Contoh Soal 6 Matriks

$\begin{array}{ll}\\ 26.&(\textbf{UM UGM 2004})\\ &\textrm{Nilai-nilai}\: \: x\: \: \textrm{agar matriks}\\ &\qquad\quad\quad\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\: \: \textrm{atau}\: \: 5\\ \color{red}\textrm{b}.&-2\: \: \textrm{atau}\: \: 2\\ \textrm{c}.&-4\: \: \textrm{atau}\: \: 5\\ \textrm{d}.&-6\: \: \textrm{atau}\: \: 4\\ \textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{supaya matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}\\ &\textrm{tidak memiliki invers},\: \textrm{maka}\\ &\textrm{determinan matriks}\: \: \color{black}\begin{pmatrix} 5x & 5\\ 4 & x \end{pmatrix}=0\\ &\color{red}\textrm{Sehingga}\\ &\begin{vmatrix} 5x & 5\\ 4 & x \end{vmatrix}=0\\ &\Leftrightarrow 5x^{2}-20=0\\ &\Leftrightarrow x^{2}=\color{red}4\\ &\Leftrightarrow x=\color{red}\pm 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&(\textbf{UM UGM 2005})\\ &\textrm{Matriks}\: \: \begin{pmatrix} x & 1\\ -2 & 1-x \end{pmatrix}\\ &\textrm{tidak memiliki invers untuk}\\ &\textrm{nilai}\: \: x=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{atau}\: \: -2\\ \textrm{b}.&-1\: \: \textrm{atau}\: \: 0\\ \textrm{c}.&-1\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{d}.&-1\: \: \textrm{atau}\: \: 2\\ \textrm{e}.&1\: \: \textrm{atau}\: \: 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Mirip dengan pembahasan no. 26}\\ &\color{blue}\begin{aligned}&\textrm{Nilai}\: \: \color{black}\begin{vmatrix} x & 1\\ -2 & 1-x \end{vmatrix}=0\\ &\Leftrightarrow x-x^{2}-(-2)=0\\ &\Leftrightarrow 2+x-x^{2}=0\\ &\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x-2)(x+1)=0\\ &\Leftrightarrow \color{red}x=2\: \: \color{blue}\textrm{atau}\: \: \color{red}x=-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&(\textbf{Mat Das SIMAK UI 2014})\\ &\textrm{Jika matriks}\: \: \textrm{A}\: \: \textrm{adalah invers}\\ &\textrm{dari matriks}\: \: \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\: \: \textrm{dan}\\ &\textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\: \: \textrm{maka nilai}\: \: 2x+y\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{10}{3}\\ \color{red}\textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \textrm{d}.&\displaystyle \frac{9}{7}\\ \textrm{e}.&\displaystyle \frac{20}{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Misalkan diketahui matriks}\\ &\textrm{B}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix},\\ &\textrm{maka}\: \: \textrm{A}=\left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1}\\ &\textrm{selanjutnya}\: \: \textrm{A}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=A^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix},\\ & \textrm{ingat bahwa}\: \: \left (\textbf{A}^{-1} \right )^{-1}=\textbf{A}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\left ( \left ( \displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix} \right )^{-1} \right )^{-1}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}.\begin{pmatrix} -1 & -3\\ 4 & 5 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{3}\begin{pmatrix} -1-9\\ 4+15 \end{pmatrix}=\begin{pmatrix} \displaystyle -\frac{10}{3}\\ \displaystyle \frac{19}{3} \end{pmatrix}\\ &2x+y=2\left ( -\displaystyle \frac{10}{3} \right )+\frac{19}{3}\\ &\qquad\: \: \: \, =\color{red}\displaystyle \frac{-20+19}{3}=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

Contoh Soal 5 Matriks

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

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

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

$\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}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&5\\ \textrm{d}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\begin{pmatrix} 1 & 2\\ 2 & 2 \end{pmatrix}&\color{black}\times A_{2\times 2}=\color{red}\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}\\ \begin{vmatrix} 1 & 2\\ 2 & 2 \end{vmatrix}&\times \color{black}\left | A_{2\times 2} \right |=\color{red}\begin{vmatrix} \alpha & \beta \\ \gamma & \delta \end{vmatrix}\\ &\color{black}\left | A_{2\times 2} \right |=\displaystyle \frac{\color{red}\begin{vmatrix} \alpha & \beta \\ \gamma & \delta \end{vmatrix}}{\color{blue}\begin{vmatrix} 1 & 2\\ 2 & 2 \end{vmatrix}}\\ &\color{black}\left | A_{2\times 2} \right |=\displaystyle \frac{\color{red}(\alpha \delta -\beta \gamma )}{\color{blue}-2}\\ &\color{black}\left | A_{2\times 2} \right |=\displaystyle \frac{\color{red}(\beta \gamma -\alpha \delta )}{\color{blue}2}\\ \textrm{Karena}&\: \: \color{black}\left | A_{2\times 2} \right |\: \: \textrm{bilangan prima}\\ \textrm{akan m}&\textrm{engakibatkan}\: \: \color{red}( \beta \gamma -\alpha \delta)\\ \textrm{harus h}&\textrm{abis dibagi}\: \: \color{red}2,\: \: \color{blue}\textrm{oleh karenanya}\\ \textrm{menyeb}&\textrm{abkan}\: \: \color{red}( \beta \gamma -\alpha \delta)\: \: \color{blue}\textrm{berupa bilangan}\\ \color{blue}\textrm{genap.}\, \, \, &\textrm{Dan karena}\: \: \color{red}( \beta \gamma -\alpha \delta)\: \: \textrm{genap},\\ \textrm{maka p}&\textrm{astilah}\: \: \color{black}\left | A_{2\times 2} \right |\: \: \color{blue}\textrm{juga bernilai genap}\\ \textrm{sehingg}&\textrm{a nilai}\: \: \color{black}\left | A_{2\times 2} \right |\: \: \textrm{pastilah 2} \end{aligned} \end{array}$

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

Contoh Soal 4 Matriks

$\begin{array}{ll}\\ 16.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-17\\ \textrm{b}.&-13\\ \textrm{c}.&11\\ \color{red}\textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Determinan}&\: \textrm{dari matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -5\\ 3 & -1 \end{vmatrix}=2(-1)-3(-5)\\ &=-2+15\\ &=13 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Determinan untuk matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&10\\ \textrm{b}.&18\\ \textrm{c}.&22\\ \textrm{d}.&30\\ \textrm{e}.&36 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Determinan}\: \: \textrm{dari matriks}\\ &\begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}\\ &=\begin{vmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{vmatrix}\\ &=+(2.4.3)+(-1.-1.1)+(-1.1.-2)\\ &\quad -(1.4.-1)-(-2.-1.2)-(3.1.-1)\\ &=24+1+2+4-24+3\\ &=10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika diketahu matriks}\\ &\textrm{A}=\begin{pmatrix} x+3&-2\\ -16&2x-6 \end{pmatrix},\\ &\textrm{maka nilai dari}\: \: \: x\: \: \textrm{supaya matriks}\\ &\textrm{A tidak memiliki invers adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \color{red}\textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{Invers}&\: \textrm{dari matriks A adalah}\: \: \: \textrm{A}^{-1}.\\ \textrm{A}^{-1}&=\displaystyle \frac{1}{det\: \textrm{A}}\times \color{red}Adjoin\: \textrm{A}.\\ \textrm{Karen}&\textrm{a}\: \textrm{tidak memiliki invers},\\ \textrm{maka}\: \, & det\: \textrm{A}=0,\: \textrm{sehingga}\\ det\: \textrm{A}&=\begin{vmatrix} x+3 & -2\\ -16 & 2x-6 \end{vmatrix}=0\\ &\Leftrightarrow (x+3)(2x-6)-(-16.-2)=0\\ &(\color{red}\textrm{masing-masing ruas dibagi 2})\\ &\Leftrightarrow (x+3)(x-3)-16=0\\ &\Leftrightarrow x^{2}-9-16=0\\ &\Leftrightarrow x^{2}-25=0\\ &\Leftrightarrow (x+5)(x-5)=0\\ &\Leftrightarrow x+5=0\quad \textrm{atau}\quad x-5=0\\ &\Leftrightarrow \: \: \: \, \, \color{red}x=-5\quad \textrm{atau}\quad x=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \begin{vmatrix} 5^{2x} & -5\\ 1 & 1 \end{vmatrix}=6.5^{x}\\ &\textrm{maka}\: \: 5^{2x}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&625\: \: \textrm{atau}\: \: 1\\ \color{red}\textrm{b}.&25\: \: \textrm{atau}\: \: 1\\ \textrm{c}.&25\: \: \textrm{atau}\: \: 0\\ \textrm{d}.&5\: \: \textrm{atau}\: \: 1\\ \textrm{e}.&5\: \: \textrm{atau}\: \: 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&5^{2x}+5=6.5^{x}\\ &5^{2x}-6.5^{x}+5=0\\ &\left ( 5^{x}-1 \right )\left ( 5^{x}-5 \right )=0\\ &5^{x}-1=0\: \: \textrm{atau}\: \: 5^{x}-5=0\\ &5^{x}=1\: \: \textrm{atau}\: \: 5^{x}=5\\ &5^{x}=5^{0}\: \: \textrm{atau}\: \: 5^{x}=5^{1}\\ &x=0\: \: \textrm{atau}\: \: x=1\\ &\color{red}\textrm{maka}\\ &5^{2x}=\begin{cases} 5^{2.1} &=5^{2}=25 \\ 5^{2.0} &=5^{0}=1 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahu determinan suatu}\\ &\textrm{matriks adalah}\: \: \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}=0.\\ &\textrm{Jika}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{adalah akar-akar}\\ &\textrm{yang memenuhi persamaan tersebut}\\ &\textrm{maka nilai dari}\: \: \: p+q\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3\\ \textrm{b}.&-\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Diketahui ba}&\textrm{hwa}:\\ \begin{vmatrix} x & 1 & 2\\ x & 1 & x\\ 5 & -3 & 7 \end{vmatrix}&=0\\ +(x.1.7)+&(1.x.5)+(2.x.-3)\\ -(5.1.2)&-(-3.x.x)-(7.x.1)=0\\ 7x+5x-6x&-10+3x^{2}-7x=0\\ 3x^{2}-x-10&=0\begin{cases} p & \textrm{salah satu akar} \\ q & \textrm{salah satu akar yang lain}, \end{cases}\\ \color{red}\textrm{dengan}\: \: \: &\begin{cases} a &=3 \\ b &=-1 \\ c &=-10 \end{cases}.\\ \textrm{maka}\: \: \: p+q\: \: &=-\displaystyle \frac{b}{a}=-\displaystyle \frac{-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$

