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Question
If A =⎡⎢⎣2−3532−411−2⎤⎥⎦ , find A−1. Hence using A−1 solve the system of equations
2x−3y+5z=11,3x+2y−4z=−5,x+y−2z=−3.
2x−3y+5z=11,3x+2y−4z=−5,x+y−2z=−3.
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Solution
Here, |A|=⎡⎢⎣2−3532−411−2⎤⎥⎦=2(−4+4)+3(−6+4)+5(3−2)=0−6+5=−1≠0∴A−1existsNow,A11=(−1)1+1∣∣∣2−41−2∣∣∣=+(−4+4)=0A12=(−1)1+2∣∣∣3−41−2∣∣∣=−(−6+4)=2A13=(−1)1+3∣∣∣3211∣∣∣=+(3−2)=1A21=(−1)2+1∣∣∣−351−2∣∣∣=−(6−5)=−1A22=(−1)2+2∣∣∣251−2∣∣∣=+(−4−5)=−9A23=(−1)2+3∣∣∣2−311∣∣∣=−(2+3)=−5A31=(−1)3+1∣∣∣−352−4∣∣∣=+(12−10)=2A32=(−1)3+2∣∣∣253−4∣∣∣=−(−8−15)=23A33=(−1)3+3∣∣∣2−332∣∣∣=+(4+9)=13∴A−1=1|A|(Adj A)=1−1⎡⎢⎣021−1−9−522313⎤⎥⎦T=1−1⎡⎢⎣0−122−9231−513⎤⎥⎦=⎡⎢⎣01−2−29−23−15−13⎤⎥⎦
The given system of equations can be written as:⎡⎢⎣2−3532−411−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣11−5−3⎤⎥⎦i.e.,AX=B
∴ Solution of the given system is given by X=A−1B.⇒ ⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣01−2−29−23−15−13⎤⎥⎦ ⎡⎢⎣11−5−3⎤⎥⎦=⎡⎢⎣0−5+6−22−45+69−11−25+39⎤⎥⎦=⎡⎢⎣123⎤⎥⎦ Hence, x=1,y=2 and z=3
The given system of equations can be written as:⎡⎢⎣2−3532−411−2⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣11−5−3⎤⎥⎦i.e.,AX=B
∴ Solution of the given system is given by X=A−1B.⇒ ⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣01−2−29−23−15−13⎤⎥⎦ ⎡⎢⎣11−5−3⎤⎥⎦=⎡⎢⎣0−5+6−22−45+69−11−25+39⎤⎥⎦=⎡⎢⎣123⎤⎥⎦ Hence, x=1,y=2 and z=3
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