The above system of equations can be written as

or AX = B

Where A = B = and X =

|A| = 9 – 5 = 4

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

X = A ^{– 1}B

Let C_ij be the cofactor of a_ijin A, then

C₁₁ = (– 1)^{1 + 1} 3 = 3

C₁₂ = (– 1)^{1 + 2} 5 = – 5

C₂₁ = (– 1)^{2 + 1} 1 = – 1

C₂₂ = (– 1)^{2 + 2} 3 = 3

Also, adj A =

=

A ^{– 1} =

A ^{– 1} =

Now, X = A ^{– 1}B

Hence, X = Y =