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Question
Solve the following equations by using a reduction method
2x−y+z=1,x+2y+3z=8,3x+y−4z=1
2x−y+z=1,x+2y+3z=8,3x+y−4z=1
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Solution
Given equation:2x−y+z=1x+2y+3z=83x+1y−4z=1Matrix form of a system of linear equations⎡⎢⎣2−1112331−4⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣181⎤⎥⎦R3→R3−8R2⎡⎢⎣2−1112−30−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣18−23⎤⎥⎦
R2→R2−12R1⎡⎢
⎢
⎢⎣2−11052520−5−13⎤⎥
⎥
⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣115/2−23⎤⎥⎦
R2→25R2⎡⎢⎣2−110110−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−23⎤⎥⎦
R3→R3+5R2⎡⎢⎣2−1101100−8⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−8⎤⎥⎦
R1→R1+R2R2→R3/(−8)⎡⎢⎣202011001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣431⎤⎥⎦
R1→R1×(12)R2→R2−R3⎡⎢⎣101010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
R1→R1−R3⎡⎢⎣100010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦⇒x=1y=2 and z=1
2x−y+z=1
x+2y+3z=8
3x+1y−4z=1
Matrix form of a system of linear equations
⎡⎢⎣2−1112331−4⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣181⎤⎥⎦
R3→R3−8R2
⎡⎢⎣2−1112−30−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣18−23⎤⎥⎦
R2→R2−12R1
⎡⎢
⎢
⎢⎣2−11052520−5−13⎤⎥
⎥
⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣115/2−23⎤⎥⎦
R2→25R2
⎡⎢⎣2−110110−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−23⎤⎥⎦
R3→R3+5R2
⎡⎢⎣2−1101100−8⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−8⎤⎥⎦
R1→R1+R2R2→R3/(−8)
⎡⎢⎣202011001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣431⎤⎥⎦
R1→R1×(12)R2→R2−R3
⎡⎢⎣101010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
R1→R1−R3
⎡⎢⎣100010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
⇒x=1y=2 and z=1
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