Question

If A = `[(2,3),(1,2)]`, B = `[(1,0),(3,1)]`, find AB and (AB)^-1 . Verify that (AB)^-1 = B^-1.A^-1.

Answer 1

AB = `[(2,3),(1,2)] [(1,0),(3,1)]`

`= [(2+9,0+3),(1+6,0+2)] = [(11,3),(7,2)]`

∴ |AB| = `|(11,3),(7,2)| = 22 - 21 = 1 ne 0`

∴ (AB)^-1 exists.

Now, (AB)(AB)^-1 = I

∴ `[(11,3),(7,2)]("AB")^-1 = [(1,0),(0,1)]`

By 2R₁, we get,

`[(22,6),(7,2)] ("AB")^-1 = [(2,0),(0,1)]`

By R₁ - 3R₂, we get,

`[(1,0),(7,2)] ("AB")^-1 = [(2,-3),(0,1)]`

By R₂ - 7R₁, we get,

`[(1,0),(0,2)] ("AB")^-1 = [(2,-3),(-14,22)]`

By `(1/2)"R"_2,` we get,

`[(1,0),(0,1)] ("AB")^-1 = [(2,-3),(-7,11)]`

∴ (AB)^-1 = `[(2,-3),(-7,11)]`     ....(1)

|A| = `|(2,3),(1,2)| = 4 - 3 = 1 ne 0`

∴ A^-1 exists.

Consider, AA^-1 = I

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

By R₁ ↔ R₂, we get,

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

By `"R"_2 - 2"R"_1,`we get,

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

By `(-1)"R"_2`, we get,

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

By `"R"_1 - 2"R"_2`, we get,

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

∴ A^-1 = `[(2,-3),(-1,2)]`

|B| = `|(1,0),(3,1)| = 1 - 0 = 1 ne 0`

∴ B^-1 exists.

Consider, BB^-1 = I 

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

By `"R"_2 - 3"R"_1`, we get,

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

∴ `"B"^-1 = [(1,0),(-3,1)]`

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

`= [(2 - 0, -3+0),(-6-1,9+2)]`

`= [(2,-3),(-7,11)]`      ....(2)

From (1) and (2), `("AB")^-1 = "B"^-1."A"^-1.`