# Solve the following system of equations by matrix method:   (CBSE 2011)

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

Now,

|A| = 2(75) – 3(– 110) + 10(72)

= 150 + 330 + 720

= 1200

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

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 120 – 45 = 75

C21 = (– 1)2 + 1 – 60 – 90 = 150

C31 = (– 1)3 + 1 15 + 60 = 75

C12 = (– 1)1 + 2 – 80 – 30 = 110

C22 = (– 1)2 + 1 – 40 – 60 = – 100

C32 = (– 1)3 + 1 10 – 40 = 30

C13 = (– 1)1 + 2 36 + 36 = 72

C23 = (– 1)2 + 1 18 – 18 = 0

C33 = (– 1)3 + 1 – 12 – 12 = – 24

adj A = = A – 1 = Now, X = A – 1B = X =  Hence, X = 2,Y = 3 and Z = 5

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