Diketahui matriks A=(1 -1 2 2 k 1), B=(0 3 2 1 1 -1), dan (AB^T)^-1=(a b c d) dengan B^T menyatakan tranpose B. Jika det(AB^T)=-2 maka a+b+c+d=...

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Saluranedukasi Admin · Accepted Answer

Diketahui matriks:
A = [1 -1 2]
    [2  k 1]

B = [0 3 2]
    [1 1 -1]

B^T (transpose dari B) adalah:
B^T = [0 1]
      [3 1]
      [2 -1]

Perkalian AB^T:
AB^T = [1  -1  2] [0  1]
       [2   k  1] [3  1]
                [2 -1]

AB^T = [(1*0 + -1*3 + 2*2)  (1*1 + -1*1 + 2*-1)]
       [(2*0 + k*3 + 1*2)  (2*1 + k*1 + 1*-1)]

AB^T = [(0 - 3 + 4)  (1 - 1 - 2)]
       [(0 + 3k + 2)  (2 + k - 1)]

AB^T = [1  -2]
       [3k+2  k+1]

Diketahui (AB^T)^-1 = [a b]
                      [c d]

Kita tahu bahwa jika M = [p q]
                      [r s]
M^-1 = 1/(ps-qr) * [s -q]
                   [-r p]

Jadi, (AB^T)^-1 = 1/((1)*(k+1) - (-2)*(3k+2)) * [k+1  -(-2)]
                                           [-(3k+2)  1]

(AB^T)^-1 = 1/(k+1 + 6k+4) * [k+1  2]
                                [-3k-2  1]

(AB^T)^-1 = 1/(7k+5) * [k+1  2]
                       [-3k-2  1]

(AB^T)^-1 = [(k+1)/(7k+5)   2/(7k+5)]
           [(-3k-2)/(7k+5)  1/(7k+5)]

Dari sini kita dapatkan:
a = (k+1)/(7k+5)
b = 2/(7k+5)
c = (-3k-2)/(7k+5)
d = 1/(7k+5)

Diketahui det(AB^T) = -2.
Dari perhitungan AB^T, determinannya adalah (1)*(k+1) - (-2)*(3k+2) = k+1 + 6k+4 = 7k+5.
Jadi, 7k+5 = -2.

Ini berarti 1/(7k+5) = 1/(-2).

Sekarang kita bisa mencari a+b+c+d:
a + b + c + d = (k+1)/(7k+5) + 2/(7k+5) + (-3k-2)/(7k+5) + 1/(7k+5)

a + b + c + d = (k+1 + 2 - 3k - 2 + 1) / (7k+5)

a + b + c + d = (k - 3k + 1 + 2 - 2 + 1) / (7k+5)

a + b + c + d = (-2k + 2) / (7k+5)

Karena kita tahu 7k+5 = -2, maka kita substitusikan:
a + b + c + d = (-2k + 2) / (-2)

Untuk mencari -2k+2, kita perlu nilai k dari 7k+5 = -2:
7k = -2 - 5
7k = -7
k = -1

Maka, -2k + 2 = -2(-1) + 2 = 2 + 2 = 4.

Jadi, a + b + c + d = 4 / (-2)
a + b + c + d = -2

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