Solve the following system of equations by matrix method:

3x + 4y + 2z = 8


2y – 3z = 3


x – 2y + 6z = – 2

The given system can be written in matrix form as:


or A X = B


A = , X = and B =


Now, |A| = 3


= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)


= 18 – 12 – 4


= 2


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


X = A – 1B


Cofactors of A are:


C11 = (– 1)1 + 1 (12 – 6) = 6


C21 = (– 1)2 + 1(24 + 4) = – 28


C31 = (– 1)3 + 1(– 12 – 4) = – 16


C12 = (– 1)1 + 2 (0 + 3) = – 3


C22 = (– 1)2 + 1 18 – 2 = 16


C32 = (– 1)3 + 1 – 9 – 0 = 9


C13 = (– 1)1 + 2 (0 – 2) = – 2


C23 = (– 1)2 + 1 (– 6 – 4) = 10


C33 = (– 1)3 + 1 6 – 0 = 6


adj A =


=


A – 1 =


Now, X = A – 1B =


X =


X =


X =


Hence, X = – 2,Y = 3 and Z = 1


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