carilah invers dari matriks pada gambar. gunakan metode OBE

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

Carilah invers dari matriks pada gambar. gunakan metode OBE

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Invers matriks tersebut adalah:
$\begin{pmatrix}\vphantom{\bigg|}\ \bf\dfrac{5}{3} & \bf0 & \bf{-}\dfrac{10}{3}\ \\\vphantom{\Bigg|}\ \bf\dfrac{7}{3} & \bf1 & \bf\dfrac{10}{3}\ \\\vphantom{\Big|}\ \bf1 & \bf{-}1 & \bf0\ \end{pmatrix}$
 

Pembahasan

Mencari Invers Matriks dengan metode Operasi Baris Elementer (OBE)

Diketahui

Matriks:
$\begin{aligned}&\begin{pmatrix}\vphantom{\bigg|}\dfrac{1}{5} & \dfrac{1}{5} & \dfrac{1}{5} \\\vphantom{\Bigg|}\dfrac{1}{5} & \dfrac{1}{5} & -\dfrac{4}{5} \\\vphantom{\bigg|}-\dfrac{1}{5} & \dfrac{1}{10} & \dfrac{1}{10}\end{pmatrix}\end{aligned}$

Ditanyakan
Invers matriks tersebut dengan menggunakan metode OBE.

PENYELESAIAN

Kita konstruksi augmented matrix-nya terlebih dahulu, lalu selesaikan dengan OBE.

$\begin{aligned}&\left(\begin{array}{c|c}\begin{matrix}\vphantom{\bigg|}\dfrac{1}{5} & \dfrac{1}{5} & \dfrac{1}{5} \\\vphantom{\Bigg|}\dfrac{1}{5} & \dfrac{1}{5} & {-}\dfrac{4}{5} \\\vphantom{\bigg|}{-}\dfrac{1}{5} & \dfrac{1}{10} & \dfrac{1}{10}\end{matrix}&\begin{matrix}\vphantom{\bigg|}1 & 0 & 0 \\\vphantom{\Bigg|}0 & 1 & 0 \\\vphantom{\bigg|}0 & 0 & 1\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}5R_1\to R_1\\5R_2\to R_2\\10R_3\to R_3\\\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 1 \\\vphantom{\Big|}1 & 1 & -4 \\\vphantom{\Big|}{-}2 & 1 & 1\end{matrix}&\begin{matrix}\vphantom{\Big|}5 & 0 & 0 \\\vphantom{\Big|}0 & 5 & 0 \\\vphantom{\Big|}0 & 0 & 10\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}R_2-R_1\to R_2\\R_3+2R_1\to R_3\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 1 \\\vphantom{\Big|}0 & 0 & -5 \\\vphantom{\Big|}0 & 3 & 3\end{matrix}&\begin{matrix}\vphantom{\Big|}5 & 0 & 0 \\\vphantom{\Big|}{-}5 & 5 & 0 \\\vphantom{\Big|}10 & 0 & 10\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}R_2\longleftrightarrow R_3\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 1\\\vphantom{\Big|}0 & 3 & 3\\\vphantom{\Big|}0 & 0 & -5\end{matrix}&\begin{matrix}\vphantom{\Big|}5 & 0 & 0\\\vphantom{\Big|}10 & 0 & 10\\\vphantom{\Big|}{-}5 & 5 & 0\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}\frac{1}{3}R_2\to R_2\\\left(-\frac{1}{5}\right)R_3\to R_3\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 1\\\vphantom{\Bigg|}0 & 1 & 1\\\vphantom{\Big|}0 & 0 & 1\end{matrix}&\begin{matrix}\vphantom{\Big|}5 & 0 & 0\\\vphantom{\Bigg|}\dfrac{10}{3} & 0 & \dfrac{10}{3}\\\vphantom{\Big|}1 & {-}1 & 0\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}R_2-R_3\to R_2\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 1\\\vphantom{\Bigg|}0 & 1 & 0\\\vphantom{\Big|}0 & 0 & 1\end{matrix}&\begin{matrix}\vphantom{\Big|}5 & 0 & 0\\\vphantom{\Bigg|}\dfrac{7}{3} & 1 & \dfrac{10}{3}\\\vphantom{\Big|}1 & {-}1 & 0\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}R_1-R_3\to R_1\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\Big|}1 & 1 & 0\\\vphantom{\Bigg|}0 & 1 & 0\\\vphantom{\Big|}0 & 0 & 1\end{matrix}&\begin{matrix}\vphantom{\Big|}4 & 1 & 0\\\vphantom{\Bigg|}\dfrac{7}{3} & 1 & \dfrac{10}{3}\\\vphantom{\Big|}1 & {-}1 & 0\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

$\begin{aligned}&\begin{aligned}R_1-R_2\to R_1\end{aligned}\quad\left(\begin{array}{c|c}\begin{matrix}\vphantom{\bigg|}1 & 0 & 0\\\vphantom{\Bigg|}0 & 1 & 0\\\vphantom{\Big|}0 & 0 & 1\end{matrix}&\begin{matrix}\vphantom{\bigg|}\bf\dfrac{5}{3} & \bf0 & \bf{-}\dfrac{10}{3}\\\vphantom{\Bigg|}\bf\dfrac{7}{3} & \bf1 & \bf\dfrac{10}{3}\\\vphantom{\Big|}\bf1 & \bf{-}1 & \bf0\end{matrix}&\end{array}\right)\\&.........................................................................\\\end{aligned}$

KESIMPULAN

Dengan demikian, invers matriks tersebut adalah:
$\begin{pmatrix}\vphantom{\bigg|}\ \bf\dfrac{5}{3} & \bf0 & \bf{-}\dfrac{10}{3}\ \\\vphantom{\Bigg|}\ \bf\dfrac{7}{3} & \bf1 & \bf\dfrac{10}{3}\ \\\vphantom{\Big|}\ \bf1 & \bf{-}1 & \bf0\ \end{pmatrix}$
$\blacksquare$