Lanjutan 4 Materi Matriks (Matematika Wajib Kelas XI)

$\color{blue}\textrm{F. Invers Matriks ordo 2x2}$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\color{blue}\begin{array}{|c|}\hline \begin{aligned}&\color{black}\textrm{Jika matriks}\: \: \color{red}A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\ &\color{black}\textrm{maka determinan matriks}\: \: \color{red}A\\ &\textrm{ditentukan dengan}\\ &det\: \: \color{red}A=\begin{vmatrix} a & b\\ c & d \end{vmatrix}=ad-bc \end{aligned}\\\hline \end{array}$

Jika  $\color{blue}det\: \: \color{red}A$  bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan  $\color{red}A^{\color{black}-1}$  dapat ditentukan dengan formula:

$\LARGE\color{blue}\boxed{\color{red}A^{\color{black}-1}=\color{blue}\frac{1}{\color{red}ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \: \color{red}A^{^{\color{black}-1}}\: \: \color{black}\textrm{adalah}:\\ &\color{red}A^{^{\color{black}-1}}=\displaystyle \frac{1}{\color{black}\begin{vmatrix} \color{red}-3 & 2\\ -1 & \color{red}4 \end{vmatrix}}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\displaystyle \frac{1}{\color{red}12\color{black}-(-2)}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad=\displaystyle \frac{1}{14}\color{blue}\begin{pmatrix} 4 & -2\\ 1 & -3 \end{pmatrix}\\ &\: \qquad =\color{blue}\begin{pmatrix} \displaystyle \frac{4}{14} & \displaystyle \frac{-2}{14}\\ \displaystyle \frac{1}{14} & \displaystyle \frac{-3}{14} \end{pmatrix}\\ &\: \qquad=\color{blue}\begin{pmatrix} \displaystyle \frac{2}{7} & -\displaystyle \frac{1}{7}\\ \displaystyle \frac{1}{14} & -\displaystyle \frac{3}{14} \end{pmatrix} \end{aligned}$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah invers matriks berikut}\\ &\textrm{a}.\quad B=\begin{pmatrix} 5 & -3\\ 4 & -2 \end{pmatrix}\\ &\textrm{b}.\quad C=\begin{pmatrix} -3 & -5\\ 6 & 9 \end{pmatrix}\\ &\textrm{c}.\quad P=\begin{pmatrix} -1 & 2\\ -3 & 6 \end{pmatrix}\\ &\textrm{d}.\quad Q=\begin{pmatrix} 6 & 9\\ 2 & 3 \end{pmatrix}\\\\ &\textrm{Jawab}:\: \color{purple}\textrm{yang dibahas poin a saja}\\ &B^{-1}=\displaystyle \frac{1}{det\: B}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad=\displaystyle \frac{1}{-10-(-12)}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}=\displaystyle \frac{1}{2}\begin{pmatrix} -2 & 3\\ -4 & 5 \end{pmatrix}\\ &\qquad =\color{blue}\begin{pmatrix} -1 & \displaystyle \frac{3}{2}\\ -2 & \displaystyle \frac{5}{2} \end{pmatrix}\\ &\color{purple}\textrm{b. Silahkan dicoba sendiri}\\ &\color{purple}\textrm{c. Silahkan dicoba sendiri}\\ &\color{purple}\textrm{d. Silahkan dicoba sendiri} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui matriks}\\ & E=\begin{pmatrix} 4 & 2\\ 5 & 3 \end{pmatrix}\\ &\textrm{a}.\quad \textrm{Tentukanlah}\\ &\qquad (\textrm{i})\: \: E^{-1}\qquad\qquad\qquad (\textrm{iii})\: \: \left ( E^{-1} \right )^{t}\\ &\qquad (\textrm{i})\: \: E^{t}\qquad\qquad\qquad\: \: \: (\textrm{iv})\: \: \left ( E^{t} \right )^{-1}\\ &\textrm{b}.\quad \textrm{Dengan menggunakan hasil-hasil}\\ &\qquad \textrm{pada a. apakah}\: \: \left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{a}.\quad(\textrm{i})\: \: E^{-1}&=\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -2\\ -5 & 4 \end{pmatrix}=\color{red}\begin{pmatrix} \displaystyle \frac{3}{2} & -1\\ -\displaystyle \frac{5}{2} & 2 \end{pmatrix}\\ (\textrm{ii})\: \: \: \: E^{t}&=\color{purple}\begin{pmatrix} 4 & 5\\ 2 & 3 \end{pmatrix}\\ (\textrm{iii})\: \: \: \quad&\left (E^{-1} \right )^{t}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ (\textrm{iv})\: \: \: \quad&\left (E^{t} \right )^{-1}=\color{purple}\displaystyle \frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \end{pmatrix}=\begin{pmatrix} \displaystyle \frac{3}{2} & -\displaystyle \frac{5}{2}\\ -1 & 2 \end{pmatrix}\\ \textrm{b}.\quad \textrm{Dari ha}&\textrm{sil yang didapat dapat disimpulkan}\\ &\color{red}\left ( E^{-1} \right )^{t}=\left ( E^{t} \right )^{-1} \end{aligned} \end{array}$

DAFTAR PUSTAKA

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

Lanjutan 3 Materi Matriks (Matematika Wajib Kelas XI)

$\color{blue}\textrm{E. Determinan Matriks}$

$\color{red}\textrm{1. Ordo 2x2}$

Misalkan A adalah matriks persegi berordo 2x2 dan dituliskan dengan  $A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$  dengan  $\color{red}a_{11}\: \: \color{purple}\textrm{dan}\: \: \color{red}a_{22}$ sebagai elemen dari diagonal utama dan $\color{blue}a_{12}\: \: \color{purple}\textrm{dan}\: \: \color{blue}a_{21}$ adalah elemen yang menempati diagonal samping, perhatikan lagi matriks A berikut:

$A=\begin{pmatrix} \color{red}a_{11} & \color{blue}a_{12}\\ \color{blue}a_{21} & \color{red}a_{22} \end{pmatrix}$

maka determinan dari matriks A yang berordo 2x2 adalah perkalian elemen pada diagonal utama dikurangi dengan hasil kali perkalian diagonal samping dan di tuliskan dengan det A atau tanda |...|. Sehingga dari pengertian tersebut kita dapat menuliskan  bahwa determinan dari matriks A dalah:

$\textrm{det}.\: A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$ sama dengan

$\color{blue}\textrm{det}.\: A=\begin{vmatrix} \color{red}a_{11} & \color{black}a_{12}\\ \color{black}a_{21} & \color{red}a_{22} \end{vmatrix}=\color{red}a_{11}\times a_{22}-\color{blue}a_{12}\times a_{21}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 2x2,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} -3 & 2\\ -1 & 4 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: A\: \: \color{black}\textrm{adalah}:\\ &\color{blue}det\: A=\color{black}\begin{vmatrix} \color{red}-3 & 2\\ -1 & \color{red}4 \end{vmatrix}=\color{red}(-3)\times (-4)\color{black}-(-1)\times (2)\\ &\: \qquad =\color{red}12\color{black}-(-2)=\color{red}12\color{black}+2=\color{blue}14 \end{aligned}$

$\color{red}\textrm{2. Ordo 3x3}$

Ada dua buah cara minimal dalam menentukan determinan matriks ordo 3x3, yaitu:

• cara menjabarkan mengikuti baris atau kolom(ekspansi kofaktor)
• aturan Sarrus

Adapun penjelasan lebih lanjut adalah sebagai berikut

$\begin{aligned}&\color{black}\textrm{Misalkan diberikan matriks ordo}\: 3x3\\ &\color{blue}A=\color{black}\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}\\ \end{aligned}$

$\color{red}\textrm{2.1 Menjabarkan mengikuti baris atau kolom}$

$\begin{aligned}\textrm{det A}&=a_{11}\begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33} \end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33} \end{vmatrix}+a_{13}\begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a_{32} \end{vmatrix}\\ &\\ &\textbf{Catatan}:\\ &\textrm{tanda}\: a_{ij}=\color{blue}\textrm{positif jika}\: i+j\: \textrm{genap}\\ &\textrm{tanda}\: a_{ij}=\color{red}\textrm{negatif jika}\: i+j\: \textrm{ganjil} \end{aligned}$

Anda juga bisa menjabarkan mengikuti baris yang lain termasuk juga menjabarkan mengikuti kolom. Sehingga total cara menjabarkan ini, karena ada 3 baris dan 3 kolom total akan ada sebanyak 6 cara menentukan determinan dari matriks A tersebut.

