Selesaikan menggunakan augmented matriks!ngasal ? → report!​

Berikut ini adalah pertanyaan dari invaliduser121 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Selesaikan menggunakan augmented matriks!

ngasal ? → report!​
Selesaikan menggunakan augmented matriks!ngasal ? → report!​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

  1. x = 1, y = 2, z = 3.
  2. x = 1, y = 7/3, z = 5/3.
  3. x = 11/12, y = –71/12, z = –5/12.

Pembahasan

Nomor 1

\begin{aligned}\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\2 & 4 & -3 & 1\\3 & 6 & -5 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_2-2B_1\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 2 & -7 & -17\\3 & 6 & -5 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_3-3B_1\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 2 & -7 & -17\\0 & 3 & -11 & -27\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_2/2\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 1 & -7/2 & -17/2\\0 & 3 & -11 & -27\end{array}\right)\\\xrightarrow{\begin{matrix}B_3-3B_2\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 1 & -7/2 & -17/2\\0 & 0 & -1/2 & -3/2\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_3\times(-2)\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 1 & -7/2 & -17/2\\0 & 0 & 1 & 3\end{array}\right)\\\xrightarrow{\begin{matrix}B_2+\frac{7}{2}B_3\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 2 & 9 \\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_1-B_2\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\end{array}\right)\\\xrightarrow{\begin{matrix}B_1-2B_3\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 0 & 1 \\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\end{array}\right)\\\end{aligned}

Solusi: x = 1, y = 2, z = 3.
______________

Nomor 2

\begin{aligned}\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\1 & 1 & 4 & 10\\-4 & 1 & 1 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_2-B_1\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\0 & 0 & 3 & 5\\-4 & 1 & 1 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_2/3\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\0 & 0 & 1 & 5/3\\-4 & 1 & 1 & 0\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_2\leftrightarrow B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\-4 & 1 & 1 & 0 \\0 & 0 & 1 & 5/3\end{array}\right)\\\xrightarrow{\begin{matrix}B_2+4B_1\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\0 & 5 & 5 & 20 \\0 & 0 & 1 & 5/3\end{array}\right)\\\xrightarrow{\begin{matrix}B_2/5\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\0 & 1 & 1 & 4 \\0 & 0 & 1 & 5/3\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_2-B_3\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 1 & 1 & 5 \\0 & 1 & 0 & 7/3 \\0 & 0 & 1 & 5/3\end{array}\right)\\\xrightarrow{\begin{matrix}B_1-B_2\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 8/3 \\0 & 1 & 0 & 7/3 \\0 & 0 & 1 & 5/3\end{array}\right)\\\xrightarrow{\begin{matrix}B_1-B_3\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 0 & 1 \\0 & 1 & 0 & 7/3 \\0 & 0 & 1 & 5/3\end{array}\right)\\\end{aligned}

Solusi: x = 1, y = 7/3, z = 5/3.

______________

Nomor 3

\begin{aligned}\left(\begin{array}{ccc|c}2 & 0 & 2 & 1 \\3 & -1 & 4 & 7\\6 & 1 & -1 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_1/2\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\3 & -1 & 4 & 7\\6 & 1 & -1 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_2-3B_1\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & -1 & 1 & 11/2\\6 & 1 & -1 & 0\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_2\times(-1)\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & 1 & -1 & -11/2\\6 & 1 & -1 & 0\end{array}\right)\\\xrightarrow{\begin{matrix}B_3-6B_1\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & 1 & -1 & -11/2\\0 & 1 & -7 & -3\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_3-B_2\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & 1 & -1 & -11/2\\0 & 0 & -6 & 5/2\end{array}\right)\\\xrightarrow{\begin{matrix}B_3/(-6)\to B_3\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & 1 & -1 & -11/2\\0 & 0 & 1 & -5/12\end{array}\right)\end{aligned}
\begin{aligned}\xrightarrow{\begin{matrix}B_2+B_3\to B_2\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 1 & 1/2 \\0 & 1 & 0 & -71/12\\0 & 0 & 1 & -5/12\end{array}\right)\\\xrightarrow{\begin{matrix}B_1-B_3\to B_1\end{matrix}}&\left(\begin{array}{ccc|c}1 & 0 & 0 & 11/12 \\0 & 1 & 0 & -71/12\\0 & 0 & 1 & -5/12\end{array}\right)\end{aligned}

Solusi: x = 11/12, y = –71/12, z = –5/12.

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Last Update: Thu, 22 Dec 22