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Question

# Solve the system of equations, using matrix methodx−y+z=4,2x+y−3z=0,x+y+z=2

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Solution

## Given system of equationsx−y+z=42x+y−3z=0x+y+z=2This can be written as AX=Bwhere A=⎡⎢⎣1−1121−3111⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣402⎤⎥⎦Here, |A|=1(1+3)+1(2+3)+1(2−1)⇒|A|=4+5+1=10Since, |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∣∣∣1−311∣∣∣⇒C11=1+3=4C12=(−1)1+2∣∣∣2−311∣∣∣⇒C12=−(2+3)=−5C13=(−1)1+3∣∣∣2111∣∣∣⇒C13=2−1=1C21=(−1)2+1∣∣∣−1111∣∣∣⇒C21=−(−1−1)=2C22=(−1)2+2∣∣∣1111∣∣∣⇒C22=1−1=0C23=(−1)2+3∣∣∣1−111∣∣∣⇒C23=−(1+1)=−2C31=(−1)3+1∣∣∣−111−3∣∣∣⇒C31=3−1=2C32=(−1)3+2∣∣∣112−3∣∣∣⇒C32=−(−3−2)=5C33=(−1)3+3∣∣∣1−121∣∣∣⇒C33=1+2=3Hence, the co-factor matrix is C=⎡⎢⎣4−5120−2253⎤⎥⎦⇒adjA=CT=⎡⎢⎣422−5051−23⎤⎥⎦⇒A−1=adjA|A|=110⎡⎢⎣422−5051−23⎤⎥⎦Solution is given by X=A−1B⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣422−5051−23⎤⎥⎦⎡⎢⎣402⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣16+4−20+104+6⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=110⎡⎢⎣20−1010⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣2−11⎤⎥⎦Hence, x=2,y=−1,z=1

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