Q24 of 29 Page 1

If , then verify that A . adj A = |A|I. Hence, find A-1 also.


Now,


A11 = (-1)1+1 M11 = 7


A12 = (-1)1+2 M12 = -1


A13 = (-1)1+3 M13 = -1


A21 = (-1)2+1 M21 = -3


A22 = (-1)2+2 M22 = 1


A23 = (-1)2+3 M23 = 0


A31 = (-1)3+1 M31 = -3


A32 = (-1)3+2 M32 = 0


A33 = (-1)3+3 M33 = 1


Thus,





Taking RHS (i.e., |A| I),


Calculating |A|,



Now,



Thus, A(adj A) = |A| I


Hence, proved.


Finding A-1,



Here |A| = 1 ≠ 0. Thus A-1 exists.


So,



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