Soal-Soal Matematika/Matriks

Bentuk dan sifat matriks

bentuk

• ordo 2x2: ${\begin{bmatrix}a&b\\c&d\\\end{bmatrix}}$

• ordo 3x3: ${\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\\end{bmatrix}}$

sifat

1. komutatif

• A + B = B + A

1. asosiatif

• (A + B) + C = A + (B + C)
• (A . B) . C = A. (B x C)

1. distributif

• A . (B + C) = A . B + A . C
• A . (B - C) = A . B - A . C

1. (k . A) . B = k. (A . B)
2. A . B ≠ B . A
3. A . I = A
4. (A^T)^T = A
5. A . A^-1 = A^-1 . A = I
6. (A . B)^T = B^T . A^T
7. (A . B)^-1 = B^-1 . A^-1
8. (A . B . C)^-1 = C^-1 . B^-1 . A^-1
9. det (A . B) = det A . det B
10. A = B . C <=> det A = det B . det C
11. det (A^T) = det A
12. det (A^-1) . det A = 1 (invers bukan pangkat)
13. det (A^-1) . det A^T = 1 (karena det (A^T) = det A)
14. det (Aⁿ) = (det A)ⁿ
15. det (k . A) = k² . det A

vektor baris

${\begin{bmatrix}3&7&2\end{bmatrix}}$

vektor kolom

${\begin{bmatrix}4\\1\\8\\\end{bmatrix}}$

matriks persegi

${\begin{bmatrix}9&13&5\\1&11&7\\2&6&3\\\end{bmatrix}}$

${\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{bmatrix}}$

• baris:

1. pertama: a11, a12 dan a13
2. kedua: a21, a22 dan a23
3. ketiga: a31, a32 dan a33

• kolom:

1. pertama: a11, a21 dan a31
2. kedua: a12, a22 dan a32
3. ketiga: a13, a23 dan a33

• diagonal

1. sisi kiri ke kanan: a11, a22 dan a33
2. sisi kanan ke kiri: a13, a22 dan a31

• Matriks diagonal

${\begin{bmatrix}a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\\\end{bmatrix}}$

• Matriks segitiga bawah

${\begin{bmatrix}a_{11}&0&0\\a_{21}&a_{22}&0\\a_{31}&a_{32}&a_{33}\\\end{bmatrix}}$

• Matriks segitiga atas

${\begin{bmatrix}a_{11}&a_{12}&a_{13}\\0&a_{22}&a_{23}\\0&0&a_{33}\\\end{bmatrix}}$

matriks perkalian

${\begin{bmatrix}a&b\\c&d\\\end{bmatrix}}\cdot {\begin{bmatrix}e&f\\g&h\\\end{bmatrix}}$ = ${\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\\\end{bmatrix}}$

ordo 2x2

bentuk

${\begin{bmatrix}a&b\\c&d\\\end{bmatrix}}$

Matriks transpos (A^T)

${\begin{bmatrix}a&c\\b&d\end{bmatrix}}$

• matriks simetris

A = A^T

A = ${\begin{bmatrix}a&d\\c&b\\\end{bmatrix}}$

A^T = ${\begin{bmatrix}a&d\\c&b\\\end{bmatrix}}$

• matriks simetris miring

A = -A^T

A = ${\begin{bmatrix}a&d\\c&b\\\end{bmatrix}}$

A^T = ${\begin{bmatrix}-a&-d\\-c&-b\\\end{bmatrix}}$

Determinan (Det)

ad - bc

Matriks singular adalah matriks yang hasil determinan bernilai nol sedangkan matriks nonsingular adalah matriks yang hasil determinan bernilai bukan nol.

Matriks nonsingular memiliki ciri yang khas yaitu kolom pertama dan kolom kedua merupakan kelipatan yang sama. contoh: ${\begin{bmatrix}2&-4\\4&-8\\\end{bmatrix}}$, ${\begin{bmatrix}5&15\\11&33\\\end{bmatrix}}$, dst

Adjoint (Adj)

${\begin{bmatrix}d&-b\\-c&a\\\end{bmatrix}}$

Matriks inverse (A^-1)

${\frac {1}{\text{det A}}}$${\begin{bmatrix}d&-b\\-c&a\\\end{bmatrix}}$

ordo 3x3

bentuk

${\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\\end{bmatrix}}$

Matriks transpos (A^T)

