If , find A–1. Using A–1 solve the system of equations

2x – 3y + 5z = 11


3x + 2y – 4z = – 5


x + y – 2z = – 3



|A| = 2(-4+4) + 3(-6+4) + 5(3-2) = -1 ≠ 0


So inverse of A exists,


Now A11 = 0, A12 = 2, A13 = 1, A21 = -1, A22 = -9, A23 = -5, A31 = 2, A32 = 23, A33 = 13


So AdjA =



So .


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


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