(ii)":color(white)(1/1)x+3y+3z=-6#
#"(iii)":color(white)(1/1)x+4y-2z=12#
Note that we can eliminate
#"(iv)" = "(ii)" - 3"(i)":color(white)(1/1)-8x+6y=-21#
#"(v)" = "(iii)" + 2"(i)":color(white)(1/1)7x+2y=22#
Then we can eliminate
#"(vi)" = "(iv)"-3"(v)":color(white)(1/1)-29x = -87#
Dividing both sides of
#x = 3#
Substituting this value of
#7(color(blue)(3))+2y = 22#
Hence:
#y = 1/2#
Then from
#z = 5-3x+y = 5-3(color(blue)(3))+color(blue)(1/2) = -7/2#
Matrix formulation
We can represent the original equations as an augmented matrix:
#((3, -1, 1, |, 5), (1, 3, 3, |, -6), (1, 4, -2, |, 12))#
The operations we performed above correspond to various row operations:
Subtract
#((3, -1, 1, |, 5), (-8, 6, 0, |, -21), (7, 2, 0, |, 22))#
Subtract
#((3, -1, 1, |, 5), (-29, 0, 0, |, -87), (7, 2, 0, |, 22))#
Divide
#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (7, 2, 0, |, 22))#
Subtract
#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 2, 0, |, 1))#
Divide
#((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 1, 0, |, 1/2))#
Add
#((0, 0, 1, |, -7/2), (1, 0, 0, |, 3), (0, 1, 0, |, 1/2))#
Permute the rows, moving
#((1, 0, 0, |, 3), (0, 1, 0, |, 1/2), (0, 0, 1, |, -7/2))#
Now the left hand