$\color{red}\textrm{2.2 Aturan Sarrus}$

$\begin{aligned}\textrm{det A}&=\color{blue}a_{11}.a_{22}.a_{33}\\ &\quad\: \color{blue}+a_{12}.a_{23}.a_{31}\\ &\quad\: \color{blue}+a_{13}.a_{21}.a_{32}\\ &\quad \color{red}-a_{31}a_{22}.a_{13}\\ &\quad \color{red}-a_{32}.a_{23}.a_{11}\\ &\quad \color{red}-a_{33}.a_{21}.a_{12} \end{aligned}$

$\begin{aligned}&\textbf{Sebagai}\: \: \color{blue}\textrm{CONTOH MENJABARKAN}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &\color{red}A=\color{black}\begin{pmatrix} 1 & 2&3\\ 1 &3& 4\\ 1&4&3 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: A\: \: \color{black}\textrm{dengan}\\ &\textrm{menjabarkan baris pertama adalah}:\\ &\color{blue}det\: A=1\begin{vmatrix} 3 & 4\\ 4 & 3 \end{vmatrix}-2\begin{vmatrix} 1 & 4\\ 1 & 3 \end{vmatrix}+3\begin{vmatrix} 1 & 3\\ 1 & 4 \end{vmatrix}\\ &\: \qquad=(9-16)-2(3-4)+3(4-3)\\ &\: \qquad=-7+2+3\\ &\: \qquad=\color{red}-2 \end{aligned}$

$\begin{aligned}&\textbf{Dan berikut}\: \: \color{blue}\textrm{CONTOH aturan SARRUS}\\ &\textrm{Diketahui sebuah matrik ordo}\: \: 3x3,\: \: \textrm{yaitu}:\\ &\color{red}B=\color{black}\begin{pmatrix} 2 & 1&3\\ 3 &1& 4\\ 4&1&3 \end{pmatrix},\: \: \textrm{maka}\: \: \color{blue}det\: B\: \: \color{black}\textrm{dengan}\\ &\textrm{metode SARRUS adalah}:\\ &\color{blue}det\: B=(2.1.3)+(1.4.4)+(3.1.3)\\ &\qquad\: \: \: \: \: \: \: -(4.1.3)-(1.4.2)-(3.1.3)\\ &\qquad\: \, =6+16+9-12-8-9=\color{red}2 \end{aligned}$

$\LARGE\color{purple}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui matriks-matriks persegi berikut}\\ &\textrm{a}.\: \: \begin{pmatrix} 2 & 3\\ 6 & 7 \end{pmatrix}\qquad\qquad \textrm{c}.\: \: \begin{pmatrix} -2 & -3\\ 6 & 7 \end{pmatrix}\\ &\textrm{b}.\: \: \begin{pmatrix} 0 & 4\\ -3 & 6 \end{pmatrix}\: \quad\quad\quad \textrm{d}.\: \: \begin{pmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{pmatrix}\\ &\\ &\textrm{Tentukanlah determinan dari}\\ &\textrm{matriks-matriks persegi di atas}\\\\ &\color{black}\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\color{red}\begin{vmatrix} 2 & 3\\ 6 & 7 \end{vmatrix}\\ &=(2).(7)-(3).(6)\\ &=14-18\\ &=-4\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad &\color{blue}\begin{vmatrix} 0 & 4\\ -3 & 6 \end{vmatrix}\\ &=(0).(6)-(4).(-3)\\ &=0-(-12)\\ &=12\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad &\color{red}\begin{vmatrix} -2 & -3\\ 6 & 7 \end{vmatrix}\\ &=(-2).(7)-(-3).(6)\\ &=(-14)-(-18)\\ &=-14+18\\ &=4 \end{aligned}&\begin{aligned}\textrm{d}.\quad &\color{blue}\begin{vmatrix} \sqrt{3} & 3\sqrt{3}\\ \sqrt{2} & -2\sqrt{2} \end{vmatrix}\\ &=(\sqrt{3}).(-2\sqrt{2})\\ &\quad-(3\sqrt{3}).(\sqrt{2})\\ &=-2\sqrt{6}-3\sqrt{6}\\ &=-5\sqrt{6}\\ & \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}=2\\\\ &\color{black}\textrm{Jawab}:\\ &\begin{aligned}\color{blue}\begin{vmatrix} 1-x & 3\\ 2 & 3-x \end{vmatrix}&=\color{red}2\\ \left ( 1-x \right )\left ( 3-x \right )-(3)(2)&=\color{red}2\\ 3-x-3x+x^{2}-6&=\color{red}2\\ x^{2}-4x-3&=\color{red}2\\ x^{2}-4x-5&=\color{red}0\\ \left ( x-5 \right )\left ( x+1 \right )&=0\\ x-5=0\: \: \textrm{atau}\: \:x+1&=0\\ \color{purple}x=5\: \: \textrm{atau}\: \: x=-1& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui matriks-matriks persegi berikut}\\ &(\textrm{i}).\: \: \begin{pmatrix} 1 & 2&3\\ 2 & 4&5\\ 3&5&4 \end{pmatrix}\quad\quad\quad\quad\: \: \, (\textrm{iii}).\: \: \begin{pmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{pmatrix}\\ &(\textrm{ii}).\: \: \begin{pmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{pmatrix}\quad\quad (\textrm{iv}).\: \: \begin{pmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{pmatrix}\\ &\\ &\textrm{Tentukanlah determinan matriks-matriks}\\ &\textrm{di atas dengan cara}\\ &\textrm{a}.\quad Sarrus\\ &\textrm{b}.\quad \textrm{Menjabarkan baris pertama}\\ &\textrm{c}.\quad \textrm{Menjabarkan baris kedua}\\ &\textrm{d}.\quad \textrm{Menjabarkan baris ketiga}\\ &\textrm{e}.\quad \textrm{Menjabarkan kolom pertama}\\ &\textrm{f}.\quad \textrm{Menjabarkan kolom kedua}\\ &\textrm{g}.\quad \textrm{Menjabarkan kolom ketiga} \end{array}$

$.\qquad\:  \begin{aligned}&\color{purple}\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline (\textrm{i}).\quad \begin{pmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{pmatrix}&(\textrm{ii}).\quad \begin{pmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{pmatrix}\\\hline (\textrm{iii}).\quad \begin{pmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{pmatrix}&(\textrm{iv}).\quad \begin{pmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{pmatrix}\\\hline \begin{aligned}(\textrm{i}).\quad &\begin{vmatrix} 1 & 1&3\\ 2 & 4&5\\ 3&5&4 \end{vmatrix}\\ &=(1)(4)(4)+\\ &\: \: \quad (1)(5)(3)+\\ &\: \: \quad (3)(2)(5)+\\ &\: \: \quad -(3)(4)(3)\\ &\: \: \quad -(5)(5)(1)\\ &\: \: \quad -(4)(2)(1)\\ &=16+15+30\\ &\: \: \: -36-25-8\\ &=-8 \\\end{aligned} &\begin{aligned}(\textrm{ii}).\quad &\begin{vmatrix} -1 & -2&-3\\ -2 & 6&0\\ -3&0&6 \end{vmatrix}\\ &=(-1)(6)(6)+\\ &\: \: \quad (-2)(0)(-3)+\\ &\: \: \quad (-3)(-2)(0)+\\ &\: \: \quad -(-3)(6)(-3)\\ &\: \: \quad -(0)(0)(-1)\\ &\: \: \quad -(6)(-2)(-2)\\ &=-36+0+0\\ &\: \: \: -54-0-24\\ &=-114 \\\end{aligned} \\\hline \begin{aligned}(\textrm{iii}).\quad &\begin{vmatrix} 1 & 2&3\\ 4 & 5&6\\ 7&8&9 \end{vmatrix}\\ &=(1)(5)(9)+\\ &\: \: \quad (2)(6)(7)+\\ &\: \: \quad (3)(4)(8)+\\ &\: \: \quad -(7)(5)(3)\\ &\: \: \quad -(8)(6)(1)\\ &\: \: \quad -(9)(4)(2)\\ &=45+84+96\\ &\: \: \: -105-48-72\\ &=0 \\\end{aligned} &\begin{aligned}(\textrm{iv}).\quad &\begin{vmatrix} 2 & 1&1\\ 1 & 2&1\\ 1&1&2 \end{vmatrix}\\ &=(2)(2)(2)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad (1)(1)(1)+\\ &\: \: \quad -(1)(2)(1)\\ &\: \: \quad -(1)(1)(2)\\ &\: \: \quad -(2)(1)(1)\\ &=8+1+1\\ &\: \: \: -2-2-2\\ &=4 \\\end{aligned} \\\hline \end{array}\\ &\color{purple}\textrm{yang belum dibahas silahkan dibuat latihan}  \end{aligned}$

DAFTAR PUSTAKA

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

Contoh Soal 3 Matriks

$\begin{array}{ll}\\ 11.&\textrm{Diketahui matriks}\\ &\textrm{A}=\begin{pmatrix} ^{a}\log 6 & 2\\ 1 & a+3b \end{pmatrix},\\ & \textrm{B}=\begin{pmatrix} 0 &-5 \\ -6 & 3a-5b \end{pmatrix},\: \: \textrm{dan}\\ & \textrm{C}=\begin{pmatrix} ^{a}\log 2 & -\displaystyle \frac{1}{2}\\ -2(b+c) & 3 \end{pmatrix},\\ &\textrm{serta}\: \: \textrm{I}\: \: \textrm{adalah matriks identitas}.\\ &\textrm{Jika}\: \: 2\textrm{A}+\textrm{B}-2\textrm{C}=2\textrm{I},\\ &\textrm{maka nilai}\: \: 4a+b+c\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&5\\ \textrm{c}.&7\\ \textrm{d}.&11\\ \color{red}\textrm{e}.&13 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&2\textrm{A}+\textrm{B}-2\textrm{C}=2\textrm{I}\\ &2\begin{pmatrix} ^{a}\log 6 & 2\\ 1 & a+3b \end{pmatrix}+\begin{pmatrix} 0 &-5 \\ -6 & 3a-5b \end{pmatrix}\\ &-2\begin{pmatrix} ^{a}\log 2 & -\displaystyle \frac{1}{2}\\ -2(b+c) & 3 \end{pmatrix}=2\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\\ &\begin{pmatrix} \color{black}2.\: ^{a}\log 6-2.\: ^{2}\log 2 & 2.2-5-2\left ( -\displaystyle \frac{1}{2} \right )\\ \color{red}2.1 -6-2(-2(b+c))& \color{purple}2(a+3b)+3a-5b-2.3 \end{pmatrix}=\begin{pmatrix} \color{black}2 & 0\\ \color{red}0 & \color{purple}2 \end{pmatrix}\\ &\begin{cases} \color{black}2 &=\color{black}2.\: ^{a}\log 6-2.\: ^{2}\log 2 \\ \color{red}0 & =\color{red}2.1 -6-2(-2(b+c)) \\ \color{purple}2 & =\color{purple}2(a+3b)+3a-5b-2.3 \end{cases}\\ &\begin{array}{|c|c|}\hline \textrm{dari persamaan}\: \: (1)&\textrm{dari persamaan}\: \: (2)\\\hline \color{red}\begin{aligned}2.\: ^{a}\log 6-2.\: ^{2}\log 2&=2\\ ^{a}\log 6^{2}-\: ^{2}\log 2^{2}&=2\\ ^{a}\log \displaystyle \frac{6^{2}}{2^{2}}&=2\\ ^{a}\log 9&=2\\ 9&=a^{2}\\ 3&=a\\ 12&=4a \end{aligned}&\color{purple}\begin{aligned}2.1 -6-2(-2(b+c))&=0\\ 2-6+4(b+c)&=0\\ 4(b+c)&=4\\ b+c&=1\\ &\\ \textrm{sehingga diperoleh}&,\\ 4a+b+c=12+1&\\ =13\: \: \: \quad& \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: \begin{pmatrix} -4x & 2y\\ y & x \end{pmatrix}\begin{pmatrix} 2\\ -3 \end{pmatrix}=\begin{pmatrix} 2\\ -12 \end{pmatrix},\\ & \textrm{maka nilai}\: \: xy=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&-6\\ \textrm{b}.&-3\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{array}{|c|c|}\hline \begin{aligned}\begin{pmatrix} -4x & 2y\\ y & x \end{pmatrix}\begin{pmatrix} 2\\ -3 \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\ \begin{pmatrix} -4x.2+2y.-3\\ y.2+x.-3 \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\ \begin{pmatrix} -8x-6y\\ -3x+2y \end{pmatrix}&=\begin{pmatrix} 2\\ -12 \end{pmatrix}\\\\ \textbf{SPLDV}& \end{aligned} &\color{red}\begin{aligned}-8x-6y&=2\: \qquad (\times 1)\\ -3x+2y&=-12\quad (\times 3)\\ \textrm{menjadi}&\\ -8x-6y&=2\\ -9x+6y&=-36\quad _{+}\\ ----&---\\ -17x&=-34\\ x&=2 \end{aligned}\\\hline \color{black}\begin{aligned}-8x-6y&=2\\ -8(2)-6y&=2\\ -16-6y&=2\\ -6y&=2+16\\ -6y&=18\\ y&=-3\\ \textrm{sehingga}&\\ xy&=2.(-3)=-6 \end{aligned}&\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Diketahui}\: \: \textrm{N}=\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}\\ & \textrm{dan}\: \: \textrm{M}=\begin{pmatrix} -1&3\\ -1&5 \end{pmatrix}.\\ &\textrm{Jika}\: \: \textrm{N}^{2}=p\textrm{N}-q\textrm{M},\\ &: \textrm{maka nilai}\: \: p-q=....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{N}^{2}=p\textrm{N}-q\textrm{M}\\ &\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}\times \begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}=p\begin{pmatrix} -2&3\\ -1 & 4 \end{pmatrix}-q\begin{pmatrix} -1&3\\ -1&5 \end{pmatrix}\\ &\begin{pmatrix} -2.-2+3.-1 & -2.3+3.4\\ -1.-2+4.-1 & -1.3+4.4 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\ &\begin{pmatrix} 4-3 & -6+12\\ 2-4 & -3+16 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\ &\begin{pmatrix} 1 & 6\\ -2 & 13 \end{pmatrix}=\begin{pmatrix} -2p+q & 3p-3q\\ -p+q & 4p-5q \end{pmatrix}\\\\ &\begin{array}{|c|c|}\hline \color{purple}\begin{aligned}-2p+q&=1\\ -p+q&=-2\quad _{-}\\ ----&---\\ -p\qquad&=3\\ p&=-3\\ &\\ & \end{aligned}&\color{red}\begin{aligned}-p+q&=-2\\ -(-3)+q&=-2\\ q&=-2-3\\ q&=-5\\ \textrm{sehingga}&\: \textrm{didapatkan}\\ p-q&=-3-(-5)\\ &=2 \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Diketahui matriks}\: \: \textrm{Z}=\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\\ & \textrm{dan}\: \: f(x)=x^{2}-x.\\ &\textrm{Jika}\: \: f(\textrm{Z})=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix},\\ & \textrm{maka nilai}\: \: p^{2}-q^{2}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&5\\ \color{red}\textrm{b}.&7\\ \textrm{c}.&9\\ \textrm{d}.&12\\ \textrm{e}.&15 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}f(\color{red}\textrm{Z})&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \color{red}\textrm{Z}^{2}-\textrm{Z}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}\times \begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}-\begin{pmatrix} -2&6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} 4-18 & -12+30\\ 6-15 & -18+25 \end{pmatrix}-\begin{pmatrix} -2 & 6\\ -3 & 5 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \begin{pmatrix} -12 & 12\\ -6 & 2 \end{pmatrix}&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \end{aligned} \\ &\color{blue}\begin{aligned} \color{black}\textrm{Sehingga}&\\ -12&=-3p-8q\quad.................(1)\\ -1&=p+q\quad......................(2)\\ \textrm{persamaan}&\: (2)\: \: \textrm{ke persamaan}\: \: (1)\\ -12&=-3p-3q-5q=-3(p+q)-5q\\ -12&=-3(-1)-5q\\ -12&=3-5q\\ 5q&=3+12\\ q&=3\quad........................(3)\\ \textrm{persamaan}&\: \: (3)\: \: \textrm{ke persamaan}\: \: (2)\\ \color{red}p+q&=-1\\ p&=-1-q=-1-3=-4\\ \color{red}p^{2}-q^{2}&=(-4)^{2}-3^{2}=16-9\\ &=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&(\textbf{SBMPTN 2013})\\ &\textrm{Jika}\: \: A=\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix},\\ &B=\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}\: \: \textrm{dan}\\ &AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\textrm{maka nilai}\: \: 2c-a=\: ....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{purple}\begin{aligned}&AB=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} 2 & -1 & 1\\ a & b & c \end{pmatrix}\begin{pmatrix} -2 & 1\\ 1 & -1\\ 0 & 2 \end{pmatrix}=\begin{pmatrix} -5 & 5\\ 3 & -3 \end{pmatrix}\\ &\begin{pmatrix} -5 & 5\\ \color{black}-2a+b&\color{black}a-b+2c \end{pmatrix}=\begin{pmatrix} -5 & 5\\ \color{red}3 & \color{red}-3 \end{pmatrix}\\ &\begin{array}{lllll}\\ -2a+b&=3&\\ a-b+2c&=-3&+\\\hline \qquad \color{red}2c-a&=0 \end{array} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA
2. Kanginan, M., Terzalgi, Z. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU.
3. Sharma, S. N. 2017. Jelajah Matematika 2 SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
4. Suparmin, S. Malau, A. 2014. Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok IPA. Bandung: YRAMA WIDYA.

Contoh Soal 2 Matriks

$\begin{array}{ll}\\ 6.&\textrm{Diketahui matriks}\\ &\textrm{M}=\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{N}=\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}.\\ &\textrm{Nilai}\: \: k\: \: \textrm{yang memenuhi jika}\\ &\textrm{M}=k\textrm{N}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&-3\\ \textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahu bahwa}\\ &\textrm{M}=k\textrm{N}\\ &(\color{red}\textrm{perkalian suatu matrik dengan skalar})\\ &\begin{pmatrix} -6 & 9&-15\\ 3 & -6&12 \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-3 & \color{red}-3.\color{black}5\\ \color{red}-3.\color{black}-1 & \color{red}-3.\color{black}2 & \color{red}-3.\color{black}-4 \end{pmatrix}\\ &=\color{red}-3\color{black}\begin{pmatrix} 2 & -3 & 5\\ -1 & 2 & -4 \end{pmatrix}\\ &=k\begin{pmatrix} 2 & -3&5\\ -1 & 2&-4 \end{pmatrix}\\ &\textrm{sehingga dari kesamaan tersebut}\\ &\textrm{maka}\quad \color{red}k=-3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Hasil dari}\: \: \begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}\\ & \textrm{adalah}...\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} 22 & 28\\ 49&64 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 22&49\\ 28&64 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 64&28\\ 49&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 2 & 8&18\\ 4&15 & 30 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&4&6\\ 4&15&30 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\begin{pmatrix} 1&2&3\\ 4&5&6 \end{pmatrix}_{\color{red}2\times \color{black}3}\times \begin{pmatrix} 1 &2 \\ 3 &4 \\ 5 & 6 \end{pmatrix}_{\color{black}3\times \color{red}2}\\ &=\begin{pmatrix} 1.1+2.3+3.5 & 1.2+2.4+3.6\\ 4.1+5.3+6.5 &4.2+5.4+6.6 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 1+6+15 & 2+8+18\\ 4+15+30 & 8+20+36 \end{pmatrix}_{\color{red}2\times 2}\\ &=\begin{pmatrix} 22 & 28\\ 49 & 64 \end{pmatrix}_{\color{red}2\times 2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika diketahui matriks}\\ & \textrm{A}=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}.\\ &\textrm{maka hasil dari}\: \: \textrm{A}^{3}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 5 & 8\\ 20&22 \end{pmatrix}\\ \color{red}\textrm{b}.&\begin{pmatrix} 6&7\\ 21&20 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 6&7\\ 20&22 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 7 & 8\\ 20 & 23 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 7&9\\ 20&23 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{A}&=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ \textrm{A}^{2}&=\textrm{A}\times \textrm{A}\\ &=\begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\times \begin{pmatrix} 0&1\\ 3&2 \end{pmatrix}\\ &=\begin{pmatrix} 0+3&0+2\\ 0+6&3+4 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} 3 & 2\\ 6 & 7 \end{pmatrix}\\ \textrm{A}^{3}&=\textrm{A}^{2}\times \textrm{A}\\ &=\begin{pmatrix} 3 & 2\\ 6 &7 \end{pmatrix}\times \begin{pmatrix} 0 & 1\\ 3 & 2 \end{pmatrix}\\ &=\begin{pmatrix} 0+6 & 3+4\\ 0+21 & 6+14 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 6 & 7\\ 21 & 20 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&(\textbf{SBMPTN Mat IPA 2014})\\ &\textrm{Jika}\: \: \textrm{A}\: \: \textrm{adalah matriks yang berordo}\\ & 2\times 2\: \: \textrm{dan memenuhi}\\ &\: \: \begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}=x^{2}-5x+8,\\ & \textrm{maka matriks A yang mungkin adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} 1 & -5\\ 8&0 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1&5\\ 8&0 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1&8\\ -5&0 \end{pmatrix}\\ \color{red}\textrm{d}.&\begin{pmatrix} 1 & 3\\ -8&8 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 1&-3\\ 8&-8 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\begin{pmatrix} x & 1 \end{pmatrix}\times \textrm{A}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x & 1 \end{pmatrix}\times \begin{pmatrix} p & q\\ r & s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} xp+r & xq+s \end{pmatrix}\times \begin{pmatrix} x\\ 1 \end{pmatrix}&=\color{red}x^{2}-5x+8\\ \begin{pmatrix} x^{2}p+xr+xq+s \end{pmatrix}&=\color{red}x^{2}-5x+8\\ px^{2}+(q+r)x+s&=\color{red}x^{2}-5x+8\\ \end{aligned}\\ &\color{blue}\begin{aligned}&\begin{cases} \color{red}p &=1 \\ q+r &=-5 \\ \color{red}s &=8 \end{cases}\quad\Rightarrow\quad \begin{pmatrix} 1 & ...\\ ... & 8 \end{pmatrix}\\ &\textrm{Sehingga yang paling mungkin}\\ & \textrm{adalah}\: \: \color{red}\begin{pmatrix} 1 & 3\\ -8 & 8 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Diketahui}\\ &\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&6\\ \color{red}\textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\begin{pmatrix} ^{x}\log a & \log (2a-6)\\ \log (b-2) & 1 \end{pmatrix}=\begin{pmatrix} \log b & 1\\ \log a & 1 \end{pmatrix}\\ &\color{black}\textrm{maka}\\ &\begin{cases} ^{x}\log a & =\log b \quad.........\color{red}(1)\\ \log (2a-6) &=1\quad..............\color{red}(2) \\ \log (b-2) &=\log a\quad.........\color{red}(3) \end{cases}\\ &\textrm{Sehingga}\: \textrm{dari persamaan}\: \: (2)\\ &\color{black}\textrm{akan didapatkan}\\ &\log (2a-6)=1=\log 10\\ &(2a-6)=10\\ &a=8\quad...........................(4)\\ &\textrm{persamaan}\: (4)\: \: \textrm{ke persamaan}\: \: (3),\\ & \color{black}\textrm{maka}\\ &\log (b-2) =\log a\\ &b-2=a=8\\ &b=10\quad.................................(5)\\ &\textrm{Selanjutnya dari persamaan}\: \: (5)\\ &\color{black}\textrm{akan diperoleh}\\ &^{x}\log a =\log b\\ &^{x}\log 8 =\log 10=1\\ &\qquad x^{1}=8\\ &\Leftrightarrow \: \: \color{red}x=8 \end{aligned} \end{array}$

