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Question
Solve the system of equations, using matrix method2x+y+z=1,x−2y−z=32,3y−5z=9
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Solution
Given system of equations2x+y+z=1
x−2y−z=32
3y−5z=9This can be written as AX=Bwhere A=⎡⎢⎣2111−2−103−5⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢
⎢
⎢⎣1329⎤⎥
⎥
⎥⎦
Here, |A|=2(10+3)−1(−5−0)+1(3−0)⇒|A|=26+5+3=34Since, |A|≠0Hence, the system of equations is consistent and has a unique solution given by X==A−1B
A−1=adjA|A| and adjA=CT
C11=(−1)1+1∣∣∣−2−13−5∣∣∣⇒C11=10+3=13
C12=(−1)1+2∣∣∣1−10−5∣∣∣⇒C12=−(−5−0)=5
C13=(−1)1+3∣∣∣1−203∣∣∣⇒C13=3−0=3
C21=(−1)2+1∣∣∣113−5∣∣∣⇒C21=−(−5−3)=8
C22=(−1)2+2∣∣∣210−5∣∣∣⇒C22=−10−0=−10
C23=(−1)2+3∣∣∣2103∣∣∣⇒C23=−(6−0)=−6
C31=(−1)3+1∣∣∣11−2−1∣∣∣⇒C31=−1+2=1
C32=(−1)3+2∣∣∣211−1∣∣∣⇒C32=−(−3−0)=3
C33=(−1)3+3∣∣∣211−2∣∣∣⇒C33=−4−1=−5
Hence, the co-factor matrix is C=⎡⎢⎣13538−10−613−5⎤⎥⎦
⇒adjA=CT=⎡⎢⎣13815−1033−6−5⎤⎥⎦
⇒A−1=adjA|A|=134⎡⎢⎣13815−1033−6−5⎤⎥⎦
Solution is given by ⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13815−1033−6−5⎤⎥⎦⎡⎢⎣13/29⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13+12+95−15+273−9−45⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣3417−51⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣11/2−3/2⎤⎥⎦
Hence, x=1,y=12,z=−32
2x+y+z=1
x−2y−z=32
x−2y−z=32
3y−5z=9
This can be written as
AX=B
where A=⎡⎢⎣2111−2−103−5⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢
⎢
⎢⎣1329⎤⎥
⎥
⎥⎦
Here, |A|=2(10+3)−1(−5−0)+1(3−0)
⇒|A|=26+5+3=34
Since, |A|≠0
Hence, the system of equations is consistent and has a unique solution given by X==A−1B
A−1=adjA|A| and adjA=CT
C11=(−1)1+1∣∣∣−2−13−5∣∣∣
⇒C11=10+3=13
C12=(−1)1+2∣∣∣1−10−5∣∣∣
⇒C12=−(−5−0)=5
C13=(−1)1+3∣∣∣1−203∣∣∣
⇒C13=3−0=3
C21=(−1)2+1∣∣∣113−5∣∣∣
⇒C21=−(−5−3)=8
C22=(−1)2+2∣∣∣210−5∣∣∣
⇒C22=−10−0=−10
C23=(−1)2+3∣∣∣2103∣∣∣
⇒C23=−(6−0)=−6
C31=(−1)3+1∣∣∣11−2−1∣∣∣
⇒C31=−1+2=1
C32=(−1)3+2∣∣∣211−1∣∣∣
⇒C32=−(−3−0)=3
C33=(−1)3+3∣∣∣211−2∣∣∣
⇒C33=−4−1=−5
Hence, the co-factor matrix is C=⎡⎢⎣13538−10−613−5⎤⎥⎦
⇒adjA=CT=⎡⎢⎣13815−1033−6−5⎤⎥⎦
⇒A−1=adjA|A|=134⎡⎢⎣13815−1033−6−5⎤⎥⎦
Solution is given by
⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13815−1033−6−5⎤⎥⎦⎡⎢⎣13/29⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣13+12+95−15+273−9−45⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=134⎡⎢⎣3417−51⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣11/2−3/2⎤⎥⎦
Hence, x=1,y=12,z=−32
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