Given that

X = , A = and B =

Also AX = B and we have to find the value of X ,

Pre-multiplying A^-1 both sides we get,

A^-1AX = A^-1B

IX = A^-1B ( A^-1A = I )

X = A^-1B ( IX = X) ……(i)

Now,

|A| = = 1(0-2) +1(2-3) +2(4-0) = -2-1+8 = 5

And adjA =

A^-1 = adjA =

From (i),

= = =

=

On comparing both sides we get,