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