Komputasi Aljabar Linear

Perintah:

Kerjaka soal berikut dengan dan hasilnya di tulis di webstatis dan dikumpulkan di webstatis github
Dijelaskan dengan proses-proses untuk mendapatkanya sesuai dengan rumus

Soal

A. Hitunglah determinan matrik berikut dengan menggunakan rumus expansi baris

\sum^{n}_{k=1}(-1)^{i+k}a_{ik}M_{ik}

(1)

dengan $M_{ij}$ adalah minor dari matrik A dan

M_{ij}=\det{A}_{ij}

(2)

$A_{ij}$ adalah submatrik dengan menghapus baris i dan kolom kolom j dari matrix $A_{m \times n}$ dengan $1 \le i, j \le n$

1.

A = \begin{bmatrix} -7 & -5 \\ 1 & 4 \end{bmatrix}

(3)

2.

A = \begin{bmatrix} 0 & 2 & -3 \\ 1 & -2 & -1 \\ 0 & 0 & 1 \end{bmatrix}

(4)

3.

A = \begin{bmatrix} 1 & -3 & 1 & 1 \\ -3 & 1 & 1 & 1 \\ 1 & 1 & -3 & 1 \\ 1 & 1 & 1 & -3 \end{bmatrix}

(5)

Jawab

$\begin{aligned} \textbf{1. } \det{A} &= (-1)^{1+1}a \text{ det }A_{11} + (-1)^{1+2}b\text{ det }A_{12} \\ &= a M_{11} - b M_{12}\\ &= (-7)[4]- (-5)[1] \\ &= (-7)4 - (-5)1\\ &= -28 - (-5)\\ &= -28 + 5\\ &= -23 \end{aligned}$

$\begin{aligned} \textbf{2. } \det{A} &= (-1)^{1+1}a \text{ det }A_{11} + (-1)^{1+2}b\text{ det }A_{12} + (-1)^{1+3}c\text{ det }A_{13}\\ &= a M_{11} - b M_{12} + c M_{13}\\ &= 0 \begin{bmatrix} -2 & -1 \\ 0 & 1 \end{bmatrix} \text{- } 2 \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \text{+ } (-3) \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix}\\ &= 0((-2)1 - (-1)0) - 2(1.1 - (-1)0) + (-3)(1.0 - (-2)0)\\ &= 0((-2)) - 2(1) + (-3)(0)\\ &= 0 - 2 + 0\\ &= 2 \end{aligned}$

$\begin{aligned} \textbf{3. } \det{A} &= (-1)^{1+1} a \det{A}_{11} + (-1)^{1+2} b \det{A}_{12} + (-1)^{1+3} c \det{A}_{13} + (-1)^{1+4} d \det{A}_{14} \\ &= a M_{11} - b M_{12} + c M_{13} - d M_{14}\\ &= 1 \begin{vmatrix} 1 & 1 & 1\\ 1 & -3 & 1\\ 1 & 1 & -3 \end{vmatrix} - (-3) \begin{vmatrix} -3 & 1 & 1\\ 1 & -3 & 1\\ 1 & 1 & -3 \end{vmatrix} + 1 \begin{vmatrix} -3 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & -3 \end{vmatrix} - 1 \begin{vmatrix} -3 & 1 & 1\\ 1 & 1 & -3\\ 1 & 1 & 1 \end{vmatrix}\\ &= 1(1((-3)(-3)-1.1)-1(1(-3)-1.1)+1(1.1-(-3)1)) - (-3)((-3)((-3)(-3)-1.1)-1(1(-3)-1.1)+1(1.1-(-3)1)) + 1((-3)(1(-3)-1.1)-1(1(-3)-1.1)+1(1.1-1.1)) - 1((-3)(1.1-(-3)1)-1(1.1-(-3)1)+1(1.1-1.1))\\ &= 1(1.(-3).(-3)-1.1.1-1.1.(-3)+1.1.1+1.1.1-1.(-3).1) - (-3)((-3)(-3)(-3)-(-3)1.1-1.1.(-3)+1.1.1+1.1.1-1.(-3).1) + 1((-3).1.(-3)-(-3).1.1-1.1.(-3)+1.1.1+1.1.1-1.1.1) - 1((-3).1.1-(-3).(-3).1-1.1.1+1.(-3).1+1.1.1-1.1.1)\\ &= 1(9-1-(-3)+1+1-(-3))-(-3)((-27)-(-3)-(-3)+1+1-(-3))+1(9-(-3)-(-3)+1+1-1)-1((-3)-9-1+(-3)+1-1)\\ &= 1(9-1+3+1+1+3)+3((-27)+3+3+1+1+3)+1(9+3+3+1+1-1)-1((-3)-9-1-3+1-1)\\ &= 1(16)+3(-16)+1(16)-1(-16)\\ &= 16-48+16-(-16)\\ &= 16-48+16+16\\ &= 0 \end{aligned}$

B. Gunakan rumus matriks adjoin untuk menghitung invers dari matriks berikut dengan rumus

(\text{adj} A)_{ij} = (-1)^{i+j} M_{ji}

(6)

dan

A^{-1} = \frac{1}{\det A} \text{adj} A

(7)

1.

A = \begin{bmatrix} -7 & -5 \\ 1 & 4 \end{bmatrix}

(8)

2.

A = \begin{bmatrix} 0 & 2 & -3 \\ 1 & -2 & -1 \\ 0 & 0 & 1 \end{bmatrix}

(9)

3.

A = \begin{bmatrix} 1 & -3 & 1 & 1 \\ -3 & 1 & 1 & 1 \\ 1 & 1 & -3 & 1 \\ 1 & 1 & 1 & -3 \end{bmatrix}

(10)

Jawab

$\begin{aligned} \textbf{1. } \text{adj} (A)_{11} &= (-1)^{1+1} M_{11}\\ &= (-1)^2 [4]\\ &= 1.4\\ &= 4\\ \text{adj} (A)_{12} &= (-1)^{1+2} M_{21}\\ &= (-1)^3 [-5]\\ &= (-1).(-5)\\ &= 5\\ \text{adj} (A)_{21} &= (-1)^{2+1} M_{12}\\ &= (-1)^3 [1]\\ &= (-1).1\\ &= -1\\ \text{adj} (A)_{22} &= (-1)^{2+2} M_{22}\\ &= (-1)^4 [-7]\\ &= 1.(-7)\\ &= -7 \end{aligned}$

\begin{aligned} A^{-1} &= \frac{1}{\det A} \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix}\\ &= \frac{1}{-23} \begin{bmatrix} 4 & 5 \\ -1 & -7 \end{bmatrix}\\ &= \begin{bmatrix} \dfrac{4}{-23} & \dfrac{5}{-23} \\ \dfrac{-1}{-23} & \dfrac{-7}{-23} \end{bmatrix}\\ \end{aligned}

(11)

$\begin{aligned} \textbf{1. } \text{adj} (A)_{11} &= (-1)^{1+1} M_{11}\\ &= (-1)^2 \begin{bmatrix} -2 & -1 \\ 0 & 1 \end{bmatrix}\\ &= 1((-2)1-(-1)0)\\ &= 1(-2)\\ &= -2\\ \text{adj} (A)_{12} &= (-1)^{1+2} M_{21}\\ &= (-1)^3 \begin{bmatrix} 2 & -3 \\ 0 & 1 \end{bmatrix}\\ &= (-1)(2.1-(-3)0)\\ &= (-1)(2)\\ &= -2\\ \text{adj} (A)_{13} &= (-1)^{1+3} M_{31}\\ &= (-1)^4 \begin{bmatrix} -2 & -1 \\ 0 & 1 \end{bmatrix}\\ &= 1(2.1-(-3)(-2))\\ &= 1(2-6)\\ &= 1(-4)\\ &= -4\\ \text{adj} (A)_{21} &= (-1)^{2+1} M_{12}\\ &= (-1)^3 \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\\ &= (-1)(1.1-(-1)0)\\ &= (-1)(1)\\ &= -1\\ \text{adj} (A)_{22} &= (-1)^{2+2} M_{22}\\ &= (-1)^4 \begin{bmatrix} 0 & -3 \\ 0 & 1 \end{bmatrix}\\ &= 1(0.1-(-3)0)\\ &= 1(0)\\ &= 0\\ \text{adj} (A)_{23} &= (-1)^{2+3} M_{32}\\ &= (-1)^5 \begin{bmatrix} 0 & -3 \\ 1 & -1 \end{bmatrix}\\ &= (-1)(0(-1)-(-3)1)\\ &= 1(3)\\ &= 3\\ \text{adj} (A)_{31} &= (-1)^{3+1} M_{13}\\ &= (-1)^4 \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\\ &= 1(1.0-(-2)0)\\ &= 1(0)\\ &= 0\\ \text{adj} (A)_{32} &= (-1)^{3+2} M_{23}\\ &= (-1)^5 \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}\\ &= (-1)(0.0-2.0)\\ &= (-1)(0)\\ &= 0\\ \text{adj} (A)_{33} &= (-1)^{3+3} M_{33}\\ &= (-1)^6 \begin{bmatrix} 0 & 2 \\ 1 & -2 \end{bmatrix}\\ &= 1(0.(-2)-2.1)\\ &= 1(-2)\\ &= -2 \end{aligned}$

\begin{aligned} A^{-1} &= \frac{1}{\det A} \begin{bmatrix} A_{11} & A_{21} & A_{31}\\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{bmatrix}\\ &= \frac{1}{-2} \begin{bmatrix} -2 & -1 & 0 \\ -2 & 0 & 0\\ -4 & 3 &-2 \end{bmatrix}\\ &= \begin{bmatrix} \dfrac{-2}{-2} & \dfrac{-1}{-2} & \dfrac{0}{-2}\\ \dfrac{-2}{-2} & \dfrac{0}{-2} & \dfrac{0}{-2}\\ \dfrac{-4}{-2} & \dfrac{3}{-2} & \dfrac{-2}{-2}\end{bmatrix}\\ &= \begin{bmatrix} 1 & \frac{1}{2} & 0\\ 1 & 0 & 0\\ 2 & -\frac{3}{2} & 1 \end{bmatrix}\\ \end{aligned}

(12)

3. $\det{A} = 0$ , maka matriks persegi tidak memiliki invers (singuler)