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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=1x−2y−z=323y−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−1BA−1=adjA|A| and adjA=CTC11=(−1)1+1∣∣∣−2−13−5∣∣∣⇒C11=10+3=13C12=(−1)1+2∣∣∣1−10−5∣∣∣⇒C12=−(−5−0)=5C13=(−1)1+3∣∣∣1−203∣∣∣⇒C13=3−0=3C21=(−1)2+1∣∣∣113−5∣∣∣⇒C21=−(−5−3)=8C22=(−1)2+2∣∣∣210−5∣∣∣⇒C22=−10−0=−10C23=(−1)2+3∣∣∣2103∣∣∣⇒C23=−(6−0)=−6C31=(−1)3+1∣∣∣11−2−1∣∣∣⇒C31=−1+2=1C32=(−1)3+2∣∣∣211−1∣∣∣⇒C32=−(−3−0)=3C33=(−1)3+3∣∣∣211−2∣∣∣⇒C33=−4−1=−5Hence, 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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