Jika A=(2 1 3 1 -1 2 1 2 1), tunjukkan bahwa: A^3-2A^2-9A=0

Terbukti bahwa A^3-2A^2-9A=0 dan A^2-2A-9I≠0 melalui perhitungan matriks.

Pembahasan

Untuk menunjukkan bahwa A^3 - 2A^2 - 9A = 0 dan A^2 - 2A - 9I ≠ 0, kita perlu melakukan perhitungan matriks.

Diketahui matriks A = [[2, 1, 3], [1, -1, 2], [1, 2, 1]].

Langkah 1: Hitung A^2

A^2 = A * A = [[2, 1, 3], [1, -1, 2], [1, 2, 1]] * [[2, 1, 3], [1, -1, 2], [1, 2, 1]]

A^2 = [[(2*2+1*1+3*1), (2*1+1*(-1)+3*2), (2*3+1*2+3*1)],
       [(1*2+(-1)*1+2*1), (1*1+(-1)*(-1)+2*2), (1*3+(-1)*2+2*1)],
       [(1*2+2*1+1*1), (1*1+2*(-1)+1*2), (1*3+2*2+1*1)]]

A^2 = [[(4+1+3), (2-1+6), (6+2+3)],
       [(2-1+2), (1+1+4), (3-2+2)],
       [(2+2+1), (1-2+2), (3+4+1)]]

A^2 = [[8, 7, 11],
       [3, 6, 3],
       [5, 1, 8]]

Langkah 2: Hitung A^3

A^3 = A^2 * A = [[8, 7, 11], [3, 6, 3], [5, 1, 8]] * [[2, 1, 3], [1, -1, 2], [1, 2, 1]]

A^3 = [[(8*2+7*1+11*1), (8*1+7*(-1)+11*2), (8*3+7*2+11*1)],
       [(3*2+6*1+3*1), (3*1+6*(-1)+3*2), (3*3+6*2+3*1)],
       [(5*2+1*1+8*1), (5*1+1*(-1)+8*2), (5*3+1*2+8*1)]]

A^3 = [[(16+7+11), (8-7+22), (24+14+11)],
       [(6+6+3), (3-6+6), (9+12+3)],
       [(10+1+8), (5-1+16), (15+2+8)]]

A^3 = [[34, 23, 49],
       [15, 3, 24],
       [19, 20, 25]]

Langkah 3: Hitung A^3 - 2A^2 - 9A

A^3 - 2A^2 - 9A = [[34, 23, 49], [15, 3, 24], [19, 20, 25]] - 2 * [[8, 7, 11], [3, 6, 3], [5, 1, 8]] - 9 * [[2, 1, 3], [1, -1, 2], [1, 2, 1]]

A^3 - 2A^2 - 9A = [[34, 23, 49], [15, 3, 24], [19, 20, 25]] - [[16, 14, 22], [6, 12, 6], [10, 2, 16]] - [[18, 9, 27], [9, -9, 18], [9, 18, 9]]

A^3 - 2A^2 - 9A = [[(34-16-18), (23-14-9), (49-22-27)],
               [(15-6-9), (3-12-(-9)), (24-6-18)],
               [(19-10-9), (20-2-18), (25-16-9)]]

A^3 - 2A^2 - 9A = [[0, 0, 0],
               [0, 0, 0],
               [0, 0, 0]]

Ini menunjukkan bahwa A^3 - 2A^2 - 9A = 0.

Langkah 4: Hitung A^2 - 2A - 9I

Di mana I adalah matriks identitas [[1, 0, 0], [0, 1, 0], [0, 0, 1]].

A^2 - 2A - 9I = [[8, 7, 11], [3, 6, 3], [5, 1, 8]] - 2 * [[2, 1, 3], [1, -1, 2], [1, 2, 1]] - 9 * [[1, 0, 0], [0, 1, 0], [0, 0, 1]]

A^2 - 2A - 9I = [[8, 7, 11], [3, 6, 3], [5, 1, 8]] - [[4, 2, 6], [2, -2, 4], [2, 4, 2]] - [[9, 0, 0], [0, 9, 0], [0, 0, 9]]

A^2 - 2A - 9I = [[(8-4-9), (7-2-0), (11-6-0)],
               [(3-2-0), (6-(-2)-9), (3-4-0)],
               [(5-2-0), (1-4-0), (8-2-9)]]

A^2 - 2A - 9I = [[-5, 5, 5],
               [1, -1, -1],
               [3, -3, -3]]

Karena matriks hasil A^2 - 2A - 9I bukan matriks nol, maka A^2 - 2A - 9I ≠ 0.