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Berikut ini adalah pertanyaan dari windy2120 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

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Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Penyajian SPL dalam bentuk operasi matriks :

$\left[\begin{array}{ccc}1&-2&1\\3&1&-2\\-2&1&1\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc}-5\\11\\-2\end{array}\right]$

$\purple{\huge{a~)}}~$ Dengan metode Cramer :

$\text{D}=\left|\begin{array}{ccc}1&-2&1\\3&1&-2\\-2&1&1\end{array}\right|$

$\text{D}=$ (1 × 1 × 1) + (–2 × –2 × –2) + (1 × 3 × 1) – (1 × 1 × –2) – (1 × –2 × 1) – (–2 × 3 × 1)

$\text{D}=$ 1 + (–8) + 3 – (–2) – (–2) – (–6) = 6

$\text{D}_{x_1}=\left|\begin{array}{ccc}-5&-2&1\\11&1&-2\\-2&1&1\end{array}\right|$

$\text{D}_{x_1}=$ (–5 × 1 × 1) + (–2 × –2 × –2) + (1 × 11 × 1) – (1 × 1 × –2) – (–5 × –2 × 1) – (–2 × 11 × 1)

$\text{D}_{x_1}=$ –5 + (–8) + 11 – (–2) – 10 – (–22) = 12

$x_1=\frac{\text{D}_{x_1}}{\text{D}}=\frac{12}{6}~\to \red{\huge{x_1=2}}$

$\text{D}_{x_2}=\left|\begin{array}{ccc}1&-5&1\\3&11&-2\\-2&-2&1\end{array}\right|$

$\text{D}_{x_2}=$ (1 × 11 × 1) + (–5 × –2 × –2) + (1 × 3 × –2) – (1 × 11 × –2) – (1 × –2 × –2) – (–5 × 3 × 1)

$\text{D}_{x_2}=$ 11 + (–20) + (–6) – (–22) – 4 – (–15) = 18

$x_2=\frac{\text{D}_{x_2}}{\text{D}}=\frac{18}{6}~\to \red{\huge{x_2=3}}$

$\text{D}_{x_3}=\left|\begin{array}{ccc}1&-2&-5\\3&1&11\\-2&1&-2\end{array}\right|$

$\text{D}_{x_3}=$ (1 × 1 × –2) + (–2 × 11 × –2) + (–5 × 3 × 1) – (–5 × 1 × –2) – (1 × 11 × 1) – (–2 × 3 × –2)

$\text{D}_{x_3}=$ –2 + 44 + (–15) – 10 – 11 – 12 = –6

$x_3=\frac{\text{D}_{x_3}}{\text{D}}=\frac{-6}{6}~\to \red{\huge{x_3=-1}}$

$\\$

$\purple{\huge{b~)}}~$ Dengan metode invers matriks :

$\text{A}=\left[\begin{array}{ccc}1&-2&1\\3&1&-2\\-2&1&1\end{array}\right]$

$\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+\left|\begin{array}{ccc}1&-2\\1&1\end{array}\right|&-\left|\begin{array}{ccc}3&-2\\-2&1\end{array}\right|&+\left|\begin{array}{ccc}3&1\\-2&1\end{array}\right|\\-\left|\begin{array}{ccc}-2&1\\1&1\end{array}\right|&+\left|\begin{array}{ccc}1&1\\-2&1\end{array}\right|&-\left|\begin{array}{ccc}1&-2\\-2&1\end{array}\right|\\+\left|\begin{array}{ccc}-2&1\\1&-2\end{array}\right|&-\left|\begin{array}{ccc}1&1\\3&-2\end{array}\right|&+\left|\begin{array}{ccc}1&-2\\3&1\end{array}\right|\end{array}\right]$

$\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+(~(1\times 1)-(-2\times 1)~)&-(~(3\times 1)-(-2\times -2)~)&+(~(3\times 1)-(1\times -2)~)\\-(~(-2\times 1)-(1\times 1)~)&+(1\times 1)-(1\times -2)~)&-(~(1\times 1)-(-2\times -2)~)\\+(~(-2\times -2)-(1\times 1)~)&-(~(1\times -2)-(1\times 3)~)&+(~(1\times 1)-(-2\times 3)~)\end{array}\right]$

$\text{Kofaktor~(A)}=\left[\begin{array}{ccc}3&1&5\\3&3&3\\3&5&7\end{array}\right]$

$Adj.\text{~(A)}=\text{Kofaktor~(A)}^\text{T}$

$Adj.\text{~(A)}=\left[\begin{array}{ccc}3&3&3\\1&3&5\\5&3&7\end{array}\right]$

$\text{A}^{-1}=\frac{1}{6}\left[\begin{array}{ccc}3&3&3\\1&3&5\\5&3&7\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}3&3&3\\1&3&5\\5&3&7\end{array}\right]\left[\begin{array}{ccc}-5\\11\\-2\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}(3\times -5)+(3\times 11)+(3\times -2)\\(1\times -5)+(3\times 11)+(5\times -2)\\(5\times -5)+(3\times 11)+(7\times -2)\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}12\\18\\-6\end{array}\right]$

$\red{\huge{\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc}2\\3\\-1\end{array}\right]}}$

$\purple{\huge{c~)}}~$ Dengan metode Gauss-Jordan :

$\left[\begin{array}{ccc}1&-2&1\\3&1&-2\\-2&1&1\end{array}\right|\left.\begin{array}{ccc}-5\\11\\-2\end{array}\right]$

$\begin{array}{rrr}~\\b_2-3b_1\to\\b_3+2b_1\to\end{array}\left[\begin{array}{ccc}1&-2&1\\0&7&-5\\0&-3&3\end{array}\right|\left.\begin{array}{ccc}-5\\26\\-12\end{array}\right]$

$\begin{array}{rrr}~\\b_2\div 7\to\\~\end{array}\left[\begin{array}{ccc}1&-2&1\\0&1&-\frac{5}{7}\\0&-3&3\end{array}\right|\left.\begin{array}{ccc}-5\\\frac{26}{7}\\-12\end{array}\right]$

$\begin{array}{rrr}b_1+2b_2\to\\~\\b_3+3b_2\to\end{array}\left[\begin{array}{ccc}1&0&-\frac{3}{7}\\0&1&-\frac{5}{7}\\0&0&\frac{6}{7}\end{array}\right|\left.\begin{array}{ccc}\frac{17}{7}\\\frac{26}{7}\\-\frac{6}{7}\end{array}\right]$

$\begin{array}{rrr}~\\~\\b_3\times \frac{7}{6}\to\end{array}\left[\begin{array}{ccc}1&0&-\frac{3}{7}\\0&1&-\frac{5}{7}\\0&0&1\end{array}\right|\left.\begin{array}{ccc}\frac{17}{7}\\\frac{26}{7}\\-1\end{array}\right]$

$\begin{array}{rrr}b_1+\frac{3}{7}b_3\to\\b_2+\frac{5}{7}b_3\to\\~\end{array}\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|\left.\begin{array}{ccc}2\\3\\-1\end{array}\right]$

$\red{\huge{\begin{array}{ccc}x_1=2&x_2=3&x_3=-1\end{array}}}$

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Last Update: Tue, 03 Aug 21

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