${\begin{bmatrix}a&d&g\\b&e&h\\c&f&i\\\end{bmatrix}}$

• matriks simetris

A = A^T

A = ${\begin{bmatrix}a&d&e\\d&b&f\\e&f&c\\\end{bmatrix}}$

A^T = ${\begin{bmatrix}a&d&e\\d&b&f\\e&f&c\\\end{bmatrix}}$

• matriks simetris miring

A = -A^T

A = ${\begin{bmatrix}a&d&e\\d&b&f\\e&f&c\\\end{bmatrix}}$

A^T = ${\begin{bmatrix}-a&-d&-e\\-d&-b&-f\\-e&-f&-c\\\end{bmatrix}}$

Determinan (Det)

• dengan aturan sarrus

${\begin{aligned}{\begin{array}{rrr|rr}a&b&c&a&b\\d&e&f&d&e\\g&h&i&g&h\\-&-&-+&+&+\\\end{array}}\\\end{aligned}}$

det A = aei + bfg + cdh - bdi - afh - ceg

• dengan minor-kofaktor

untuk minor M_ij = det A_ij

untuk kofaktor C_ij = (-1)^i+j . M_ij

det A = $\sum _{j=1}^{n}i=a_{ij}\cdot C_{ij}$ dimana sembarang baris i atau kolom j (i atau j = 1, 2, 3, ..., n)

Matriks nonsingular memiliki ciri yang khas yaitu kolom pertama, kolom kedua dan/atau kolom tiga merupakan kelipatan yang sama. contoh: ${\begin{bmatrix}3&-7&6\\4&1&8\\-5&2&-10\\\end{bmatrix}}$, ${\begin{bmatrix}-6&-1&-4\\12&2&8\\18&3&12\\\end{bmatrix}}$, dst

Adjoint (Adj)

kof (A) = (-1)^i+j . M_ij

kof (A) = ${\begin{bmatrix}M_{11}&M_{12}&M_{13}\\M_{21}&M_{22}&M_{23}\\M_{31}&M_{32}&M_{33}\\\end{bmatrix}}$

adj A = (kof (A))^T

Matriks inverse (A^-1)

• dengan adjoint

${\frac {adjA}{\text{det A}}}$

• dengan elementer

A | I diubah menjadi I | A^-1

matriks persamaan linear (aturan cramer)

ordo 2x2 (dua variabel)

persamaan linear yaitu a₁x + b₁y = c₁ serta a₂x + b₂y = c₂.

maka sebagai berikut:

D = ${\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\\\end{bmatrix}}=a_{1}\cdot b_{2}-a_{2}\cdot b_{1}$

D_x = ${\begin{bmatrix}c_{1}&b_{1}\\c_{2}&b_{2}\\\end{bmatrix}}=c_{1}\cdot b_{2}-c_{2}\cdot b_{1}$

D_y = ${\begin{bmatrix}a_{1}&c_{1}\\a_{2}&c_{2}\\\end{bmatrix}}=a_{1}\cdot c_{2}-a_{2}\cdot c_{1}$

x = ${\frac {D_{x}}{D}}$ serta y = ${\frac {D_{y}}{D}}$

ordo 3x3 (tiga variabel)

persamaan linear yaitu a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂ serta a₃x + b₃y + c₃z = d₃.

maka sebagai berikut:

D = ${\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\\\end{bmatrix}}=a_{1}\cdot b_{2}\cdot c_{3}+b_{1}\cdot c_{2}\cdot a_{3}+c_{1}\cdot a_{2}\cdot b_{3}-a_{3}\cdot b_{2}\cdot c_{1}-b_{3}\cdot c_{2}\cdot a_{1}-c_{3}\cdot a_{2}\cdot b_{1}$

D_x = ${\begin{bmatrix}d_{1}&b_{1}&c_{1}\\d_{2}&b_{2}&c_{2}\\d_{3}&b_{3}&c_{3}\\\end{bmatrix}}=d_{1}\cdot b_{2}\cdot c_{3}+b_{1}\cdot c_{2}\cdot d_{3}+c_{1}\cdot d_{2}\cdot b_{3}-d_{3}\cdot b_{2}\cdot c_{1}-b_{3}\cdot c_{2}\cdot d_{1}-c_{3}\cdot d_{2}\cdot b_{1}$

D_y = ${\begin{bmatrix}a_{1}&d_{1}&c_{1}\\a_{2}&d_{2}&c_{2}\\a_{3}&d_{3}&c_{3}\\\end{bmatrix}}=a_{1}\cdot d_{2}\cdot c_{3}+d_{1}\cdot c_{2}\cdot a_{3}+c_{1}\cdot a_{2}\cdot d_{3}-a_{3}\cdot d_{2}\cdot c_{1}-d_{3}\cdot c_{2}\cdot a_{1}-c_{3}\cdot a_{2}\cdot d_{1}$

