Answer :

This can be written as:


|A| = 1(2) – 1(4) + 1(2)


= 2 – 4 + 2


|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 14 – 12 = 2


C21 = (– 1)2 + 1 7 – 4 = – 3


C31 = (– 1)3 + 1 3 – 2 = 1


C12 = (– 1)1 + 2 7 – 3 = – 4


C22 = (– 1)2 + 1 7 – 1 = 6


C32 = (– 1)3 + 1 3 – 1 = 2


C13 = (– 1)1 + 2 4 – 2 = 2


C23 = (– 1)2 + 1 4 – 1 = – 3


C33 = (– 1)3 + 1 2 – 1 = 1


adj A =


=


Adj A x B =


Now, AX = B has infinite many solution


Let z = k


Then, x + y = 6 – k


X + 2y = 14 – 3k


This can be written as:



|A| = 1


Adj A =


Now, X = A – 1B =


=


=



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


Hence, x = k – 2, y = 8 – 2k, z = k


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