Answer :

The given system can be written in matrix form as:



AX = B


Now,


|A| = 2(75) – 3(– 110) + 10(72)


= 150 + 330 + 720


= 1200


So, the above system has a unique solution, given by


X = A – 1B


Cofactors of A are:


C11 = (– 1)1 + 1 120 – 45 = 75


C21 = (– 1)2 + 1 – 60 – 90 = 150


C31 = (– 1)3 + 1 15 + 60 = 75


C12 = (– 1)1 + 2 – 80 – 30 = 110


C22 = (– 1)2 + 1 – 40 – 60 = – 100


C32 = (– 1)3 + 1 10 – 40 = 30


C13 = (– 1)1 + 2 36 + 36 = 72


C23 = (– 1)2 + 1 18 – 18 = 0


C33 = (– 1)3 + 1 – 12 – 12 = – 24


adj A =


=


A – 1 =


Now, X = A – 1B =


X =



Hence, X = 2,Y = 3 and Z = 5


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