Question

Find the inverse of the following matrix (if they exist):

`[(2,-3,3),(2,2,3),(3,-2,2)]`

Answer 1

Let A = `[(2,−3,3),(2,2,3),(3,−2,2)]`

∴ |A| = `|(2,-3,3),(2,2,3),(3,−2,2)|`

= 2(4 + 6) + 3(4 − 9) + 3(− 4 − 6)

= 20 − 15 − 30

= − 25 ≠ 0

∴ A^-1 exists.

Consider AA^-1 = I

∴ `[(2,-3,3),(2,2,3),(3,-2,2)] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]`

By R₁↔R₃, we get,

`[(3,-2,2),(2,2,3),(2,-3,3)] "A"^-1 = [(0,0,1),(0,1,0),(1,0,0)]`

By R₁ → R₁ − R₂, we get

`[(1,-4,-1),(2,2,3),(2,-3,3)] "A"^-1 = [(0,-1,1),(0,1,0),(1,0,0)]`

By R₂ → R₂ − 2R₁ and R₃ → R₃ − 2R₁, we get,

`[(1,-4,-1),(0,10,5),(0,5,5)] "A"^-1 = [(0,-1,1),(0,3,-2),(1,2,-2)]`

By R₂ → `(1/10) "R"_2`, we get,

`[(1,-4,-1),(0,1,1/2),(0,5,5)] "A"^-1 = [(0,-1,1),(0,3/10,-1/5),(1,2,-2)]`

By R₁ → R₁ + 4R₂ and R₃ → R₃ - 5R_2, we get,

`[(1,0,1),(0,1,1/2),(0,0,5/2)] "A"^-1 = [(0,1/5,1/5),(0,3/10,-1/5),(1,1/2,-1)]`

By R₃ → `(2/5)"R"_3`, we get,

`[(1,0,1),(0,1,1/2),(0,0,1)] "A"^-1 = [(0,1/5,1/5),(0,3/10,-1/5),(2/5,1/5,-2/5)]`

By R₁ →R₁ − R₃ and R₂ → `"R"_2 - 1/2"R"_3`, we get,

`[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(-2/5,0,3/5),(-1/5,1/5,0),(2/5,1/5,-2/5)]`

∴ `"A"^-1 = [(-2/5,0,3/5),(-1/5,1/5,0),(2/5,1/5,-2/5)]`