Solve the following system of equations by matrix method:

3x + 4y + 2z = 8

2y – 3z = 3

x – 2y + 6z = – 2

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 3

= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)

= 18 – 12 – 4

= 2

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (12 – 6) = 6

C₂₁ = (– 1)^{2 + 1}(24 + 4) = – 28

C₃₁ = (– 1)^{3 + 1}(– 12 – 4) = – 16

C₁₂ = (– 1)^{1 + 2} (0 + 3) = – 3

C₂₂ = (– 1)^{2 + 1} 18 – 2 = 16

C₃₂ = (– 1)^{3 + 1} – 9 – 0 = 9

C₁₃ = (– 1)^{1 + 2} (0 – 2) = – 2

C₂₃ = (– 1)^{2 + 1} (– 6 – 4) = 10

C₃₃ = (– 1)^{3 + 1} 6 – 0 = 6

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = – 2,Y = 3 and Z = 1