D_z = ${\begin{bmatrix}a_{1}&b_{1}&d_{1}\\a_{2}&b_{2}&d_{2}\\a_{3}&b_{3}&d_{3}\\\end{bmatrix}}=a_{1}\cdot b_{2}\cdot d_{3}+b_{1}\cdot d_{2}\cdot a_{3}+d_{1}\cdot a_{2}\cdot b_{3}-a_{3}\cdot b_{2}\cdot d_{1}-b_{3}\cdot d_{2}\cdot a_{1}-d_{3}\cdot a_{2}\cdot b_{1}$

x = ${\frac {D_{x}}{D}}$, y = ${\frac {D_{y}}{D}}$ serta z = ${\frac {D_{z}}{D}}$

contoh

1. tentukan hasil determinan serta matriks invers dari ${\begin{bmatrix}2&1\\6&4\\\end{bmatrix}}$!

Jawaban

$${\begin{aligned}*detA&=2\cdot 4-1\cdot 6=8-6=2\\*A^{-1}&={\frac {1}{detA}}\cdot {\begin{bmatrix}4&-1\\-6&2\\\end{bmatrix}}\\&={\frac {1}{2}}\cdot {\begin{bmatrix}4&-1\\-6&2\\\end{bmatrix}}\\&={\begin{bmatrix}2&-{\frac {1}{2}}\\-3&1\\\end{bmatrix}}\\\end{aligned}}$$

1. tentukan hasil determinan serta matriks invers dari ${\begin{bmatrix}1&2&3\\2&8&7\\1&5&6\\\end{bmatrix}}$!