Contoh Soal 1 Matriks

$\begin{array}{l}\\ 1.&\textrm{Diketahui matriks}\\ &\textrm{A}=\begin{pmatrix} 2020 & -4&-3&2\\ 2020 & -6&-7&1\\ 2020&4&-3&0\\ 2020&6&-7&8 \end{pmatrix}\\ &\textrm{Ordo dari matriks}\: \: \textrm{A}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\times 2&&&\\ \textrm{b}.&3\times 3\\ \textrm{c}.&3\times 4\\ \textrm{d}.&4\times 3\\ \color{red}\textrm{e}.&4\times 4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\textrm{Cukup jelas}\\ &\color{blue}\textrm{Karena matriknya mengandung}\\ &\color{red}\textrm{4 baris}\color{blue}\times \color{red}\textrm{4 kolom} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui matriks}\\ &\textrm{B}=\begin{pmatrix} 1 & 2&3&2020\\ 5 & 13&7&2019\\ 11&14&3&-2018\\ -15&6&17&2017 \end{pmatrix}\\ &\textrm{Jika}\: \: \textrm{b}_{ij}\: \: \textrm{menunjukkan elemen}\\ &\textrm{yang terletak pada baris ke}-i\\ &\textrm{dan kolom ke}-j\: \: \textrm{pada matriks B}\\ &\textrm{ di atas, maka}\: \: b_{43}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\\ \textrm{b}.&9\\ \textrm{c}.&-1\\ \textrm{d}.&3\\ \color{red}\textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{Perh}&\textrm{atikan bahwa}\\ \color{black}\textrm{B}_{4\times 4}&=\begin{pmatrix} b_{11} & b_{12} & b_{13} & b_{14}\\ b_{21} & b_{22} & b_{23} & b_{24}\\ b_{31} & b_{32} & b_{33} & b_{34}\\ b_{41} & b_{42} & \color{red}b_{43} & b_{44} \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2&3&2020\\ 5 & 13&7&2019\\ 11&14&3&-2018\\ -15&6&\color{red}17&2017 \end{pmatrix}\\ \color{black}\textrm{sehi}&\color{black}\textrm{ngga entri}\: \: \color{red}b_{43}=17 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui matriks}\: \: \textrm{C}\: \: \textrm{adalah matriks}\\ &\textrm{berordo}\: \: 3\times 3.\: \: \textrm{Jika}\: \: \textrm{c}_{ij}=4j-5i,\\ &\textrm{maka matriks C tersebut adalah}....\\ &\begin{array}{llllllll}\\ \color{red}\textrm{a}.&\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} -1 & 7 & 3\\ -6 & 2 & -2\\ -7 & -11 & -3 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} -1 & -7 & -11\\ -6 & 7 & 3\\ -2 & 2 & -3 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} -1 &-6 & -11\\ 3 & -2 & 2\\ 7 & 2 & -3 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} -1 & -2 & -3\\ 3 & -6 & -11\\ 7 & -7 & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\: \: \color{red}c_{ij}=4j-5i,\\ \textrm{mak}&\textrm{a}\\ \textrm{C}_{3\times 3}&=\begin{pmatrix} c_{\color{red}11} & c_{\color{red}12} & c_{\color{red}13} \\ c_{\color{red}21} & c_{\color{red}22} & c_{\color{red}23} \\ c_{\color{red}31} & c_{\color{red}32} & c_{\color{red}33} \end{pmatrix}\\ &=\begin{pmatrix} 4.1-5.1 & 4.2-5.1&4.3-5.1\\ 4.1-5.2 & 4.2-5.2&4.3-5.2\\ 4.1-5.3&4.2-5.3&4.3-5.3 \end{pmatrix}\\ &=\begin{pmatrix} 4-5 & 8-5 & 12-5\\ 4-10 & 8-10 & 12-10\\ 4-15 & 8-15 & 12-15 \end{pmatrix}\\ &=\begin{pmatrix} -1 & 3 & 7\\ -6 & -2 & 2\\ -11 & -7 & -3 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Jika diketahui matriks}\\ &\textrm{X}=\begin{pmatrix} -7 & 9&-1\\ 4 & -6&15 \end{pmatrix}.\\ &\textrm{maka transpose matriks}\: \: \textrm{X}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -6 & 15\\ -7 & 9 & -1 \end{pmatrix}\\ \textrm{b}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -6 & 4\\ -1 & 9 & -7 \end{pmatrix}\\ \color{red}\textrm{c}.&\textrm{X}^{t}=\begin{pmatrix} -7 & 4\\ 9 & -6\\ -1 & 15 \end{pmatrix}\\ \textrm{d}.&\textrm{X}^{t}=\begin{pmatrix} 4 & -7\\ -6 & 9\\ 15 & -1 \end{pmatrix}\\ \textrm{e}.&\textrm{X}^{t}=\begin{pmatrix} 15 & -1\\ -6 & 9\\ 4 & -7 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \textrm{X}_{2\times 3}&=\begin{pmatrix} x_{\color{red}11} & x_{\color{red}12} & x_{\color{red}13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}\\ &=\begin{pmatrix} \color{red}-7 & \color{red}9&\color{red}-1\\ 4 & -6&15 \end{pmatrix}\\ \color{black}\textrm{maka}&\\ \textrm{X}_{3\times 2}^{t}&=\begin{pmatrix} x_{\color{red}11} & x_{21}\\ x_{\color{red}12} & x_{22}\\ x_{\color{red}13} & x_{23} \end{pmatrix}=\begin{pmatrix} \color{red}-7 & 4\\ \color{red}9 & -6\\ \color{red}-1 & 15 \end{pmatrix} \\ \textrm{adal}&\textrm{ah sebuah}\: \color{red}\textrm{matriks baru} \\ \textrm{deng}&\textrm{an ordo}\: \: \color{red}3\times 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui matriks}\: \: \textrm{P}=\begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}\\ &\textrm{dan}\: \: \textrm{Q}=\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}.\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi jika}\\ & \textrm{P}=2\textrm{Q}^{t}\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-2\\ \textrm{b}.&3\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{P}&=2\textrm{Q}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & 2a+1\\ a & b+7 \end{pmatrix}^{\color{red}t}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=2\begin{pmatrix} 2c-3b & a\\ 2a+1 & b+7 \end{pmatrix}\\ \begin{pmatrix} a & 4\\ 2b & 3c \end{pmatrix}&=\begin{pmatrix} 4c-6b & 2a\\ 4a+2 & 2b+14 \end{pmatrix}\\ &(\color{red}\textrm{kesamaan 2 buah matriks})\\ \color{black}\textrm{akibat}&\color{black}\textrm{nya}\\ &\begin{cases} a &= 4c-6b \quad ..................(1)\\ 4 &=2a \quad ........................(2)\\ 2b &=4a+2 \quad ......................(3)\\ 3c &=2b+14 \quad ......................(4) \end{cases}\\ \textrm{dari}&\: \textrm{persamaan}\: \: (2)\\ & 2a=4\Rightarrow a=2\quad....(5)\\ \textrm{pers}&\textrm{amaan}\: \: (5)\: \: \textrm{hasilnya}\\ \textrm{disu}&\textrm{bstitusikan ke persamaan}\: \: (3),\\ \color{black}\textrm{yait}&\color{black}\textrm{u}\\ 2b&=4a+2\Rightarrow 2b=4(2)+2=10\\ b&=5\quad.....................(6)\\ \textrm{pers}&\textrm{amaan}\: \: (6)\: \: \textrm{hasilnya disbstitusikan}\\ \textrm{ke p}&\textrm{ersamaan}\\ (4),&\: \textrm{dan akan mendapatkan}\\ 3c&=2b+14\Rightarrow 3c=2(5)+14=24\\ \color{red}c&\color{red}=8 \end{aligned} \end{array}$

