Q. 294.7( 3 Votes )

Solve the f

Answer :

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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