[tex]\color{aqua} \underline{ \small{ \colorbox{black}{ \gray{ \boxed{{ \sf \: ✨ \color{fdff28}{QUIZ.... Question from @Gusti2601

August 2022 1 19 Report

[tex]\color{aqua} \underline{ \small{ \colorbox{black}{ \gray{ \boxed{{ \sf \: ✨ \color{fdff28}{QUIZ}✨{ { \pink{ \boxed{ \color{fdff28}{ \: - \tt \: MATEMATIKA - \: }}}}}}}}}}} \\ \\[/tex]
Tentukan invers matriks berikut
[tex] \bf{A = } \begin{aligned} \left( \begin{array}{ccc} \tt{1}& \tt{2}& \tt{3} \\ \tt{2}& \tt{8}& \tt{7} \\ \tt{1}& \tt{5}& \tt{6} \end{array} \right) \end{aligned}[/tex]

henriyulianto

Invers dari matriks A adalah:
$\begin{aligned}A^{-1}&=\bf\begin{pmatrix}\bf\vphantom{\Bigg|}\dfrac{13}{9} &\bf\dfrac{1}{3} &\bf-\dfrac{10}{9} \\\bf\vphantom{\Bigg|}{-}\dfrac{5}{9} &\bf\dfrac{1}{3} &\bf{-}\dfrac{1}{9} \\\bf\vphantom{\Bigg|}\dfrac{2}{9} &\bf{-}\dfrac{1}{3} &\bf\dfrac{4}{9} \\\end{pmatrix}\\\end{aligned}$
 

Pembahasan

Kita akan menentukan invers dari matriks:

$\begin{aligned}A&=\begin{pmatrix}1 & 2 & 3 \\2 & 8 & 7 \\1 & 5 & 6\end{pmatrix}\end{aligned}$

Kita gunakan metode Minor-Kofaktor.

$\begin{aligned}A^{-1}&=\frac{1}{\det(A)}\cdot C^{T}\\&=\frac{1}{\det(A)}\begin{pmatrix}C_{1,1} & C_{2,1} & C_{3,1}\\ C_{1,2} & C_{2,2} & C_{3,2}\\ C_{1,3} & C_{2,3} & C_{3,3}\end{pmatrix}\end{aligned}$
dengan syarat determinan matriks A ≠ 0.

$\begin{aligned}\det(A)&=48+14+30-(24+35+24)\\&=92-83\\&=\bf9\:\ne\:0\end{aligned}$
maka, proses dapat kita lanjutkan.

$\begin{aligned}C_{1,1}&=(-1)^{1+1}\begin{vmatrix}* & * & * \\ * & 8 & 7 \\ * & 5 & 6\end{vmatrix}\\&=48-35=\bf13\\C_{1,2}&=(-1)^{1+2}\begin{vmatrix}* & * & * \\2 & * & 7 \\1 & * & 6\end{vmatrix}\\&=(-1)(12-7)=\bf-5\\C_{1,3}&=(-1)^{1+3}\begin{vmatrix}* & * & * \\2 & 8 & * \\1 & 5 & *\end{vmatrix}\\&=10-8=\bf2\end{aligned}$
$\begin{aligned}C_{2,1}&=(-1)^{2+1}\begin{vmatrix}* & 2 & 3 \\ * & * & * \\ * & 5 & 6\end{vmatrix}\\&=(-1)(12-15)=\bf3\\C_{2,2}&=(-1)^{2+2}\begin{vmatrix}1 & * & 3 \\ * & * & * \\1 & * & 6\end{vmatrix}\\&=6-3=\bf3\\C_{2,3}&=(-1)^{2+3}\begin{vmatrix}1 & 2 & * \\ * & * & * \\1 & 5 & *\end{vmatrix}\\&=(-1)(5-2)=\bf-3\end{aligned}$
$\begin{aligned}C_{3,1}&=(-1)^{3+1}\begin{vmatrix}* & 2 & 3 \\ * & 8 & 7 \\ * & * & *\end{vmatrix}\\&=14-24=\bf{-}10\\C_{3,2}&=(-1)^{3+2}\begin{vmatrix}1 & * & 3 \\2 & * & 7 \\ * & * & *\end{vmatrix}\\&=(-1)(7-6)=\bf-1\\C_{3,3}&=(-1)^{3+3}\begin{vmatrix}1 & 2 & * \\2 & 8 & * \\ * & * & *\end{vmatrix}\\&=8-4=\bf4\\\end{aligned}$

Lalu, substitusi ke rumus $A^{-1}$ di atas.

$\begin{aligned}A^{-1}&=\frac{1}{\det(A)}\begin{pmatrix}C_{1,1} & C_{2,1} & C_{3,1}\\ C_{1,2} & C_{2,2} & C_{3,2}\\ C_{1,3} & C_{2,3} & C_{3,3}\end{pmatrix}\\&=\frac{1}{9}\begin{pmatrix}13 & 3 & -10 \\-5 & 3 & -1 \\2 & -3 & 4\end{pmatrix}\end{aligned}$

$\begin{aligned}\therefore\ A^{-1}&=\bf\begin{pmatrix}\bf\vphantom{\Bigg|}\dfrac{13}{9} &\bf\dfrac{1}{3} &\bf-\dfrac{10}{9} \\\bf\vphantom{\Bigg|}{-}\dfrac{5}{9} &\bf\dfrac{1}{3} &\bf{-}\dfrac{1}{9} \\\bf\vphantom{\Bigg|}\dfrac{2}{9} &\bf{-}\dfrac{1}{3} &\bf\dfrac{4}{9} \\\end{pmatrix}\\\end{aligned}$