Lanjutan 2 Materi Matriks (Matematika Wajib Kelas XI)

 $\color{blue}\textrm{C. Tarnspose dan Kesamaan Dua Buah Matriks}$

$\color{purple}\begin{array}{|c|l|}\hline 1.&\color{blue}\textrm{Transpose Matriks}\\\hline &\begin{aligned}&\textrm{Membentuk matriks baru dari matriks}\\ &\textrm{dengan cara mengubah baris matriks ke}-i\\ &\textrm{menjadi kolom ke}-i,\: \textrm{pada matriks baru}\\ &\textrm{dan demikian pula untuk kolomnya}.\: \textrm{Jika}\\ &\textrm{matriks pertama adalah A maka matriks}\\ &\textrm{transposenya adalah}\: \: \textrm{A}'\: \: \textrm{atau}\: \: \textrm{A}^{t} \end{aligned}\\\hline 2.&\color{red}\textrm{Kesamaan Duan Buah Matriks}\\\hline &\begin{aligned}&\textrm{Misalkan matriks}\: \: \textrm{A}=\left ( a_{ij} \right )\: \: \textrm{dan}\: \: \textrm{B}=\left ( b_{ij} \right )\\ &\textrm{adalah dua buah matriks berordo sama},\\ &\textrm{maka matriks A dikatakan sama dengan matriks B}\\ &\textrm{jika elemen-elemen yang seletak sama pada}\\ &\textrm{kedua matriks tersebut bernilai sama}\\ & \end{aligned}\\\hline \end{array}$

$\begin{array}{|l|}\hline \color{blue}\textrm{Berikut contoh transpose}\\ \textrm{A}=\displaystyle \begin{pmatrix} 1 & 2\\ -7 & 0\\ 5 & 4 \end{pmatrix}\Rightarrow \textrm{A}^{t}=\begin{pmatrix} 1 & -7 & 5\\ 2 & 0 & 4 \end{pmatrix}\\\hline \color{blue}\textrm{DAn berikut contoh kesamaan dua matriks}\\ \textrm{A}=\displaystyle \begin{pmatrix} 1 & 2\\ -7 & 0\\ 5 & 4 \end{pmatrix},\quad \textrm{D}=\displaystyle \begin{pmatrix} 1 & 2\\ -7 & 0\\ 5 & 4 \end{pmatrix},\quad \Rightarrow \textrm{A}=\textrm{D}\\\hline \end{array}$

 $\color{blue}\textrm{D. Operasi Matriks}$

$\begin{array}{|c|l|l|l|}\hline \textrm{No}&\qquad\textrm{Operasi}&\quad\textrm{Ketentuan}&\qquad\qquad\textrm{Contoh}\\ &\qquad\textrm{Matriks}&&\\\hline 1&\textrm{Penjumlahan}\: \&&\textrm{ordo sama}&A=\begin{pmatrix} 1\\ 2 \end{pmatrix},\: B=\begin{pmatrix} 8\\ 9 \end{pmatrix},\: \textrm{maka}\\2&\textrm{Pengurangan}&\textrm{ordo sama}&\color{red}A+B=\color{blue}\begin{pmatrix} 1+8\\ 2+9 \end{pmatrix}=\begin{pmatrix} 9\\ 11 \end{pmatrix}\\\hline 3&\textrm{Perkalian}&\textrm{Dengan}&\color{red}k\color{black}\begin{pmatrix} p & q\\ r & s \end{pmatrix}=\color{red}\begin{pmatrix} kp & kq\\ kr & ks \end{pmatrix}\\ &\textrm{Skalar}&\textrm{mengalikan}&\\ &&\color{red}\textrm{ke setiap elemen}&\\\hline 4&\textrm{Perkalian}&\begin{aligned}&\textrm{Dua matriks }\\ &\textrm{dapat dikalikan }\\ &\color{red}\textrm{jika}\\ &\textrm{banyaknya kolom}\\ &\textrm{matriks pertama}\\ &\textrm{sama dengan}\\ &\textrm{banyaknya baris}\\ &\textrm{matriks kedua} \end{aligned}&\begin{aligned}E=&\begin{pmatrix} 1 & 2\\ 3 & -1 \end{pmatrix},\: F=\begin{pmatrix} 5\\ 0 \end{pmatrix},\\ &\textrm{maka}\\ E&\times F\\ &=\begin{pmatrix} 1 & 2\\ 3 & -1 \end{pmatrix}_{2\times 2}\times \begin{pmatrix} 5\\ 0 \end{pmatrix}_{2\times 1}\\ &\color{red}\textrm{syarat memenuhi yaitu}:\\ &\textrm{kolom matriks 1}\\ &= \textrm{baris matriks 2}\\ &\textrm{dan hasilnya adalah }\\ &\textrm{matriks baru}\\ &\color{red}\textrm{dengan ordo }\\ &\textrm{banyak baris matriks 1}\\ &\textrm{kali banyak}\\ &\textrm{kolom matriks 2}\\ &\textrm{Dan aturan perkaliannya }\\ &\color{red}\textrm{adalah}\\ &\textrm{elemen baris matriks 1 kali}\\ &\textrm{elemen kolom matriks 2}\\ &\textrm{sehingga}\\ &=\begin{pmatrix} 1(5)+2(0)\\ 3(5)+-1(0) \end{pmatrix}_{2\times 1}\\ &=\begin{pmatrix} 5+0\\ 15-0 \end{pmatrix}=\begin{pmatrix} 5\\ 15 \end{pmatrix}_{2\times 1} \end{aligned} \\\hline \end{array}$

$\LARGE{\color{blue}\fbox{CONTOH SOAL}}$

$\begin{array}{ll}\\ \bullet &\textrm{Penjumlahan}\\ &\begin{aligned}&\begin{pmatrix} 1 & 2\\ 3 & -4 \end{pmatrix}+\begin{pmatrix} 5 & 6\\ 7 & -8 \end{pmatrix}\\ &=\begin{pmatrix} 1+5 & 2+6\\ 3+7 & (-4)+(-8) \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 6 & 8\\ 10 & -12 \end{pmatrix} \end{aligned}\\ \bullet &\textrm{Lawan suatu matriks}\\ &\begin{aligned}&\textrm{Jika}\: A=\begin{pmatrix} 1 & 2\\ 3 & -4 \end{pmatrix},\\ &\textrm{maka lawan matriks A adalah -A,} \\ &\textrm{Sehingga} \color{red}-A=\begin{pmatrix} -1 & -2\\ -3 & 4 \end{pmatrix} \end{aligned}\\ \bullet &\textrm{Pengurangan}\\ &\begin{aligned}&\begin{pmatrix} 1 & 2\\ 3 & -4 \end{pmatrix}-\begin{pmatrix} 5 & 6\\ 7 & -8 \end{pmatrix}\\ &=\begin{pmatrix} 1-5 & 2-6\\ 3-7 & (-4)-(-8) \end{pmatrix}\\ &\color{red}=\begin{pmatrix} -4 & -4\\ -4 & 4 \end{pmatrix} \end{aligned}\\ \bullet &\textrm{Perkalian}\\ &\begin{aligned}&(1)\: \: \textrm{Perkalian suatu matriks dengan skalar}\: \color{red}k\\ &\: \: \: \: \: \: \: 2\times \begin{pmatrix} 1 & 2\\ 3 & -4 \end{pmatrix}=\begin{pmatrix} 2\times 1 & 2\times 2\\ 2\times 3 & 2\times (-4) \end{pmatrix}\\ &\color{red}=\begin{pmatrix} 2 & 4\\ 6 & -8 \end{pmatrix} \end{aligned}\\ &\begin{aligned}&(2)\: \: \textrm{Perkalian antara dua buah matriks}\\ &\begin{pmatrix} 1 & 2\\ 3 & -4 \end{pmatrix}\times \begin{pmatrix} 5 & 6\\ 7 & -8 \end{pmatrix}\\ &\color{red}\textrm{perhatikan syarat memenuhi}\\ &=\begin{pmatrix} 1\times 5+2\times 7 & 1\times 6+2\times (-8)\\ 3\times 5 +(-4)\times 7&3\times 6+(-4)\times (-8) \end{pmatrix}\\ &=\begin{pmatrix} 5+14 & 6+(-16)\\ 15+(-28) & 18+32 \end{pmatrix}\\ &\color{red}=\begin{pmatrix} 19 & -10\\ -13 & 50 \end{pmatrix} \end{aligned} \end{array}$

Lanjutan 1 Materi Matriks (Matematika Wajib Kelas XI)

$\color{blue}\textrm{B. Jenis-Jenis Matriks}$

$\color{purple}\begin{array}{|l|l|l|l|l|}\hline \textrm{No}&\: \: \textrm{Jenis}&\qquad\qquad\qquad\textrm{Keterangan}&\textrm{Contoh}&\textrm{Ordo}\\\hline 1.&\textrm{Matriks}&\textrm{Matriks yang elemen} &\begin{pmatrix} 1 & 3 & -5 \end{pmatrix}&1\times 3\\ &\textrm{Baris}&\textrm{penyusunnya satu baris saja}&&\\\hline 2.&\textrm{Matriks}&\textrm{Matriks yang elemen}&\begin{pmatrix} 5\\ -5\\ 2 \end{pmatrix}&3\times 1\\ &\textrm{Kolom}&\textrm{penyusunnya tepat satu kolom saja}&&\\\hline 3.&\textrm{Matriks}&\textrm{Matriks yang semua elemennya} &\begin{pmatrix} 0 & 0&0\\ 0 & 0&0 \end{pmatrix}&2\times 3\\ &\textrm{Nol}&\textrm{adalah bilangan nol}&&\\\hline 4.&\textrm{Matriks}&\textrm{Matriks yang jumlah} &\begin{pmatrix} 2 & 8\\ 6 & 1 \end{pmatrix}&2\times 2\\ &\textrm{Persegi}&\textrm{baris dan kolomnya sama}&&\\\hline \end{array}$

