Matriks Balikan (Invers)/Lanjutan

13 Aug 2021

17 min.

Contoh 3: Matriks

A =

{\begin{bmatrix}3&1\\5&2\\\end{bmatrix}}

${\begin{bmatrix}3&1\\5&2\\\end{bmatrix}}$

Tentukan Nilai dari A⁻¹

Jawab: $A^{-1}={\frac {1}{(3)(2)-(5)(1)}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\frac {1}{6-5}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\frac {1}{1}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}$ $A^{-1}={\frac {1}{(3)(2)-(5)(1)}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\frac {1}{6-5}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\frac {1}{1}}{\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}={\begin{bmatrix}2&-1\\-5&3\\\end{bmatrix}}$

Contoh 4: Matriks

A =

{\begin{bmatrix}1&2\\1&3\\\end{bmatrix}}

${\begin{bmatrix}1&2\\1&3\\\end{bmatrix}}$ , B =

{\begin{bmatrix}3&2\\2&2\\\end{bmatrix}}

${\begin{bmatrix}3&2\\2&2\\\end{bmatrix}}$ , AB =

{\begin{bmatrix}7&6\\9&8\\\end{bmatrix}}

${\begin{bmatrix}7&6\\9&8\\\end{bmatrix}}$

Dengan menggunakan rumus, maka didapatkan

A^{-1}={\begin{bmatrix}3&-2\\-1&1\\\end{bmatrix}}

$A^{-1}={\begin{bmatrix}3&-2\\-1&1\\\end{bmatrix}}$ ,

B^{-1}={\begin{bmatrix}1&-1\\-1&{\frac {3}{2}}\\\end{bmatrix}}

$B^{-1}={\begin{bmatrix}1&-1\\-1&{\frac {3}{2}}\\\end{bmatrix}}$ ,

(AB)^{-1}={\begin{bmatrix}4&-3\\-{\frac {9}{2}}&7\\\end{bmatrix}}

$(AB)^{-1}={\begin{bmatrix}4&-3\\-{\frac {9}{2}}&7\\\end{bmatrix}}$

Maka

B^{-1}A^{-1}={\begin{bmatrix}1&-1\\-1&{\frac {3}{2}}\\\end{bmatrix}}

$B^{-1}A^{-1}={\begin{bmatrix}1&-1\\-1&{\frac {3}{2}}\\\end{bmatrix}}$

{\begin{bmatrix}3&-2\\-1&1\\\end{bmatrix}}

${\begin{bmatrix}3&-2\\-1&1\\\end{bmatrix}}$ =

{\begin{bmatrix}4&-3\\-{\frac {9}{2}}&7\\\end{bmatrix}}

${\begin{bmatrix}4&-3\\-{\frac {9}{2}}&7\\\end{bmatrix}}$

Ini membuktikan bahwa $(AB)^{-1}=B^{-1}A^{-1}$ $(AB)^{-1}=B^{-1}A^{-1}$

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Matriks Balikan (Invers)/Lanjutan

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