Jawaban

$${\begin{aligned}*{\text{cara 1}}\\{\begin{array}{rrr|rr}1&2&3&1&2\\2&8&7&2&8\\1&5&6&1&5\\-&-&-+&+&+\\\end{array}}\\detA&=1\cdot 8\cdot 6+2\cdot 7\cdot 1+3\cdot 2\cdot 5-2\cdot 2\cdot 6-1\cdot 7\cdot 5-3\cdot 8\cdot 1=48+14+30-24-35-24=9\\{\text{cara 2}}\\detA&=a_{11}\cdot c_{11}+a_{12}\cdot c_{12}+a_{13}\cdot c_{13}\\&=1\cdot (-1)^{1+1}\cdot {\begin{bmatrix}8&7\\5&6\\\end{bmatrix}}+2\cdot (-1)^{1+2}\cdot {\begin{bmatrix}2&7\\1&6\\\end{bmatrix}}+3\cdot (-1)^{1+3}\cdot {\begin{bmatrix}2&8\\1&5\\\end{bmatrix}}\\&=1\cdot 1\cdot (48-35)+2\cdot (-1)\cdot (12-7)+3\cdot 1\cdot (10-8)\\&=13-10+6=9\\*{\text{cara 1}}\\kof(A)&={\begin{bmatrix}{\begin{bmatrix}8&7\\5&6\\\end{bmatrix}}&-{\begin{bmatrix}2&7\\1&6\\\end{bmatrix}}&{\begin{bmatrix}2&8\\1&5\\\end{bmatrix}}\\-{\begin{bmatrix}2&3\\5&6\\\end{bmatrix}}&{\begin{bmatrix}1&3\\1&6\\\end{bmatrix}}&-{\begin{bmatrix}1&2\\1&5\\\end{bmatrix}}\\{\begin{bmatrix}2&3\\8&7\\\end{bmatrix}}&-{\begin{bmatrix}1&3\\2&7\\\end{bmatrix}}&{\begin{bmatrix}1&2\\2&8\\\end{bmatrix}}\\\end{bmatrix}}\\&={\begin{bmatrix}13&-5&2\\3&3&-3\\-10&-1&4\\\end{bmatrix}}\\adj(A)&=(kof(A))^{T}\\&={\begin{bmatrix}13&3&-10\\-5&3&-1\\2&-3&4\\\end{bmatrix}}\\A^{-1}&={\frac {adjA}{detA}}\\&={\frac {1}{9}}\cdot {\begin{bmatrix}13&3&-10\\-5&3&-1\\2&-3&4\\\end{bmatrix}}\\&={\begin{bmatrix}{\frac {13}{9}}&{\frac {1}{3}}&-{\frac {10}{9}}\\-{\frac {5}{9}}&{\frac {1}{3}}&-{\frac {1}{9}}\\{\frac {2}{9}}&-{\frac {1}{3}}&{\frac {4}{9}}\\\end{bmatrix}}\\{\text{cara 2}}\\{\begin{array}{rrr|rrr}1&2&3&1&0&0\\2&8&7&0&1&0\\1&5&6&0&0&1\\\end{array}}\\{\text{b2-2b1}}\\{\begin{array}{rrr|rrr}1&2&3&1&0&0\\0&4&1&-2&1&0\\1&5&6&0&0&1\\\end{array}}\\{\text{b3-b1}}\\{\begin{array}{rrr|rrr}1&2&3&1&0&0\\0&4&1&-2&1&0\\0&3&3&-1&0&1\\\end{array}}\\{\text{1/4b2}}\\{\begin{array}{rrr|rrr}1&2&3&1&0&0\\0&1&{\frac {1}{4}}&-{\frac {1}{2}}&{\frac {1}{4}}&0\\0&3&3&-1&0&1\\\end{array}}\\{\text{b1-2b2}}\\{\begin{array}{rrr|rrr}1&0&{\frac {5}{2}}&2&-{\frac {1}{2}}&0\\0&1&{\frac {1}{4}}&-{\frac {1}{2}}&{\frac {1}{4}}&0\\0&3&3&-1&0&1\\\end{array}}\\{\text{b3-3b2}}\\{\begin{array}{rrr|rrr}1&0&{\frac {5}{2}}&2&-{\frac {1}{2}}&0\\0&1&{\frac {1}{4}}&-{\frac {1}{2}}&{\frac {1}{4}}&0\\0&0&{\frac {9}{4}}&{\frac {1}{2}}&-{\frac {3}{4}}&1\\\end{array}}\\{\text{4/9b3}}\\{\begin{array}{rrr|rrr}1&0&{\frac {5}{2}}&2&-{\frac {1}{2}}&0\\0&1&{\frac {1}{4}}&-{\frac {1}{2}}&{\frac {1}{4}}&0\\0&0&1&{\frac {2}{9}}&-{\frac {1}{3}}&{\frac {4}{9}}\\\end{array}}\\{\text{b1-5/2b3}}\\{\begin{array}{rrr|rrr}1&0&0&{\frac {13}{9}}&{\frac {1}{3}}&-{\frac {10}{9}}\\0&1&{\frac {1}{4}}&-{\frac {1}{2}}&{\frac {1}{4}}&0\\0&0&1&{\frac {2}{9}}&-{\frac {1}{3}}&{\frac {4}{9}}\\\end{array}}\\{\text{b2-1/4b3}}\\{\begin{array}{rrr|rrr}1&0&0&{\frac {13}{9}}&{\frac {1}{3}}&-{\frac {10}{9}}\\0&1&0&-{\frac {5}{9}}&{\frac {1}{3}}&-{\frac {1}{9}}\\0&0&1&{\frac {2}{9}}&-{\frac {1}{3}}&{\frac {4}{9}}\\\end{array}}\\{\text{jadi }}A^{-1}&={\begin{bmatrix}{\frac {13}{9}}&{\frac {1}{3}}&-{\frac {10}{9}}\\-{\frac {5}{9}}&{\frac {1}{3}}&-{\frac {1}{9}}\\{\frac {2}{9}}&-{\frac {1}{3}}&{\frac {4}{9}}\\\end{bmatrix}}\\\end{aligned}}$$

1. tentukan nilai x dan y dari 2x + 3y = 16 serta 3x + y = 10!

cara 1

Jawaban

$${\begin{aligned}{\begin{bmatrix}2&3\\3&1\\\end{bmatrix}}\cdot {\begin{bmatrix}x\\y\\\end{bmatrix}}&={\begin{bmatrix}16\\10\\\end{bmatrix}}\\{\begin{bmatrix}x\\y\\\end{bmatrix}}&={\begin{bmatrix}2&3\\3&1\\\end{bmatrix}}^{-1}\cdot {\begin{bmatrix}16\\10\\\end{bmatrix}}\\{\begin{bmatrix}x\\y\\\end{bmatrix}}&={\frac {1}{2(1)-3(3)}}{\begin{bmatrix}1&-3\\-3&2\\\end{bmatrix}}\cdot {\begin{bmatrix}16\\10\\\end{bmatrix}}\\{\begin{bmatrix}x\\y\\\end{bmatrix}}&=-{\frac {1}{7}}{\begin{bmatrix}1&-3\\-3&2\\\end{bmatrix}}\cdot {\begin{bmatrix}16\\10\\\end{bmatrix}}\\{\begin{bmatrix}x\\y\\\end{bmatrix}}&=-{\frac {1}{7}}{\begin{bmatrix}-14\\-28\\\end{bmatrix}}\\{\begin{bmatrix}x\\y\\\end{bmatrix}}&={\begin{bmatrix}2\\4\\\end{bmatrix}}\\\end{aligned}}$$

