Find the complete solution of the system
{x−3y+2z+w=−2x−2y−2w=−10z+5w=153x+2z+w=−3
We transform the system into reduced row-echelon form
[1−321−21−20−2−100015153021−3]
R2−R1→R2
[1−321−201−2−3−80015153021−3]
R4−3R1→R4
[1−321−201−2−3−800151509−4−23]
R4−9R2→R4
[1−321−201−2−3−800151500142575]
R4−14R3→R4
[1−321−201−2−3−8001515000−45−135]
−145R4
[1−321−201−2−3−800151500013]
R3−5R4→R3
[1−321−201−2−3−80010000013]
R2+3R4→R2
[1−321−201−2010010000013]
R1−R4→R1
[1−320−501−2010010000013]
R2+2R3→R2
[1−320−5010010010000013]
R1−2R3→R1
[1−300−5010010010000013]
R1+3R2→R1
[1000−2010010010000013]
We now have an equivalent matrix in reduced row-echelon form, and the system of equations is
{x=−2y=1z=0w=3
We can write the solution as the ordered quadruple (−2,1,0,3).
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