Find A – 1, if . Hence, solve the following system of linear equations:

x + 2y + 5z = 10, x – y – z = – 2,


2x + 3y – z = – 11

A =

|A| = 1(1 + 3) + 2(– 1 + 2) + 5(3 + 2)


= 4 + 2 + 25


= 27


Now, the cofactors of A


C11 = (– 1)1 + 1 1 + 3 = 4


C21 = (– 1)2 + 1 – 2 – 15 = 17


C31 = (– 1)3 + 1 – 2 + 5 = 3


C12 = (– 1)1 + 2 – 1 + 2 = – 1


C22 = (– 1)2 + 1 – 1 – 10 = – 11


C32 = (– 1)3 + 1 – 1 – 5 = 6


C13 = (– 1)1 + 2 3 + 2 = 5


C23 = (– 1)2 + 1 3 – 4 = 1


C33 = (– 1)3 + 1 – 1 – 2 = – 3


adj A =


A – 1 =


A – 1 =


Now the given equation can be written as:



A X B


Or, X = A – 1B


=



X =


X =


Hence, x = – 1,y = – 2 and z = 3


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