Pembahasan Penerapan Matriks UTUL UGM 2017 Matematika Dasar Kode 823

Soal yang Akan Dibahas
Sistem persamaan linear
$ \, \, \, \, \, \, \begin{align} & 2x \sin a + y \cos a = -2 \\ & 2x \cos a - y \sin a = 2 \end{align} $
mempunyai solusi $ \left( \begin{matrix} x \\ y \end{matrix} \right) = .... $
A). $ \left( \begin{matrix} \sin a + \cos a \\ -2\cos a - 2 \sin a \end{matrix} \right) \, $
B). $ \left( \begin{matrix} -\sin a + \cos a \\ 2\cos a - 2 \sin a \end{matrix} \right) \, $
C). $ \left( \begin{matrix} \sin a - \cos a \\ -2\cos a - 2 \sin a \end{matrix} \right) \, $
D). $ \left( \begin{matrix} -\sin a + \cos a \\ -2\cos a - 2 \sin a \end{matrix} \right) \, $
E). $ \left( \begin{matrix} -\sin a + \cos a \\ -2\cos a + 2 \sin a \end{matrix} \right) \, $

$\spadesuit $ Konsep Dasar
*). Determinan matriks A :
$ A = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \rightarrow |A| = ad - bc $
*). Invers matriks A :
$ A = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \rightarrow A^{-1} = \frac{1}{|A|} \left( \begin{matrix} d & -b \\ -c & a \end{matrix} \right) $
*). Sifat invers matriks :
$ AX = B \rightarrow X = A^{-1}.B $
*). Sistem persamaan Linear Metode invers:
$ \, \, \, \, \, \, \begin{align} & a_1x + b_1y = c_1 \\ & a_2x + b_2y = c_2 \end{align} $
Bentuk persamaan matriksnya :
$ \left( \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} c_1 \\ c_2 \end{matrix} \right) $
*). Identitas trigonometri :
$ \sin ^2 x + \cos ^2 x = 1 $

$\clubsuit $ Pembahasan
*). Menyusun persamaan matriksnya :
$ \, \, \, \, \, \, \begin{align} & 2x \sin a + y \cos a = -2 \\ & 2x \cos a - y \sin a = 2 \end{align} $
Persamaannya :
$ \left( \begin{matrix} 2\sin a & \cos a \\ 2\cos a & -\sin a \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) $
*). Menentukan hasil akhirnya :
$ \begin{align} & \left( \begin{matrix} 2\sin a & \cos a \\ 2\cos a & -\sin a \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) \\ \left( \begin{matrix} x \\ y \end{matrix} \right) & = \left( \begin{matrix} 2\sin a & \cos a \\ 2\cos a & -\sin a \end{matrix} \right) ^{-1} . \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) \\ & = \frac{1}{-2\sin ^2 a - 2\cos ^2 a} \left( \begin{matrix} -\sin a & -\cos a \\ -2\cos a & 2\sin a \end{matrix} \right). \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) \\ & = \frac{1}{-2(\sin ^2 a + \cos ^2 a)} \left( \begin{matrix} -\sin a & -\cos a \\ -2\cos a & 2\sin a \end{matrix} \right). \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) \\ & = \frac{1}{-2.1} \left( \begin{matrix} -\sin a & -\cos a \\ -2\cos a & 2\sin a \end{matrix} \right). \left( \begin{matrix} -2 \\ 2 \end{matrix} \right) \\ & = \left( \begin{matrix} -\sin a & -\cos a \\ -2\cos a & 2\sin a \end{matrix} \right). \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \\ & = \left( \begin{matrix} -\sin a + \cos a \\ -2\cos a - 2\sin a \end{matrix} \right) \end{align} $
Jadi, nilai $ \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} -\sin a + \cos a \\ -2\cos a - 2\sin a \end{matrix} \right) . \, \heartsuit $

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