Diketahui matriks A= (1 3 2 2 6 2 5 9 4). Tentukan berikut

det A = -12, adj(A) = [[6, 6, -6], [2, -6, 2], [-12, 6, 0]], A^-1 = [[-1/2, -1/2, 1/2], [-1/6, 1/2, -1/6], [1, -1/2, 0]]

Pembahasan

Untuk matriks A= [[1, 3, 2], [2, 6, 2], [5, 9, 4]]:

a. Determinan (det A):
| 1  3  2 |
det A = | 2  6  2 |
| 5  9  4 |

det A = 1(6*4 - 2*9) - 3(2*4 - 2*5) + 2(2*9 - 6*5)
      = 1(24 - 18) - 3(8 - 10) + 2(18 - 30)
      = 1(6) - 3(-2) + 2(-12)
      = 6 + 6 - 24
      = -12

b. Adjoin (adj A):
Determinan minor:
M11 = (6*4 - 2*9) = 24-18 = 6
M12 = -(2*4 - 2*5) = -(8-10) = 2
M13 = (2*9 - 6*5) = 18-30 = -12
M21 = -(3*4 - 2*9) = -(12-18) = 6
M22 = (1*4 - 2*5) = 4-10 = -6
M23 = -(1*9 - 3*5) = -(9-15) = 6
M31 = (3*2 - 2*6) = 6-12 = -6
M32 = -(1*2 - 2*2) = -(2-4) = 2
M33 = (1*6 - 3*2) = 6-6 = 0

Matriks Kofaktor (C):
[[ 6,  2, -12]
 [ 6, -6,   6]
 [-6,  2,   0]]

Adjoin A adalah transpose dari matriks kofaktor:
adj(A) = C^T = [[ 6,  6, -6],
              [ 2, -6,  2],
              [-12, 6,  0]]

c. Invers (A^-1):

A^-1 = (1 / det A) * adj(A)
     = (1 / -12) * [[ 6,  6, -6],
                      [ 2, -6,  2],
                      [-12, 6,  0]]
     = [[-6/12, -6/12,  6/12],
        [-2/12,  6/12, -2/12],
        [ 12/12, -6/12,  0/12]]
     = [[-1/2, -1/2,  1/2],
        [-1/6,  1/2, -1/6],
        [ 1,   -1/2,  0]]