cara 2

Jawaban

$${\begin{aligned}D&={\begin{bmatrix}2&3\\3&1\\\end{bmatrix}}=2(1)-3(3)=-7\\D_{x}&={\begin{bmatrix}16&3\\10&1\\\end{bmatrix}}=16(1)-10(3)=-14\\D_{y}&={\begin{bmatrix}2&16\\3&10\\\end{bmatrix}}=2(10)-3(16)=-28\\x&={\frac {-14}{-7}}=2\\y&={\frac {-28}{-7}}=4\\\end{aligned}}$$

1. tentukan nilai x, y dan z dari 2x + 3y + 5z = 17, 3x + y + 4z = 15 serta x + 7y + z = 18!

cara 1

Jawaban

$${\begin{aligned}{\begin{array}{rrr|r}2&3&5&17\\3&1&4&15\\1&7&1&18\\\end{array}}\\{\text{b1/2}}\\{\begin{array}{rrr|r}1&{\frac {3}{2}}&{\frac {5}{2}}&{\frac {17}{2}}\\3&1&4&15\\1&7&1&18\\\end{array}}\\{\text{b2-3b1}}\\{\begin{array}{rrr|r}1&{\frac {3}{2}}&{\frac {5}{2}}&{\frac {17}{2}}\\0&-{\frac {7}{2}}&-{\frac {7}{2}}&-{\frac {21}{2}}\\1&7&1&18\\\end{array}}\\{\text{b3-b1}}\\{\begin{array}{rrr|r}1&{\frac {3}{2}}&{\frac {5}{2}}&{\frac {17}{2}}\\0&-{\frac {7}{2}}&-{\frac {7}{2}}&-{\frac {21}{2}}\\0&{\frac {11}{2}}&-{\frac {3}{2}}&{\frac {19}{2}}\\\end{array}}\\{\text{-2b2/7}}\\{\begin{array}{rrr|r}1&{\frac {3}{2}}&{\frac {5}{2}}&{\frac {17}{2}}\\0&1&1&3\\0&{\frac {11}{2}}&-{\frac {3}{2}}&{\frac {19}{2}}\\\end{array}}\\{\text{b1-3b2/2}}\\{\begin{array}{rrr|r}1&0&1&4\\0&1&1&3\\0&{\frac {11}{2}}&-{\frac {3}{2}}&{\frac {19}{2}}\\\end{array}}\\{\text{b3-11b2/2}}\\{\begin{array}{rrr|r}1&0&1&4\\0&1&1&3\\0&0&-7&-7\\\end{array}}\\{\text{-b3/7}}\\{\begin{array}{rrr|r}1&0&1&4\\0&1&1&3\\0&0&1&1\\\end{array}}\\{\text{b1-b3}}\\{\begin{array}{rrr|r}1&0&0&3\\0&1&1&3\\0&0&1&1\\\end{array}}\\{\text{b2-b3}}\\{\begin{array}{rrr|r}1&0&0&3\\0&1&0&2\\0&0&1&1\\\end{array}}\\\end{aligned}}$$

cara 2

Jawaban

$${\begin{aligned}D&={\begin{bmatrix}2&3&5\\3&1&4\\1&7&1\\\end{bmatrix}}=2(1)(1)+3(4)(1)+5(3)(7)-1(1)(5)-7(4)(2)-1(3)(3)=49\\D_{x}&={\begin{bmatrix}17&3&5\\15&1&4\\18&7&1\\\end{bmatrix}}=17(1)(1)+3(4)(18)+5(15)(7)-18(1)(5)-7(4)(17)-1(15)(3)=147\\D_{y}&={\begin{bmatrix}2&17&5\\3&15&4\\1&18&1\\\end{bmatrix}}=2(15)(1)+17(4)(1)+5(3)(18)-1(15)(5)-18(4)(2)-1(3)(17)=98\\D_{z}&={\begin{bmatrix}2&3&17\\3&1&15\\1&7&18\\\end{bmatrix}}=2(1)(18)+3(15)(1)+17(3)(7)-1(1)(17)-7(15)(2)-18(3)(3)=49\\x&={\frac {147}{49}}=3\\y&={\frac {98}{49}}=2\\z&={\frac {49}{49}}=1\\\end{aligned}}$$