Soal-Soal Matematika/Matriks

Bentuk dan sifat matriks

bentuk

• ordo 2x2: $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

• ordo 3x3: $\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

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 . A^-1 = I
5. (A . B)^-1 = B^-1 . A^-1
6. (A . B . C)^-1 = C^-1 . B^-1 . A^-1
7. det (A^-1) = det A (invers bukan pangkat)
8. det (A^T) = 1/det A
9. det (Aⁿ) = (det A)ⁿ

vektor baris

$\left[\begin{array}{ccc}3& 7& 2\end{array}\right]$

vektor kolom

$\left[\begin{array}{c}4\\ 1\\ 8\end{array}\right]$

matriks persegi

$\left[\begin{array}{ccc}9& 13& 5\\ 1& 11& 7\\ 2& 6& 3\end{array}\right]$

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

• 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

$\left[\begin{array}{ccc}{a}_{11}& 0& 0\\ 0& a_{22}& 0\\ 0& 0& a_{33}\end{array}\right]$

• Matriks segitiga bawah

$\left[\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{22}& 0\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

• Matriks segitiga atas

$\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ 0& a_{22}& a_{23}\\ 0& 0& a_{33}\end{array}\right]$

matriks perkalian

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\cdot \left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$ = $\left[\begin{array}{cc}ae+bg& af+bh\\ ce+dg& cf+dh\end{array}\right]$

ordo 2x2

bentuk

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Matriks transpos (A^T)

$\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$

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 inverse (A^-1)

$\frac{1}{\text{det A}}$$\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

ordo 3x3

bentuk

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$

Matriks transpos (A^T)

$\left[\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]$

Determinan (Det)

• dengan sarrus

$\begin{array}{c}\begin{array}{ccccc}a& b& c& a& b\\ d& e& f& d& e\\ g& h& i& g& h\\ -& -& -+& +& +\end{array}\\ \end{array}$

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)

Adjoint (Adj)

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

kof (A) = $\left[\begin{array}{ccc}{M}_{11}& M_{12}& M_{13}\\ M_{21}& M_{22}& M_{23}\\ M_{31}& M_{32}& M_{33}\end{array}\right]$

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

contoh

1. tentukan hasil determinan serta matriks invers dari $\left[\begin{array}{cc}2& 1\\ 6& 4\end{array}\right]$!

Jawaban

$$\begin{array}{cc}*detA& =2\cdot 4-1\cdot 6=8-6=2\\ *A^{-1}& =\frac{1}{detA}\cdot \left[\begin{array}{cc}4& -1\\ -6& 2\end{array}\right]\\ & =\frac{1}{2}\cdot \left[\begin{array}{cc}4& -1\\ -6& 2\end{array}\right]\\ & =\left[\begin{array}{cc}2& -\frac{1}{2}\\ -3& 1\end{array}\right]\\ \end{array}$$

1. tentukan hasil determinan serta matriks invers dari $\left[\begin{array}{ccc}1& 2& 3\\ 2& 8& 7\\ 1& 5& 6\end{array}\right]$!

Jawaban

$$\begin{array}{c}*\text{cara 1}\\ \begin{array}{ccccc}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 \left[\begin{array}{cc}8& 7\\ 5& 6\end{array}\right]+2\cdot (-1{)}^{1+2}\cdot \left[\begin{array}{cc}2& 7\\ 1& 6\end{array}\right]+3\cdot (-1{)}^{1+3}\cdot \left[\begin{array}{cc}2& 8\\ 1& 5\end{array}\right]\\ & =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)& =\left[\begin{array}{ccc}\left[\begin{array}{cc}8& 7\\ 5& 6\end{array}\right]& -\left[\begin{array}{cc}2& 7\\ 1& 6\end{array}\right]& \left[\begin{array}{cc}2& 8\\ 1& 5\end{array}\right]\\ -\left[\begin{array}{cc}2& 3\\ 5& 6\end{array}\right]& \left[\begin{array}{cc}1& 3\\ 1& 6\end{array}\right]& -\left[\begin{array}{cc}1& 2\\ 1& 5\end{array}\right]\\ \left[\begin{array}{cc}2& 3\\ 8& 7\end{array}\right]& -\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]& \left[\begin{array}{cc}1& 2\\ 2& 8\end{array}\right]\end{array}\right]\\ & =\left[\begin{array}{ccc}13& -5& 2\\ 3& 3& -3\\ -10& -1& 4\end{array}\right]\\ adj(A)& =(kof(A){)}^T\\ & =\left[\begin{array}{ccc}13& 3& -10\\ -5& 3& -1\\ 2& -3& 4\end{array}\right]\\ A^{-1}& =\frac{adjA}{detA}\\ & =\frac{1}{9}\cdot \left[\begin{array}{ccc}13& 3& -10\\ -5& 3& -1\\ 2& -3& 4\end{array}\right]\\ & =\left[\begin{array}{ccc}\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{array}\right]\\ \text{cara 2}\\ \begin{array}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}{cccccc}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}& =\left[\begin{array}{ccc}\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{array}\right]\\ \end{array}$$