1
Question
Solve the following system of linear equations by matrix method:
3x+x+z=10,2x−y−z=0,x−y+2z=1
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)]=4
A31=−1−(−1)=0,A32=−[−3−2]=5A33=−3−2=−5
Therefore,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.
Given system of equation-
3x+y+z=10
2x−y−z=0
x−y+2z=1
Let 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≠0
Thus, system of equation is consistent and has a unique solution.
AX=B
⇒X=A−1B
Now,
A−1=adj.(A)|A|
adj.(a)=⎡⎢⎣A11A12A13A21A22A23A31A32A33⎤⎥⎦
A=⎡⎢⎣3112−1−11−12⎤⎥⎦
A11=−2−1=−3,A12=−[4−(−1)]=−5A13=−2−(−1)=−1
A21=−[2−(−1)]=−3,A22=6−1=5A23=−[−3−(1)]=4
A31=−1−(−1)=0,A32=−[−3−2]=5A33=−3−2=−5
Therefore,
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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