MATRIKS-CONTOH 5

 $\begin{array}{ll}\\ 21.&(\textbf{SPMB 2003})\\ &\textrm{Diketahu matriks}\: \: \textrm{A}=\begin{pmatrix}a&b\\ c&d \end{pmatrix}.\\ &\textrm{Jika}\: \: \: \textrm{A}^{t}=\textrm{A}^{-1}\: \: \textrm{dengan}\: \: \textrm{A}^{t}\\ &\textrm{adalah transpose matriks A},\\ &\textrm{maka nilai}\: \: ad-bc=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{b}.&1\: \: \textrm{atau}\: \: \sqrt{2}\\ \textrm{c}.&-\sqrt{2}\: \: \textrm{atau}\: \: -\sqrt{2}\\ \textrm{d}.&-1\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&1\: \: \textrm{atau}\: \: -\sqrt{2}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{matriks}\: \: \textrm{A}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{A}^{t}=\textrm{A}^{-1},\: \textrm{maka}\\ &\textrm{A}^{t}=\textrm{A}^{-1}\\ &\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{t}=\displaystyle \frac{1}{ad-bc}\times \textrm{Adjoin Matriks}\: \: \textrm{A}\\ &\begin{pmatrix} a & c\\ b & d \end{pmatrix}=\displaystyle \frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix},\\ & \textrm{didapatkan hubungan}\\ &c=\displaystyle \frac{-b}{ad-bc}\quad ...............(1)\\ &b=\displaystyle \frac{-c}{ad-bc}\quad ...............(2)\\ &\textrm{Persamaan}\: \:  (2)\: \: \textrm{disubstitusikan ke persamaan}\: \: (1)\\ &c=\displaystyle \displaystyle \frac{-\displaystyle \frac{-c}{ad-bc}}{ad-bc}\\ &1=(ad-bc)^{2}\\ &(ad-bc)= -1\: \: \textrm{atau}\: \: 1 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 23.&(\textbf{UM UGM 2006})\\ &\textrm{Apabila}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{memenuhi}\\ &\textrm{persamaan matriks}\\ &\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -1\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: x+y=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ A.X&=B\\ A^{-1}.A.X&=A^{-1}.B\\ A^{0}.X&=A^{-1}.B\\ X&=A^{-1}.B\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ -1 & 3 \end{pmatrix}^{-1}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ -1 & 3 \end{vmatrix}}&\begin{pmatrix} 3 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3-2}&\begin{pmatrix} 3.(-1)+2.2 \\ 1.(-1)+1.2 \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1\\ 1 \end{pmatrix}\\ x+y&=1+1=2 \end{aligned} \end{array}$

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

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