Question 888422: use matrices to solve this system:
2x+3y+z=2
5x-2y-z=4
x-y-2z=-3
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! 2x+3y+z=2
5x-2y-z=4
x-y-2z=-3
2,3,1,2
5,-2,-1,4
1,-1,-2,-3
divide row 1 by 2/1
1,3/2,1/2,1
5,-2,-1,4
1,-1,-2,-3
add down (-5/1) *row 1 to row 2
1,3/2,1/2,1
0,-19/2,-7/2,-1
1,-1,-2,-3
add down (-1/1) *row 1 to row 3
1,3/2,1/2,1
0,-19/2,-7/2,-1
0,-5/2,-5/2,-4
divide row 2 by -19/2
1,3/2,1/2,1
0,1,-14/-38,-2/-19
0,-5/2,-5/2,-4
add down (5/2) *row 2 to row 3
1,3/2,1/2,1
0,1,7/19,2/19
0,0,-120/76,-142/38
divide row 3 by -30/19
1,3/2,1/2,1
0,1,7/19,2/19
0,0,1,-1349/-570
We now have the value for the last variable.
We will work our way up and get the other solutions.
add up (-7/19) *row 3 to row 2
1,3/2,1/2,1
0,1,0,-8303/10830
0,0,1,71/30
add up (-1/2) *row 3 to row 1
1,6/4,0,-11/60
0,1,0,-23/30
0,0,1,71/30
add up (-3/2) *row 2 to row 1
1,0,0,3480/3600
0,1,0,-23/30
0,0,1,71/30
final
1,0,0,29/30
0,1,0,-23/30
0,0,1,71/30
1,0,0,29/30 = 0.96666667
0,1,0,-23/30 = -0.76666667
0,0,1,71/30 = 2.36666667
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