Question

Find the inverse of matrix A by using adjoint method; where A = `[(1, 0, 1), (0, 2, 3), (1, 2, 1)]`

Answer 1

where A = `[(1, 0, 1), (0, 2, 3), (1, 2, 1)]`

|A| = I(2 -6) - 0(0 -3) + 1(0 - 2)

|A| = -4 -2

|A| = -6 ≠ 0

`therefore A^-1` exists.

Minors	Co-factors
M11 = -4	A11 = -4
M12 = -3	A12 = 3
M13 = -2	A13 = -2
M21 = -2	A21 = 2
M22 = 0	A22 = 0
M23 = 2	A23 = -2
M31 = -2	A31 = -2
M32 = 3	A32 = -3
M33 = 2	A33 = 2

adj (A) = `[(-4, 2, -2), (3, 0, -3), (-2, -2, 2)]`

A^-1 = `1/|A|.` adj(A)

A^-0 = `(-1)/6 [(-4, 2, -2), (3, 0, -3), (-2, -2, 2)]`

Answer 2

where A = `[(1, 0, 1), (0, 2, 3), (1, 2, 1)]`

|A| = I(2 -6) - 0(0 -3) + 1(0 - 2)

|A| = -4 -2

|A| = -6 ≠ 0

`therefore A^-1` exists.

First we have to find the cofactor matrix

`= ["A"_"ij"]_(3xx3), "where"  "A"_"ij" = (- 1)^("i" + "j") "M"_"ij"`

Now `"A"_11 = (-1)^(1 + 1)  "M"_11 = |(2,3),(2,1)| = 2 - 6 = - 4`

`"A"_12 = (-1)^(1 + 2)  "M"_12 = |(0,3),(1,1)| = 0 - 3 = 3`

`"A"_13 = (-1)^(1 + 3)  "M"_13 =  |(0,2),(1,2)| = 0 - 2 = - 2`

`"A"_21 = (-1)^(2 + 1)  "M"_21 = |(0,1),(2,1)| = - 0 - 2 = 2`

`"A"_22 = (-1)^(2 + 2)  "M"_22 = |(1,1),(1,1)| = 1 - 1 = 0`

`"A"_23 = (-1)^(2 + 3)  "M"_23 = |(1,0),(1,2)| = - 2 - 0 = - 2`

`"A"_31 = (-1)^(3 + 1)  "M"_31 = |(0,1),(2,1)| = 0 - 2 = - 2`

`"A"_32 = (-1)^(3 + 2)  "M"_32 = |(1,1),(0,3)| = - 3 - 0 = - 3`

`"A"_33 = (-1)^(3 + 3)  "M"_33 = |(1,0),(0,2)| = 2 - 0 = 2`

Matrix of co. factor is = `[(-4,  3, -2), (2, 0, -2), (-2, -3, 2)]`

adj (A) = `[(-4, 2, -2), (3, 0, -3), (-2, -2, 2)]`

A^-1 = `1/|A|.` adj(A)

A^-0 = `(1)/6 [(-4, 2, -2), (3, 0, -3), (-2, -2, 2)]`