1
Question
Solve the following system of equations by matrix method
x+y+z=6 , x−y−z=−4 , x+2y−2z=−1
x+y+z=6 , x−y−z=−4 , x+2y−2z=−1
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Solution
Consider the given system of equationsx+y+z=6x−y−z=4x+2y−2z=1Let us write above system of equations as⎡⎢⎣1111−1−112−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦⎡⎢⎣6−4−1⎤⎥⎦AX=B Let A=⎡⎢⎣1111−1−112−2⎤⎥⎦X=⎡⎢⎣xyz⎤⎥⎦B=⎡⎢⎣6−4−1⎤⎥⎦
Let us multiply A−1 on both sides, we have A−1AX=AB⇒IX=A−1B⇒X=A−1BConsider the given matrix A and we need to find A−1. We know that A=IA.⇒⎡⎢⎣1111−1−112−2⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦AR3→R2−R,R3→R3−R1⎡⎢⎣1110−2−2011⎤⎥⎦=⎡⎢⎣100−110−101⎤⎥⎦AR3→R3+R2⎡⎢⎣1110−2−201−8⎤⎥⎦=⎡⎢⎣100−110−312⎤⎥⎦A
R3→38,R2→R2−2⎡⎢⎣111011001⎤⎥⎦=⎡⎢⎣1001/2−1/203/8−1/8−2/8⎤⎥⎦AR2→R2−R3,R1→R1−R2⎡⎢⎣101010001⎤⎥⎦=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦AThus, A−1=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦We have X=A−1B⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦⎡⎢⎣6−4−1⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣248−168+068+128−28188+48+28⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣88168248⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢⎣123⎤⎥⎦Thus, x=1,y=2,z=3
Consider the given system of equations
x+y+z=6x−y−z=4
x+2y−2z=1
Let us write above system of equations as
⎡⎢⎣1111−1−112−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦⎡⎢⎣6−4−1⎤⎥⎦
AX=B Let A=⎡⎢⎣1111−1−112−2⎤⎥⎦X=⎡⎢⎣xyz⎤⎥⎦B=⎡⎢⎣6−4−1⎤⎥⎦
Let us multiply A−1 on both sides, we have A−1AX=AB⇒IX=A−1B⇒X=A−1B
Consider the given matrix A and we need to find A−1. We know that A=IA.
⇒⎡⎢⎣1111−1−112−2⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
R3→R2−R,R3→R3−R1
⎡⎢⎣1110−2−2011⎤⎥⎦=⎡⎢⎣100−110−101⎤⎥⎦A
R3→R3+R2
⎡⎢⎣1110−2−201−8⎤⎥⎦=⎡⎢⎣100−110−312⎤⎥⎦A
R3→38,R2→R2−2
⎡⎢⎣111011001⎤⎥⎦=⎡⎢⎣1001/2−1/203/8−1/8−2/8⎤⎥⎦A
R2→R2−R3,R1→R1−R2
⎡⎢⎣101010001⎤⎥⎦=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦A
Thus, A−1=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦
We have X=A−1B
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣4/84/801/8−3/82/83/8−1/8−2/8⎤⎥⎦⎡⎢⎣6−4−1⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣248−168+068+128−28188+48+28⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣88168248⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢⎣123⎤⎥⎦
Thus, x=1,y=2,z=3
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