Question

x+y+z+w=2 2x-y+2z-w=-5 3x+2y+3z+4w=7 x-2y-3z+2w=5 solve the system of equations by Gauss-Jordan method​

Answer 1

Answer:

x+y+z+w=2 2x-y+2z-w=-5 3x+2y+3z+4w=7 x-2y-3z+2w=5

Answer 2

Answer:

You really want me to solve this? Its a linear system that has more variables than equations, so unless appropriate variables cancel out during the row echelon procedure, its pretty sure there aren’t any unique solutions.

We start off by writing the system.

x+2y−z+3v+w2x+4y−2z+6v+3w−x−2y+z−v+3w=2=6=4x+2y−z+3v+w=22x+4y−2z+6v+3w=6−x−2y+z−v+3w=4

Now convert the system into matrix format…

⎛⎝⎜⎜12−124−2−1−2136−1133264⎞⎠⎟⎟⎛⎝⎜⎜100200−100302114226⎞⎠⎟⎟Operation: {R3=R1+R3R2=R2+(−2)R1⎧⎩⎨w=22v+4w=6⟹v+2w=3⟹v=−1(12−131224−2636−1−21−134)Operation: {R3=R1+R3R2=R2+(−2)R1(12−1312000012000246){w=22v+4w=6⟹v+2w=3⟹v=−1

And for the first equation… consider y=s,z=ty=s,z=t because these are our free variables

x+2y−z+3v+wx+2s−t−3+2x=2=2=t−2s+3x+2y−z+3v+w=2x+2s−t−3+2=2x=t−2s+3

This proves that the system indeed, has infinite