$\color{purple}\begin{array}{|l|l|l|l|l|}\hline \textrm{No}&\: \: \textrm{Jenis}&\qquad\qquad\qquad\textrm{Keterangan}&\textrm{Contoh}&\textrm{Ordo}\\\hline 5.&\textrm{Matriks}&\textrm{Matriks Persegi yang semua} &\begin{pmatrix} \color{red}1 & 0\\ 0 & \color{red}7 \end{pmatrix}&2\times 2\\ &\textrm{Diagonal}&\textrm{elemennya nol kecuali pada }&\begin{pmatrix} \color{red}1 & 0 & 0\\ 0 & \color{red}3 & 0\\ 0 & 0 & \color{red}6 \end{pmatrix}&3\times 3\\ &&\textrm{diagonal utama}& &\\\hline 6.&\textrm{Matriks}&\textrm{Matriks yang elemen semuanya}&\begin{pmatrix} 5\\ -5\\ 2 \end{pmatrix}&3\times 1\\ &\textrm{Identitas}&\textrm{nol kecuali pada diagonal}&&\\ &&\textrm{utama berupa angka 1}&\begin{pmatrix} \color{red}1 & 0\\ 0 & \color{red}1 \end{pmatrix}&2\times 2\\\hline \end{array}$

$\color{purple}\begin{array}{|l|l|l|l|l|}\hline \textrm{No}&\: \: \textrm{Jenis}&\qquad\qquad\qquad\textrm{Keterangan}&\textrm{Contoh}&\textrm{Ordo}\\\hline 7.&\textrm{Matriks}&\textrm{matriks persegi yang semua elemen} &\begin{pmatrix} \color{red}5 & \color{red}8\\ 0 & \color{red}7 \end{pmatrix}&2\times 3\\ &\textrm{Segitiga}&\textrm{di bawah diagonal utama berupa }&\begin{pmatrix} \color{red}1 & \color{red}2 & \color{red}3\\ 0 & \color{red}3 & \color{red}5\\ 0 & 0 & \color{red}6 \end{pmatrix}&3\times 3\\ &\textrm{atas}&\textrm{bilangan nol}& &\\\hline 8.&\textrm{Matriks}&\textrm{Matriks persegi yang semua elemen}&\begin{pmatrix} \color{red}1 & 0 & 0\\ \color{red}4 & \color{red}2 & 0\\ \color{red}5 & \color{red}6 & \color{red}3 \end{pmatrix}&3\times 3\\ &\textrm{segitiga}&\textrm{di bawah diagonal utama berupa}&&\\ &\textrm{bawah}&\textrm{angka nol}&\begin{pmatrix} \color{red}1 & 0\\ \color{red}6 & \color{red}2 \end{pmatrix}&2\times 2\\\hline \end{array}$

$\begin{array}{|l|l|l|l|l|}\hline \textrm{No}&\: \: \textrm{Jenis}&\qquad\qquad\qquad\textrm{Keterangan}&\textrm{Contoh}&\textrm{Ordo}\\\hline 9.&\textrm{Matriks}&\textrm{Suatu matriks disebut sebagai matriks} &\begin{pmatrix} \color{red}5 & 0\\ \color{red}8 & \color{red}7 \end{pmatrix}&2\times 2\\ &\textrm{Simetris}&\textrm{simetris jika dan hanya jika elemen-elemen}&&\\ &\textrm{utama}&\textrm{yang letaknya simetris terhadap diagonal }& &\\ &&\textrm{atau bernilai sama}&&\\\hline \end{array}$

Matriks (Matematika Wajib Kelas XI)

$\color{blue}\textrm{A. Pendahuluan}$

Dalam kehidupan sehari-hari kita sering menjumpai atau mendapatkan informasi yang tersaji dalam bentuk daftar atau tabel. Sebagai misal ketika seorang wali kelas merekap data absensi kelas waliannya selama satu semester sebagaimana diperlihatkan dalam tabel berikut ini

$\begin{array}{ll}\\ \begin{array}{|l|c|c|c|}\hline \textrm{Nama}&\textrm{Sakit}&\textrm{Izin}&\textrm{Tanpa}\\ \textrm{Siswa}&&&\textrm{Keterangan}\\\hline \textrm{Andi}&2&1&3\\ \textrm{Budi}&1&4&2\\ \textrm{Carli}&1&1&5\\ \textrm{Dodi}&2&2&1\\\hline \end{array}&\begin{aligned}&\Leftarrow \color{blue}\textrm{Judul baris}\\ &\\ &\\ &\\ & \end{aligned}\\ \begin{aligned}&\quad\Uparrow \\ &\color{red}\textrm{Judul kolom} \end{aligned}& \end{array}$

Dari tabel di atas, jika kita tuliskan bilangannya saja maka akan kita dapatkan bilangan yang seolah-olah tersusun berbentuk persegi atau persegi panjang dan oleh kareanya sebagaimana ilsutrasi tabel di atas bilangan terbut juga tersusun dalam baris dan kolom sebagaimana berikut ini

$\color{blue}\begin{array}{lccc}\\ &\qquad2&\qquad1&\qquad3\\\\ &\qquad1&\qquad4&\qquad2\\\\ &\qquad1&\qquad1&\qquad5\\\\ &\qquad2&\qquad2&\qquad1 \end{array}$

Selanjutnya kaitanya dengan matriks apa bila susunan bilangan-bilangan di atas, diberikan tanda kurung tertentu jadil bentuk matriks. Dan dari paparan tersebut kita dapat mengakatakan matriks adalah susunan bilangan dalam bentuk persegi atau persegi panjang yang di atur menurut baris dan kolom dalam tanda kurung tertentu.

Selanjutnya bilangan yang diatus menurut baris dan kolom disebut unsur atau elemen atau entri dari suatu matri

Perhatikanlah ilustrasi berikut

Sebagai tambahan nama sebuah matrik adalah sebuah huruf besar dan memiliki ukuran sebuah matrik(selanjutnya dapat isebut sebagai ordo) = baris x kolom.

$\LARGE{\color{blue}\fbox{CONTOH SOAL}}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \color{red}M=\begin{bmatrix} -2 & 3 & 0&0&-3\\ -1 & 2 & 1&1&2\\ 0 & 1 & -3&2&-2\\ 1 & 0 & -5&3&1 \end{bmatrix},\: \color{black}\textrm{tentukanlah}\\\\ &\textrm{a}.\quad \textrm{ordo dari matrik tersebut}\\ &\textrm{b}.\quad \textrm{elemen penyusun kolom pertama}\\ &\textrm{c}.\quad \textrm{elemen penyusun baris pertama}\\ &\textrm{d}.\quad \textrm{elemen penyusun kolom kedua}\\ &\textrm{e}.\quad \textrm{elemen penyusun baris kedua}\\ &\textrm{f}.\quad \textrm{elemen penyusun kolom ketiga}\\ &\textrm{g}.\quad \textrm{elemen penyusun baris ketiga}\\ &\textrm{h}.\quad \textrm{elemen penyusun kolom keempat}\\ &\textrm{i}.\quad \textrm{elemen penyusun baris keempat}\\ &\textrm{j}.\quad \textrm{elemen penyusun kolom kelima}\\ &\textrm{k}.\quad \textrm{elemen penyusun baris kelima}\\ &\textrm{l}.\quad \textrm{elemen yang terletak pada baris kelima dan kolom kelima}\\ &\textrm{m}.\quad \textrm{elemen yang terletak pada baris pertama dan kolom kelima}\\ &\textrm{n}.\quad \textrm{elemen yang terletak pada baris kedua dan kolom keempat}\\ &\textrm{o}.\quad \textrm{elemen yang terletak pada baris ketiga dan kolom ketiga}\\ &\textrm{p}.\quad \textrm{elemen yang terletak pada baris kedua dan kolom kedua}\\ &\textrm{q}.\quad \textrm{elemen yang terletak pada baris pertama dan kolom pertama}\\ \end{array}$

$.\: \qquad \color{blue}\begin{aligned}&\color{red}\textrm{Jawab}:\\ &\textrm{a}.\quad \color{red}\textrm{ordo matriknya}\: \: 4\times 5\\ &\begin{array}{|cc|cc|cc|cc|}\hline \textrm{b}.& -2,-1,0,1&\textrm{c}.& -2,3,0,0,-3\\\hline \textrm{d}.& 3,2,1,0&\textrm{e}.& -1,2,1,1,2\\\hline \textrm{f}.& 0,1,-3,-5&\textrm{g}.&0,1,-3,2,-2\\\hline \textrm{h}.&0,1,2,3&\textrm{i}.&1,0,-5,3,1 \\\hline \textrm{j}&-3,2,-2,1&\textrm{k}&\: \color{red}\textrm{tidak ada}\\\hline \textrm{l}.&\color{red}\textrm{tidak ada}&\textrm{m}&m_{15}=-3\\\hline \textrm{n}.&m_{24}=1&\textrm{o}.&m_{33}=-3 \\\hline \textrm{p}.&m_{22}=2&\textrm{q}.&m_{11}=-2\\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: A=\begin{bmatrix} 1\\ 2\\ 3\\ 4 \end{bmatrix},\: B=\begin{bmatrix} 3 & 7\\ 4 & 8\\ 5 & 9\\ 6 & 10 \end{bmatrix},\\ &C=\begin{bmatrix} 11 & 12 & 13\\ 14 & 15 & 16\\ 17 & 18 & 19\\ 20 &21 & 22 \end{bmatrix}\: \: \textrm{tentukanlah nilai}\\\\ &\textrm{a}.\quad a_{11},\: a_{31},\: b_{42},\: \textrm{dan}\: \: c_{43}\\ &\textrm{b}.\quad b_{42}+c_{43}\\ &\textrm{c}.\quad a_{41}-b_{31}+c_{21}-a_{11}\\ &\textrm{d}.\quad a_{11}+b_{22}+c_{33}\\ &\textrm{e}.\quad a_{11}^{2}+b_{22}^{2}+c_{33}^{2}\\ &\textrm{f}.\quad \left (a_{11}+b_{22} \right )^{2}-c_{33}^{2}\\ \end{array}$

