Question

how to solve this using matrices:

x1+2x2-2x3+x4=2
2x1-x2+2x3=6
x2-2x4=-1

Answer 1

\[x _{1}+2x _{2}-2x _{3}+x _{4}=2\]

\[2x _{1}-x _{2}+2x _{3}=6\]

\[x _{2}-2x _{4}=-1\]

Answer 2

can be written in augmented matrix form as
=1 2 -2 1 2
2 -1 2 0 6
0 1 0 -2 -1

performing r2-->r2-2r1

=1 2 -2 1 2
0 -5 6 -2 2
0 1 0 -2 -1

performing r2-->r2+5r3

=1 2 -2 1 2
0 0 6 -12 -3
0 1 0 -2 -1

performing r2<-->r3

1 2 -2 1 2
0 1 0 -2 -1
0 0 6 -12 -3

since we have three equations but four variables we have to have a free variable.. i assume that variable to be x4 over here
so i will get the answer in terms of x4 and constants

1) 6x3-12x4=-3
=>6x3=-3+12x4
=>x3=2x4-(1/2)

2)x2-2x4=-1
=>x2=2x4-1

3)x1+2x2-2x3+x4=2
=>x1=2-2(2x4-1)+2(2x4-1/2)-x4
=>x1 = 2-x4+1
=>x1=3-x4

Answer 3

thank you! :)