# Find A – 1, if . Hence, solve the following system of linear equations:x + 2y + 5z = 10, x – y – z = – 2,2x + 3y – z = – 11 (CBSE 2012)

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

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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