Answer :

The given system can be written in matrix form as:



AX = B


Now, |A| = 8


= 8(– 1) – 4(1) + 3(3)


= – 8 – 4 + 9


= – 3


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


X = A – 1B


Cofactors of A are:


C11 = (– 1)1 + 1 1 – 2 = – 1


C21 = (– 1)2 + 1 4 – 6 = 2


C31 = (– 1)3 + 1 4 – 3 = 1


C12 = (– 1)1 + 2 2 – 1 = – 1


C22 = (– 1)2 + 1 8 – 3 = 5


C32 = (– 1)3 + 1 8 – 6 = – 2


C13 = (– 1)1 + 2 4 – 1 = 3


C23 = (– 1)2 + 1 16 – 4 = – 12


C33 = (– 1)3 + 1 8 – 8 = 0


adj A =


=


A – 1 =


Now, X = A – 1B =


X =


X =


X =


Hence, X = 1,Y = 1 and Z = 2


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