Contoh Soal 5 Matriks

$\begin{array}{l}\\ 21.& (\mathbf{\text{SPMB 2003}})\\ & \text{Diketahu matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right).\\ & \text{Jika}{\text{A}}^t={\text{A}}^{-1}\text{dengan}{\text{A}}^t\\ & \text{adalah transpose matriks A},\\ & \text{maka nilai}ad-bc=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-\sqrt{2}\\ \text{b}.& 1\text{atau}\sqrt{2}\\ \text{c}.& -\sqrt{2}\text{atau}-\sqrt{2}\\ \text{d}.& -1\text{atau}1\\ \text{e}.& 1\text{atau}-\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\text{matriks}\text{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & \text{dan}{\text{A}}^t={\text{A}}^{-1},\text{maka}\\ & {\text{A}}^t={\text{A}}^{-1}\\ & {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}^t={\displaystyle \frac{1}{ad-bc}\times \text{Adjoin Matriks}\text{A}}\\ & \left(\begin{array}{cc}a& c\\ b& d\end{array}\right)={\displaystyle \frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),}\\ & \text{didapatkan hubungan}\\ & c={\displaystyle \frac{-b}{ad-bc}...............(1)}\\ & b={\displaystyle \frac{-c}{ad-bc}...............(2)}\\ & \text{Persamaan}(2)\text{disubstitusikan ke persamaan}(1)\\ & c={\displaystyle {\displaystyle \frac{-{\displaystyle \frac{-c}{ad-bc}}}{ad-bc}}}\\ & 1=(ad-bc{)}^2\\ & (ad-bc)=-1\text{atau}1\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Diketahu matriks}\text{H}\\ & \text{yang memenuhi persamaan}\\ & \text{H}\left(\begin{array}{cc}3& 2\\ 1& 4\end{array}\right)=\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right),\\ & \text{maka nilai dari}det\text{H}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -2\\ \text{c}.& -1\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{|c|}\hline \mathbf{\text{Alternatif 1}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\\ {\text{H.A.A}}^{-1}& ={\text{B.A}}^{-1}\\ \text{H}& ={\text{B.A}}^{-1}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{\left|\begin{array}{cc}3& 2\\ 1& 4\end{array}\right|}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & =\left(\begin{array}{cc}7& 8\\ 4& 6\end{array}\right).{\displaystyle \frac{1}{12-2}\left(\begin{array}{cc}4& -2\\ -1& 3\end{array}\right)}\\ & ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}28+(-8)& (-14)+24\\ 16+(-6)& (-8)+18\end{array}\right)}\\ \text{H}& ={\displaystyle \frac{1}{10}\left(\begin{array}{cc}20& 10\\ 10& 10\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)}\\ det\text{H}& =\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=2.1-1.1=2-1=1\end{array}\\ \mathbf{\text{Alternatif 2}}\\ \begin{array}{rl}\text{H.A}& =\text{B}\{\begin{array}{ll}det\text{H}& =\left|\text{H}\right|\\ det\text{A}& =\left|\text{A}\right|=\left|\begin{array}{cc}3& 4\\ 2& 1\end{array}\right|\\ & =12-2=10\\ det\text{B}& =\left|\text{B}\right|=\left|\begin{array}{cc}7& 8\\ 4& 6\end{array}\right|\\ & =42-32=10\end{array}\\ \left|\text{H}\right|.\left|\text{A}\right|& =\left|\text{B}\right|\\ \left|\text{H}\right|& ={\displaystyle \frac{\left|\text{B}\right|}{\left|\text{A}\right|}}\\ & ={\displaystyle \frac{10}{10}}\\ & =1\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 23.& (\mathbf{\text{UM UGM 2006}})\\ & \text{Apabila}x\text{dan}y\text{memenuhi}\\ & \text{persamaan matriks}\\ & \left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-1\\ 2\end{array}\right),\\ & \text{maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ A.X& =B\\ A^{-1}.A.X& =A^{-1}.B\\ A^0.X& =A^{-1}.B\\ X& =A^{-1}.B\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)}^{-1}\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|}}& \left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}-1\\ 2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3-2}}& \left(\begin{array}{c}3.(-1)+2.2\\ 1.(-1)+1.2\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}1\\ 1\end{array}\right)\\ x+y& =1+1=2\end{array}\end{array}$

$\begin{array}{l}\\ 24.& (\mathbf{\text{KSM Matematika Kabupten 2019}})\\ & \text{Matriks}A\text{dengan entri bulat dan}\\ & \text{berukuran 2x2},\text{dikalikan dengan matriks}\\ & \left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)\text{dari kanan menghasilkan matriks}\\ & \text{yang semua entrinya bilangan prima}.\\ & \text{Jika determinan dari matriks}A\text{juga}\\ & \text{bilangan prima, maka nilai minimum dari}\\ & detA\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 5\\ \text{d}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{cc}1& 2\\ 2& 2\end{array}\right)& \times A_{2\times 2}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\\ \left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|& \times \left|A_{2\times 2}\right|=\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{\left|\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\alpha \delta -\beta \gamma )}{-2}}\\ & \left|A_{2\times 2}\right|={\displaystyle \frac{(\beta \gamma -\alpha \delta )}{2}}\\ \text{Karena}& \left|A_{2\times 2}\right|\text{bilangan prima}\\ \text{akan m}& \text{engakibatkan}(\beta \gamma -\alpha \delta )\\ \text{harus h}& \text{abis dibagi}2,\text{oleh karenanya}\\ \text{menyeb}& \text{abkan}(\beta \gamma -\alpha \delta )\text{berupa bilangan}\\ \text{genap.}& \text{Dan karena}(\beta \gamma -\alpha \delta )\text{genap},\\ \text{maka p}& \text{astilah}\left|A_{2\times 2}\right|\text{juga bernilai genap}\\ \text{sehingg}& \text{a nilai}\left|A_{2\times 2}\right|\text{pastilah 2}\end{array}\end{array}$

$\begin{array}{l}\\ 25.& (\mathbf{\text{UM UGM 2005}})\text{Jika}\\ & \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \text{dan}A\text{suatu konstanta, maka}x+y=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \left(\begin{array}{cc}x\sin \alpha -y\cos \alpha & x\cos \alpha +y\sin \alpha \end{array}\right)=\left(\begin{array}{cc}\sin A& \cos A\end{array}\right)\\ & \{\begin{array}{ll}\sin A& =x\sin \alpha -y\cos \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{x}{-y}})\\ \cos A& =x\cos \alpha +y\sin \alpha =\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}})\end{array}\\ & \text{Supaya}\cos A=\sqrt{x^2+y^2}\cos (\alpha -{\tan }^{-1}{\displaystyle \frac{y}{x}}),\text{maka}\\ & \{\begin{array}{ll}\sqrt{x^2+y^2}& =1\\ {\tan }^{-1}{\displaystyle \frac{y}{x}}& =0\Rightarrow \{\begin{array}{ll}y& =0\\ x& =1\end{array}\end{array}\\ & \text{Sehingga}x+y=1+0=1\end{array}\end{array}$