If A=2−3532−411−2, find A−1. Use it to solve the system of equations 2x−3y+5z=113x+2y−4z=−5x+y−2z=−3.ORUsing elementary row transformations, find the inverse of the matrix A=123257−2−4−5.

Question

If A=2−3532−411−2, find A−1. Use it to solve the system of equations 2x−3y+5z=113x+2y−4z=−5x+y−2z=−3.ORUsing elementary row transformations, find the inverse of the matrix A=123257−2−4−5.

Accepted Answer

Given: A=2−3532−411−2Now, |A|=2−3532−411−2=2(−4+4)+3(−6+4)+5(3−2)|A|=0−6+5=−1≠0∴ A−1 exists. Let Aij be the cofactors of elements aij in A=aij, thenA11=(−1)1+12−41−2=−4+4=0A12=(−1)1+23−41−2=−(−6+4)=2A13=(−1)1+33211=3−2=1A21=(−1)2+1−351−2=−(6−5)=−1A22=(−1)2+2251−2=−4−5=−9A23=(−1)2+32−311=−(2+3)=−5A31=(−1)3+1−352−4=(12−10)=2A32=(−1)3+2253−4=−(−8−15)=23A33=(−1)3+32−332=4+9=13∴ Aij=021−1−9−522313adj⁡(A)=AijT=0−122−9231−513A−1=adj⁡(A)|A|=1(−1)0−122−9231−513=01−2−29−23−15−13 Now, given system of equations can be written a2−3−532−411−2xyz    =11−5−3⇒        AX    =B∴        X    =A−1B    X    =01−2−29−23−15−1311−5−3⇒        xyz    =0−5+6−22−45+69−11−25+39=123∴        x    =1,y=2,z=3Therefore, the solutions are x=1, y=2 and z=3.ORGiven: A=123257−2−4−5Now, We know that A=IA1        2       32       5         7−2    −4    −5=1    0    00    1    00    0    1AOn applying, R2→R2−2R1 and R3→R3+2R1, we get1    2    30    1    10    0    1=1    0    0−2    1    02    0    1AOn applying, R1→R1−3R3 and R2→R2−R3 we get 1    2    00    1    00    0    1=−5    0    −3−4    1    −12    0    1AOn applying, R1→R1−2R2, we get    100010001    =3−2−1−41−1201A∴        A−1=    3−2−1−41−1201Therefore, the inverse of the given matrix is 3−2−1−41−1201

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