A=⎡⎢⎣−51371−51−11⎤⎥⎦,B=⎡⎢⎣112321213⎤⎥⎦
∴AB=⎡⎢⎣−5+3+6−5+2+3−10+1+97+3−107+2−514+1−151−3+21−2+12−1+3⎤⎥⎦
AB=⎡⎢⎣400040004⎤⎥⎦
AB=4⎡⎢⎣100010001⎤⎥⎦
AB=4I3
⇒B(14A)=I3
⇒B−1=14A
⇒B−1=14⎡⎢⎣−51371−51−11⎤⎥⎦
The given system of equation is
x+y+2z=1
3x+2y+z=7
2x+y+3z=2
This system can be written as
⎡⎢⎣112321213⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣172⎤⎥⎦ or BX=C
where X=⎡⎢⎣xyz⎤⎥⎦,C=⎡⎢⎣172⎤⎥⎦
As B−1 exists, the given system has a unique solution
X=B−1C
⇒X=14⎡⎢⎣−51371−51−11⎤⎥⎦⎡⎢⎣172⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=14⎡⎢⎣−5+7+67+7−101−7+2⎤⎥⎦
⇒⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣21−1⎤⎥⎦⇒x=2,y=1,z=−1
Hence, the solution of given system of equation is x=2,y=1,z=−1.