Solve the following system of equations by matrix method:

6x – 12y + 25z = 4

4x + 15y – 20z = 3

2x + 18y + 15z = 10

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 6

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)

= 3510 + 1200 + 1050

= 5760

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (225 + 360) = 585

C₂₁ = (– 1)^{2 + 1} (– 180 – 450) = 630

C₃₁ = (– 1)^{3 + 1} (240 – 375) = – 135

C₁₂ = (– 1)^{1 + 2} (60 + 40) = – 100

C₂₂ = (– 1)^{2 + 1} (90 – 50) = 40

C₃₂ = (– 1)^{3 + 1} (– 120 – 100) = 220

C₁₃ = (– 1)^{1 + 2} (72 – 30) = 42

C₂₃ = (– 1)^{2 + 1}(108 + 24) = – 132

C₃₃ = (– 1)^{3 + 1} (90 + 48) = 138

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

Hence, X = ,Y = and Z =