# Solve the f

The given system can be written in matrix form as: A X = B

Now,

|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)

= 7 + 19 – 22

= 4

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

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 12 – 5 = 7

C21 = (– 1)2 + 1 – 3 + 2 = 1

C31 = (– 1)3 + 1 5 – 8 = – 3

C12 = (– 1)1 + 2 9 + 10 = – 19

C22 = (– 1)2 + 1 3 – 4 = – 1

C32 = (– 1)3 + 1 – 5 – 6 = 11

C13 = (– 1)1 + 2 – 3 – 8 = – 11

C23 = (– 1)2 + 1 – 1 + 2 = – 1

C33 = (– 1)3 + 1 4 + 3 = 7

adj A = = A – 1 = Now, X = A – 1B = X =  Hence, X = 2,Y = 1 and Z = 3

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