tolong jawab soal ini dari nomor 1-6

Baik, mari kita selesaikan soal-soal matriks tersebut. 1. Tentukan matriks M yang berordo 5 \times 2 yang setiap elemennya mengikuti aturan m_{ij} = -3i + 2j^2. Solusi: Matriks M berordo 5 \times 2 berarti matriks tersebut memiliki 5 baris dan 2 kolom. Elemen-elemen matriks M dapat dihitung menggunakan rumus m_{ij} = -3i + 2j^2. $M = \begin{bmatrix} m_{11} & m_{12} \ m_{21} & m_{22} \ m_{31} & m_{32} \ m_{41} & m_{42} \ m_{51} & m_{52} \end{bmatrix}$ m_{11} = -3(1) + 2(1)^2 = -3 + 2 = -1 m_{12} = -3(1) + 2(2)^2 = -3 + 8 = 5 m_{21} = -3(2) + 2(1)^2 = -6 + 2 = -4 m_{22} = -3(2) + 2(2)^2 = -6 + 8 = 2 m_{31} = -3(3) + 2(1)^2 = -9 + 2 = -7 m_{32} = -3(3) + 2(2)^2 = -9 + 8 = -1 m_{41} = -3(4) + 2(1)^2 = -12 + 2 = -10 m_{42} = -3(4) + 2(2)^2 = -12 + 8 = -4 m_{51} = -3(5) + 2(1)^2 = -15 + 2 = -13 m_{52} = -3(5) + 2(2)^2 = -15 + 8 = -7 Jadi, $M = \begin{bmatrix} -1 & 5 \ -4 & 2 \ -7 & -1 \ -10 & -4 \ -13 & -7 \end{bmatrix}$   2. Jika matriks P = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} dan I adalah matriks identitas. Tentukan nilai dari -P^2 + 4I. Solusi: P^2 = P \times P = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1-2 & -1+1 \\ 2-2 & -2+1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} -P^2 = -\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} 4I = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} -P^2 + 4I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}   3. Jika matriks A = \begin{bmatrix} -1 & -1 & 0 \\ -1 & 1 & 2 \end{bmatrix} dan matriks B = \begin{bmatrix} -1 & x \\ 1 & y \\ 0 & 0 \end{bmatrix}, dan AB = \begin{bmatrix} 0 & 2 \\ 2 & 4 \end{bmatrix}. Tentukan nilai y - x. Solusi: AB = \begin{bmatrix} -1 & -1 & 0 \\ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & x \\ 1 & y \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (-1)(1) + (0)(0) & (-1)(x) + (-1)(y) + (0)(0) \\ (-1)(-1) + (1)(1) + (2)(0) & (-1)(x) + (1)(y) + (2)(0) \end{bmatrix} = \begin{bmatrix} 1-1+0 & -x-y+0 \\ 1+1+0 & -x+y+0 \end{bmatrix} = \begin{bmatrix} 0 & -x-y \\ 2 & -x+y \end{bmatrix} Diketahui AB = \begin{bmatrix} 0 & 2 \\ 2 & 4 \end{bmatrix}, maka: -x - y = 2 -x + y = 4 Menambahkan kedua persamaan: -2x = 6 x = -3 Substitusi x = -3 ke -x - y = 2: -(-3) - y = 2 3 - y = 2 y = 1 y - x = 1 - (-3) = 1 + 3 = 4   4. Misalkan A^T adalah transpos matriks A. Jika A = \begin{bmatrix} a & 1 \\ 0 & b \end{bmatrix} dan B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} yang memenuhi persamaan A^T B = \begin{bmatrix} 1 & 2 \\ 5 & 10 \end{bmatrix}. Maka nilai a + b = ... Solusi: A^T = \begin{bmatrix} a & 0 \\ 1 & b \end{bmatrix} A^T B = \begin{bmatrix} a & 0 \\ 1 & b \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} a(1) + 0(2) & a(2) + 0(4) \\ 1(1) + b(2) & 1(2) + b(4) \end{bmatrix} = \begin{bmatrix} a & 2a \\ 1+2b & 2+4b \end{bmatrix} Diketahui A^T B = \begin{bmatrix} 1 & 2 \\ 5 & 10 \end{bmatrix}, maka: a = 1 2a = 2 1 + 2b = 5 2 + 4b = 10 Dari 1 + 2b = 5: 2b = 4 b = 2 a + b = 1 + 2 = 3   5. Diketahui matriks P = \begin{bmatrix} 4 & -2 \\ 1 & -7 \end{bmatrix} dan Q = \begin{bmatrix} -1 & 2 \\ 6 & -5 \end{bmatrix}. Jika x merupakan elemen dari matriks X yang memenuhi persamaan 4Q + 4X = 2X + 3P - QP. Tentukan nilai dari \frac{x_{12} + x_{21} - x_{22}}{x_{11}}. Solusi: 4Q + 4X = 2X + 3P - QP 4X - 2X = 3P - QP - 4Q 2X = 3P - QP - 4Q X = \frac{1}{2} (3P - QP - 4Q) 3P = 3 \begin{bmatrix} 4 & -2 \\ 1 & -7 \end{bmatrix} = \begin{bmatrix} 12 & -6 \\ 3 & -21 \end{bmatrix} QP = \begin{bmatrix} -1 & 2 \\ 6 & -5 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ 1 & -7 \end{bmatrix} = \begin{bmatrix} -4+2 & 2-14 \\ 24-5 & -12+35 \end{bmatrix} = \begin{bmatrix} -2 & -12 \\ 19 & 23 \end{bmatrix} 4Q = 4 \begin{bmatrix} -1 & 2 \\ 6 & -5 \end{bmatrix} = \begin{bmatrix} -4 & 8 \\ 24 & -20 \end{bmatrix} 3P - QP - 4Q = \begin{bmatrix} 12 & -6 \\ 3 & -21 \end{bmatrix} - \begin{bmatrix} -2 & -12 \\ 19 & 23 \end{bmatrix} - \begin{bmatrix} -4 & 8 \\ 24 & -20 \end{bmatrix} = \begin{bmatrix} 12+2+4 & -6+12-8 \\ 3-19-24 & -21-23+20 \end{bmatrix} = \begin{bmatrix} 18 & -2 \\ -40 & -24 \end{bmatrix} X = \frac{1}{2} \begin{bmatrix} 18 & -2 \\ -40 & -24 \end{bmatrix} = \begin{bmatrix} 9 & -1 \\ -20 & -12 \end{bmatrix} x_{11} = 9, x_{12} = -1, x_{21} = -20, x_{22} = -12 \frac{x_{12} + x_{21} - x_{22}}{x_{11}} = \frac{-1 + (-20) - (-12)}{9} = \frac{-1 - 20 + 12}{9} = \frac{-9}{9} = -1   6. Jika diketahui A = \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix}, B memiliki invers, dan BA = \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix}. Tentukan det(B^{-1}). Solusi: BA = \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix} \det(BA) = (1)(0) - (-1)(3) = 0 + 3 = 3 \det(BA) = \det(B) \det(A) \det(A) = (2)(1) - (3)(-1) = 2 + 3 = 5 \det(B) \det(A) = 3 \det(B) (5) = 3 \det(B) = \frac{3}{5} \det(B^{-1}) = \frac{1}{\det(B)} = \frac{1}{\frac{3}{5}} = \frac{5}{3}