Solve the following system of equations by matrix method:
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12
The given system can be written in matrix form as:
A X = B
Now,
|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)
= 7 + 19 – 22
= 4
So, the above system has a unique solution, given by
X = A – 1B
Cofactors of A are:
C11 = (– 1)1 + 1 12 – 5 = 7
C21 = (– 1)2 + 1 – 3 + 2 = 1
C31 = (– 1)3 + 1 5 – 8 = – 3
C12 = (– 1)1 + 2 9 + 10 = – 19
C22 = (– 1)2 + 1 3 – 4 = – 1
C32 = (– 1)3 + 1 – 5 – 6 = 11
C13 = (– 1)1 + 2 – 3 – 8 = – 11
C23 = (– 1)2 + 1 – 1 + 2 = – 1
C33 = (– 1)3 + 1 4 + 3 = 7
adj A =
=
A – 1 =
Now, X = A – 1B =
X =
Hence, X = 2,Y = 1 and Z = 3