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Question

# Solve the following system of linear equations by matrix method:3x+x+z=10,2x−y−z=0,x−y+2z=1

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Solution

## Given system of equation-3x+y+z=102x−y−z=0x−y+2z=1Let A=⎡⎢⎣3112−1−11−12⎤⎥⎦,B=⎡⎢⎣100−1⎤⎥⎦,X=⎡⎢⎣xyz⎤⎥⎦Now,AX=B⎡⎢⎣3112−1−11−12⎤⎥⎦⎡⎢⎣XYZ⎤⎥⎦,X=⎡⎢⎣100−1⎤⎥⎦|A|=∣∣ ∣∣3112−1−11−12∣∣ ∣∣=3(−2−1)−1(4−(−1))+1(−2−(−1))=−9−5−1=−15∵|A|=−15≠0Thus, system of equation is consistent and has a unique solution.AX=B⇒X=A−1BNow,A−1=adj.(A)|A|adj.(a)=⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦A=⎡⎢⎣3112−1−11−12⎤⎥⎦A11=−2−1=−3,A12=−[4−(−1)]=−5A13=−2−(−1)=−1A21=−[2−(−1)]=−3,A22=6−1=5A23=−[−3−(1)]=4A31=−1−(−1)=0,A32=−[−3−2]=5A33=−3−2=−5Therefore,adj(A)=⎡⎢⎣−3−5−1−35405−5⎤⎥⎦′=⎡⎢⎣−3−30−555−14−5⎤⎥⎦∴A−1=1−15⎡⎢⎣−3−30−555−14−5⎤⎥⎦=⎡⎢ ⎢ ⎢⎣1515013−13−13115−41513⎤⎥ ⎥ ⎥⎦Therefore,⎡⎢⎣xyz⎤⎥⎦=⎡⎢ ⎢ ⎢⎣1515013−13−13115−41513⎤⎥ ⎥ ⎥⎦⎡⎢⎣100−1⎤⎥⎦⎡⎢⎣xyZ⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣2+0+0103+0+1323+0−13⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣211313⎤⎥ ⎥ ⎥ ⎥ ⎥⎦Hence value of x,y and z are 2,113 and 13 respectively.

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