SOLUTION: Solve by the Gauss-Jordan elimination: x+y+2z+t=1 x+2y+z+t=2 2x+y+z+t=4 x+y+z+2t=3

Question 343991: Solve by the Gauss-Jordan elimination:
x+y+2z+t=1
x+2y+z+t=2
2x+y+z+t=4
x+y+z+2t=3
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!




Put in all the 1 coefficients



Take out the variables and replace the column
of equal signs by a vertical line:



Do row operations

-1R1+R2->R2

(That means to multiply row 1 temporarily by -1
and add it to row 2, replace row 2 by that, then
restore row 1 as it was):



-2R1+R3->R3



-R1+R4->R4



-R2+R1->R1



R2+R3->R3



R3->R4
R4->R3
(That means to interchange row 3 and row 4)



-R3->R3
{That means to multiply row 3 by -1)



-3R3+R1->R1



R3+R2->R2



4R3+R4->R4



R4->R4
(That means to multiply row 4 by )



-4R4+R1->R1



R4+R2->R2



-R4+R3->R3



Place the variables and equal signs back in



Simplify:



Edwin

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