Lanjutan 4 Materi Matriks (Matematika Wajib Kelas XI)

$\text{F. Invers Matriks ordo 2x2}$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\begin{array}{|c|}\hline \begin{array}{rl}& \text{Jika matriks}A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & \text{maka determinan matriks}A\\ & \text{ditentukan dengan}\\ & detA=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc\end{array}\\ \hline\end{array}$

Jika  $detA$  bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan  $A^{-1}$  dapat ditentukan dengan formula:

$\boxed{{\displaystyle A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)}}$

$\begin{array}{rl}& \mathbf{\text{Sebagai}}\text{CONTOH}\\ & \text{Diketahui sebuah matrik ordo}2x2,\text{yaitu}:\\ & A=\left(\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right),\text{maka}{A}^{{}^{-1}}\text{adalah}:\\ & A^{{}^{-1}}={\displaystyle \frac{1}{\left|\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right|}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & ={\displaystyle \frac{1}{12-(-2)}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & ={\displaystyle \frac{1}{14}\left(\begin{array}{cc}4& -2\\ 1& -3\end{array}\right)}\\ & =\left(\begin{array}{cc}{\displaystyle \frac{4}{14}}& {\displaystyle \frac{-2}{14}}\\ {\displaystyle \frac{1}{14}}& {\displaystyle \frac{-3}{14}}\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{2}{7}}& -{\displaystyle \frac{1}{7}}\\ {\displaystyle \frac{1}{14}}& -{\displaystyle \frac{3}{14}}\end{array}\right)\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah invers matriks berikut}\\ & \text{a}.B=\left(\begin{array}{cc}5& -3\\ 4& -2\end{array}\right)\\ & \text{b}.C=\left(\begin{array}{cc}-3& -5\\ 6& 9\end{array}\right)\\ & \text{c}.P=\left(\begin{array}{cc}-1& 2\\ -3& 6\end{array}\right)\\ & \text{d}.Q=\left(\begin{array}{cc}6& 9\\ 2& 3\end{array}\right)\\ \\ & \text{Jawab}:\text{yang dibahas poin a saja}\\ & B^{-1}={\displaystyle \frac{1}{detB}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)}\\ & ={\displaystyle \frac{1}{-10-(-12)}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)={\displaystyle \frac{1}{2}\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right)}}\\ & =\left(\begin{array}{cc}-1& {\displaystyle \frac{3}{2}}\\ -2& {\displaystyle \frac{5}{2}}\end{array}\right)\\ & \text{b. Silahkan dicoba sendiri}\\ & \text{c. Silahkan dicoba sendiri}\\ & \text{d. Silahkan dicoba sendiri}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui matriks}\\ & E=\left(\begin{array}{cc}4& 2\\ 5& 3\end{array}\right)\\ & \text{a}.\text{Tentukanlah}\\ & (\text{i})E^{-1}(\text{iii}){\left(E^{-1}\right)}^t\\ & (\text{i})E^t(\text{iv}){\left(E^t\right)}^{-1}\\ & \text{b}.\text{Dengan menggunakan hasil-hasil}\\ & \text{pada a. apakah}{\left(E^{-1}\right)}^t={\left(E^t\right)}^{-1}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.(\text{i})E^{-1}& ={\displaystyle \frac{1}{2}\left(\begin{array}{cc}3& -2\\ -5& 4\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -1\\ -{\displaystyle \frac{5}{2}}& 2\end{array}\right)}\\ (\text{ii})E^t& =\left(\begin{array}{cc}4& 5\\ 2& 3\end{array}\right)\\ (\text{iii})& {\left(E^{-1}\right)}^t=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{5}{2}}\\ -1& 2\end{array}\right)\\ (\text{iv})& {\left(E^t\right)}^{-1}={\displaystyle \frac{1}{2}\left(\begin{array}{cc}3& -5\\ -2& 4\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{5}{2}}\\ -1& 2\end{array}\right)}\\ \text{b}.\text{Dari ha}& \text{sil yang didapat dapat disimpulkan}\\ & {\left(E^{-1}\right)}^t={\left(E^t\right)}^{-1}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA