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