1
Question
Solve the following equations by the method of reduction:2x−y+z=1,x+2y+3z=8,3x+y−4z=1
Solve the following equations by the method of reduction:
2x−y+z=1,x+2y+3z=8,3x+y−4z=1
Open in App
Solution
2x−y+z=1x+2y+3z=8,3x+y−4z=1⎡⎢⎣2−1112331−4⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣181⎤⎥⎦R3→R3−3R2⎡⎢⎣2−111230−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣18−23⎤⎥⎦R2→R2−12R1⎡⎢
⎢
⎢⎣2−11052520−5−13⎤⎥
⎥
⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢
⎢
⎢⎣1152−23⎤⎥
⎥
⎥⎦R2→25R2⎡⎢⎣2−110110−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−23⎤⎥⎦R3→R3+5R2⎡⎢⎣2−1101100−8⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−8⎤⎥⎦R1→R1+R2 and R3→R3×−18⎡⎢⎣202011001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣431⎤⎥⎦R2→R2−R3 and R1→R1×12⎡⎢⎣101010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣221⎤⎥⎦R1→R1−R3 ⎡⎢⎣100010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦∴x=1,y=2,z=1.
2x−y+z=1x+2y+3z=8,3x+y−4z=1
⎡⎢⎣2−1112331−4⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣181⎤⎥⎦
R3→R3−3R2
⎡⎢⎣2−111230−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣18−23⎤⎥⎦
R2→R2−12R1
⎡⎢
⎢
⎢⎣2−11052520−5−13⎤⎥
⎥
⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢
⎢
⎢⎣1152−23⎤⎥
⎥
⎥⎦
R2→25R2
⎡⎢⎣2−110110−5−13⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−23⎤⎥⎦
R3→R3+5R2
⎡⎢⎣2−1101100−8⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣13−8⎤⎥⎦
R1→R1+R2 and R3→R3×−18
⎡⎢⎣202011001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣431⎤⎥⎦
R2→R2−R3 and R1→R1×12
⎡⎢⎣101010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣221⎤⎥⎦
R1→R1−R3
⎡⎢⎣100010001⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣121⎤⎥⎦
∴x=1,y=2,z=1.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
