Diketahui matriks A = (6 2 3 3) dan B = (3 4 4 5).

a. [[26, 34], [21, 27]], b. [[-9/4, 17/6], [7/4, -13/6]], c. [[-23/12, 3/2], [13/4, -5/2]], d. [[-9/4, 17/6], [7/4, -13/6]]

Pembahasan

Diketahui matriks A = [[6, 2], [3, 3]] dan B = [[3, 4], [4, 5]].

a. AB = [[6*3 + 2*4, 6*4 + 2*5], [3*3 + 3*4, 3*4 + 3*5]] = [[18 + 8, 24 + 10], [9 + 12, 12 + 15]] = [[26, 34], [21, 27]]

b. Untuk mencari $(AB)^{-1}$, kita perlu determinan dari AB. det(AB) = $(26 \times 27) - (34 \times 21) = 702 - 714 = -12$.
Maka, $(AB)^{-1} = \frac{1}{-12} \begin{pmatrix} 27 & -34 \ -21 & 26 \end{pmatrix} = \begin{pmatrix} -27/12 & 34/12 \ 21/12 & -26/12 \end{pmatrix} = \begin{pmatrix} -9/4 & 17/6 \ 7/4 & -13/6 \end{pmatrix}$.

c. $A^{-1}$: det(A) = $(6 \times 3) - (2 \times 3) = 18 - 6 = 12$. $A^{-1} = \frac{1}{12} \begin{pmatrix} 3 & -2 \ -3 & 6 \end{pmatrix}$.
$B^{-1}$: det(B) = $(3 \times 5) - (4 \times 4) = 15 - 16 = -1$. $B^{-1} = \frac{1}{-1} \begin{pmatrix} 5 & -4 \ -4 & 3 \end{pmatrix} = \begin{pmatrix} -5 & 4 \ 4 & -3 \end{pmatrix}$.
$A^{-1}B^{-1} = \frac{1}{12} \begin{pmatrix} 3 & -2 \ -3 & 6 \end{pmatrix} \begin{pmatrix} -5 & 4 \ 4 & -3 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} (3*-5)+(-2*4) & (3*4)+(-2*-3) \ (-3*-5)+(6*4) & (-3*4)+(6*-3) \end{pmatrix} = \frac{1}{12} \begin{pmatrix} -15-8 & 12+6 \ 15+24 & -12-18 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} -23 & 18 \ 39 & -30 \end{pmatrix} = \begin{pmatrix} -23/12 & 18/12 \ 39/12 & -30/12 \end{pmatrix} = \begin{pmatrix} -23/12 & 3/2 \ 13/4 & -5/2 \end{pmatrix}$.

d. $B^{-1}A^{-1} = \begin{pmatrix} -5 & 4 \ 4 & -3 \end{pmatrix} \frac{1}{12} \begin{pmatrix} 3 & -2 \ -3 & 6 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} (-5*3)+(4*-3) & (-5*-2)+(4*6) \ (4*3)+(-3*-3) & (4*-2)+(-3*6) \end{pmatrix} = \frac{1}{12} \begin{pmatrix} -15-12 & 10+24 \ 12+9 & -8-18 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} -27 & 34 \ 21 & -26 \end{pmatrix} = \begin{pmatrix} -27/12 & 34/12 \ 21/12 & -26/12 \end{pmatrix} = \begin{pmatrix} -9/4 & 17/6 \ 7/4 & -13/6 \end{pmatrix}$.

Perhatikan bahwa $(AB)^{-1} = B^{-1}A^{-1}$.