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Question
Solve the following system of equation by matrix method.
3x−2y+3z=8
2x+y−z=1
4x−3y+2z=4
3x−2y+3z=8
2x+y−z=1
4x−3y+2z=4
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Solution
A=⎡⎢⎣3−2321−14−32⎤⎥⎦,x=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣814⎤⎥⎦
|A|=⎡⎢⎣3−2321−14−32⎤⎥⎦
=3[1−1−32]−1(−2)[2−142]+3[214−3]⇒3(2−3)−2(4+4)+3(−6−4)
=−3+16−30=−17
Thus |A|≠0∴ We have unique solution & consistentAX=BX=A−1B=1|A|adj(A)B.adj C=(Aii)TA=⎡⎢⎣m−2321−14−32⎤⎥⎦cofactor of 3=[1−132]=2−3=−1cofactor of −2=8cofactor of 3=−10cofactor of 2=5cofactor of 1=−6cofactor of −1=−1cofactor of 4−1cofactor of −3=−9cofactor of 2=7
⇒⎡⎢⎣−18−105−6−1−1−97⎤⎥⎦
⇒ adj A=(aii)T=⎡⎢⎣−1−5−1−8−6+9−10+17⎤⎥⎦
Now ⇒x=−117⎡⎢⎣−1−5−1−8−69−1017⎤⎥⎦[814]=−117⎡⎢⎣−1(8)+(−5)1+(−1)4−8(8)+(−6)1+9(4)−10(8)+1(1)+7(4)⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=−117⎡⎢⎣−17−34−51⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣123⎤⎥⎦
∴x=1,y=2,z=3.
A=⎡⎢⎣3−2321−14−32⎤⎥⎦,x=⎡⎢⎣xyz⎤⎥⎦,B=⎡⎢⎣814⎤⎥⎦
|A|=⎡⎢⎣3−2321−14−32⎤⎥⎦
=3[1−1−32]−1(−2)[2−142]+3[214−3]⇒3(2−3)−2(4+4)+3(−6−4)
=−3+16−30=−17
Thus |A|≠0
∴ We have unique solution & consistent
AX=B
X=A−1B=1|A|adj(A)B.
adj C=(Aii)T
A=⎡⎢⎣m−2321−14−32⎤⎥⎦
cofactor of 3=[1−132]=2−3=−1
cofactor of −2=8
cofactor of 3=−10
cofactor of 2=5
cofactor of 1=−6
cofactor of −1=−1
cofactor of 4−1
cofactor of −3=−9
cofactor of 2=7
⇒⎡⎢⎣−18−105−6−1−1−97⎤⎥⎦
⇒ adj A=(aii)T=⎡⎢⎣−1−5−1−8−6+9−10+17⎤⎥⎦
Now ⇒x=−117⎡⎢⎣−1−5−1−8−69−1017⎤⎥⎦[814]=−117⎡⎢⎣−1(8)+(−5)1+(−1)4−8(8)+(−6)1+9(4)−10(8)+1(1)+7(4)⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=−117⎡⎢⎣−17−34−51⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣123⎤⎥⎦
∴x=1,y=2,z=3.
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