Answer :

This can be written as:


|A| = 5(260 – 4) – 3(30 – 14) + 7(6 – 182)


= 5(256) – 3(16) + 7(176)


|A| = 0


So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:


(Adj A)x B≠0 or (Adj A)x B = 0


Cofactors of A are:


C11 = (– 1)1 + 1 260 – 4 = 256


C21 = (– 1)2 + 1 30 – 14 = – 16


C31 = (– 1)3 + 1 6 – 182 = – 176


C12 = (– 1)1 + 2 30 – 14 = – 16


C22 = (– 1)2 + 1 50 – 49 = 1


C32 = (– 1)3 + 1 10 – 21 = 11


C13 = (– 1)1 + 2 6 – 182 = – 176


C23 = (– 1)2 + 1 10 – 21 = 11


C33 = (– 1)3 + 1 130 – 9 = 121


adj A =


=


Adj A x B =


Now, AX = B has infinite many solution


Let z = k


Then, 5x + 3y = 4 – 7k


3x + 26y = 9 – 2k


This can be written as:



|A| = 121


Adj A =


Now, X = A – 1B =


=


=



There values of x,y,z satisfy the third equation


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