A =

|A| = 2(0) + 3(– 2) + 5(1)

= – 1

Now, the cofactors of A

C₁₁ = (– 1)^{1 + 1} – 4 + 4 = 0

C₂₁ = (– 1)^{2 + 1} 6 – 5 = – 1

C₃₁ = (– 1)^{3 + 1} 12 – 10 = 2

C₁₂ = (– 1)^{1 + 2} – 6 + 4 = 2

C₂₂ = (– 1)^{2 + 1} – 4 – 5 = – 9

C₃₂ = (– 1)^{3 + 1} – 8 – 15 = 23

C₁₃ = (– 1)^{1 + 2} 3 – 2 = 1

C₂₃ = (– 1)^{2 + 1} 2 + 3 = – 5

C₃₃ = (– 1)^{3 + 1} 4 + 9 = 13

adj A =

A ^{– 1 =}

A ^{– 1} =

A ^{– 1} =

Now the given equation can be written as:

A X B

Or, X = A ^{– 1}B

=

X =

Hence, x = 1,y = 2 and z = 3