$.\: \qquad\color{blue}\begin{aligned}&\color{red}\textrm{Jawab}\\ &\begin{array}{|c|l|c|l|c|l|}\hline \textrm{a}.&\begin{aligned}&\textrm{perhatikanlah}\\ &a_{11}=1,\: a_{31}=3\\ &b_{42}=10,\: c_{43}=22 \end{aligned}&\textrm{b}.&\begin{aligned}&b_{42}+c_{43}\\ &=10+22\\ &=32 \end{aligned}\\\hline \textrm{c}.&\begin{aligned}&a_{41}-b_{31}+c_{21}-a_{11}\\ &=4-5+14-1\\ &=12 \end{aligned}&\textrm{d}.&\begin{aligned}&a_{11}+b_{22}+c_{33}\\ &=1+8+19\\ &=28 \end{aligned}\\\hline \textrm{e}.&\begin{aligned}&a_{11}^{2}+b_{22}^{2}+c_{33}^{2}\\ &=1^{2}+8^{2}+19^{2}\\ &=1+64+361=426 \end{aligned}&\textrm{f}.&\begin{aligned}&\left (a_{11}+b_{22} \right )^{2}-c_{33}^{2}\\ &=(1+8)^{2}-19^{2}\\ &=81-361=-280 \end{aligned}\\\hline \end{array} \end{aligned}$

Contoh Soal 8 Statistika

$\begin{array}{ll}\\ 36.&\textrm{Diketahui nilai statistik lima serangkai}\\ &\textrm{dari empat kelompok data seperti terlihat}\\ &\textrm{dalam tabel berikut}\\ &\begin{array}{|l|c|c|c|c|c|}\hline \textrm{Data}&\textrm{min}&Q_{1}&Q_{2}&Q_{3}&\textrm{mak}\\\hline \: \: \textrm{I}&74&80&88&92,5&99\\ \: \: \textrm{II}&66&81,5&86&90,5&96\\ \: \: \textrm{III}&70&77,5&85&92,5&100\\ \: \: \textrm{IV}&55&80&88&90&97,5\\\hline \end{array}\\ &\textrm{Data yang memuat pencilan terdapat pada}\\ &\textrm{tabel}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{I dan II}\\ \textrm{b}.&\textrm{II dan III}\\ \textrm{c}.&\textrm{I dan III}\\ \textrm{d}.&\textrm{III dan IV}\\ \color{red}\textrm{e}.&\textrm{II dan IV} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: \: \color{red}\textbf{pencilan}\: \: \color{blue}\textrm{adalah datum yang}\\ \color{black}\textrm{bernil}&\textrm{ai kurang dari pagar dalam dan lebih besar}\\ \textrm{dari p}&\textrm{agar luar}\\ \textrm{Rumu}&\textrm{s pagar dalam}\: =\: Q_{1}-L\\ &=Q_{1}-\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )=\color{red}\displaystyle \frac{5}{2}Q_{1}-\frac{3}{2}Q_{3}\\ \textrm{Rumu}&\textrm{s pagar luar}\: =\: Q_{3}+L\\ &=Q_{3}+\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )=\color{red}\displaystyle \frac{5}{2}Q_{3}-\frac{3}{2}Q_{1}\\ \textrm{Data I}&\: \textrm{pagar dalamnya}=\displaystyle \frac{5}{2}(80)-\frac{3}{2}(92,5)\\ &=200-138,75=\color{black}61,25\\ \textrm{Data I}&\: \textrm{pagar luarnya}=\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(80)\\ &=231,25-120=\color{black}111,25\\ \textrm{Jadi, d}&\textrm{ata I tidak ada pencilan}\\ \textrm{Data I}&\textrm{I pagar dalamnya}=\displaystyle \frac{5}{2}(81,5)-\frac{3}{2}(90,5)\\ &=203,75-135,75=\color{black}68\\ \textrm{Data I}&\textrm{I pagar luarnya}=\displaystyle \frac{5}{2}(90,5)-\frac{3}{2}(81,5)\\ &=226,25-122,25=\color{black}104\\ \textrm{Jadi, d}&\textrm{ata II ada pencilan, yaitu}\: \: \color{red}66<\color{black}68\\ &\textrm{karena}\: \color{red}66\: \color{blue}\textrm{adalah datum terkecil data II}\\ \textrm{Data I}&\textrm{II pagar dalamnya}=\displaystyle \frac{5}{2}(77,5)-\frac{3}{2}(92,5)\\ &=193,75-138,75=\color{black}55\\ \textrm{Data I}&\textrm{II pagar luarnya}=\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(77,5)\\ &=231,25-116,25=\color{black}115\\ \textrm{Jadi, d}&\textrm{ata III tidak ada pencilan}\\ \textrm{Data I}&\textrm{V pagar dalamnya}=\displaystyle \frac{5}{2}(80)-\frac{3}{2}(90)\\ &=200-135=\color{black}65\\ \textrm{Data I}&\textrm{V pagar luarnya}=\displaystyle \frac{5}{2}(90)-\frac{3}{2}(70)\\ &=225-105=\color{black}120\\ \textrm{Jadi, d}&\textrm{ata IV ada pencilan, yaitu}\: \: \color{red}55<\color{black}65\\ &\textrm{karena}\: \color{red}55\: \color{blue}\textrm{adalah datum terkecil data IV} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, & Sulasim. 2005. Kompetensi Matematika SMA Kelas 2 Semester 1 Program IPA Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung: YRAMA WIDYA.
3. Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.
4. Tim Supermat. 2007. Cara Mudah MenghadapiS SMBB TELKOM. Jakarta: LITERATUR MEDIA SUKSES.

Contoh Soal 7 Statistika

$\begin{array}{ll}\\ 31.&(\textbf{SMBB TELKOM 06})\\ &\textrm{Berikut tidak termasuk ukuran penyebaran}\\ &\textrm{data adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&\textrm{rentang}\\ \textrm{b}.&\textrm{varians}\\ \textrm{c}.&\textrm{jarak antar kuartil}\\ \color{red}\textrm{d}.&\textrm{kuartil}\\ \textrm{e}.&\textrm{simpangan baku} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Ingat bahwa mean, median, modus}\\ &\color{red}\textrm{adalah bagian ukuran pemusatan data}\\ &\textrm{sedangkan kuartil adalah ukuran letak data} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 32.&\textrm{Perhatikanlah barisan}\: \: 1,-2,3,-4,5,-6,...\\ &\textrm{dengan suku ke}-n\: \: \textrm{adalah}\: \: (-1)^{n-1}\times n.\\ &\textrm{Rata-rata 200 suku pertama barisan tersebut}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \color{red}\textrm{b}.&-0,5\\ \textrm{c}.&0\\ \textrm{d}.&0,5\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui bahwa}\: : 1,-2,3,-4,5,-6,...,199,-200\\ &\overline{x}_{\color{red}200}=\displaystyle \frac{1-2+3-4+5-6+...+199-200}{200}\\ &\: \: \: \quad =\displaystyle \frac{-100}{200}=-0,5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&(\textbf{STT TELKOM 2005})\\ &\textrm{Seratus mahasiswa telah mengikuti ujian psikotes}\\ &\textrm{dan rata-rata skornya yang diperoleh adalah 100}.\\ &\textrm{Banyaknya mahasiswa junior yang mengikuti}\\ &\textrm{ujian psikotes}\: \: 50\%\: \: \textrm{lebih besar dari banyknya ma}-\\ &\textrm{hasiswa senior. Jika rata-rata skor dari mahasiswa}\\ &\textrm{senior}\: \: 50\%\: \: \textrm{lebih tinggi dari mahasiswa junior,}\\ &\textrm{maka rata-rata skor mahasiswa senior adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&110\\ \textrm{b}.&115\\ \textrm{c}.&120\\ \color{red}\textrm{d}.&125\\ \textrm{e}.&150 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Perhatikanlah tabel berikut}\\ &\begin{array}{|c|c|c|}\hline \textrm{Mahasiswa}&\textrm{Mahasiswa}&\color{purple}\textrm{Total}\\ \textrm{senior}&\textrm{junior}&\\\hline n_{\color{red}senior}&n_{\color{red}junior}&\color{black}100\\\hline n_{\color{red}s}&\displaystyle \frac{3}{2}n_{\color{red}s}&\color{black}100\\\hline \overline{x}_{\color{red}senior}&\overline{x}_{\color{red}junior}&\color{red}100\\\hline \overline{x}_{\color{red}s}&\displaystyle \frac{2}{3}\overline{x}_{\color{red}s}&\color{red}100\\\hline \end{array}\\ &n_{\color{red}s}+\displaystyle \frac{3}{2}n_{\color{red}s}=\color{black}100\\ &\: \: \: \: \qquad n_{\color{red}s}=\color{black}40,\\ &\textrm{maka}\: \: n_{\color{red}j}=\color{black}60.\\ &\textrm{Selanjutnya kita tentukan nilai rata-rata}\\ &\textrm{mahasiswa senior, dengan}\\ &\displaystyle \frac{\color{black}40\color{blue}\overline{x}_{\color{red}s}+\color{black}(60)\color{blue}\displaystyle \frac{2}{3}\overline{x}_{\color{red}s}}{\color{black}100}=\color{red}100\\ &\: \: \: \, \quad\quad\qquad\color{black}80\color{blue}\overline{x}_{\color{red}s}=10000\\ &\qquad\qquad\qquad \color{blue}\overline{x}_{\color{red}s}=\displaystyle \frac{10000}{\color{black}80}\\ &\qquad\qquad\qquad \color{blue}\overline{x}_{\color{red}s}=\color{red}125 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&(\textbf{SPMB 04})\\ &\textrm{Median, rata-rata, dan modus dari data yang}\\ &\textrm{terdiri atas empat bilangan asli adalah 7. Jika}\\ &\textrm{selisih antara data terbesar dan terkecil adalah}\\ &\textrm{6, maka hasil kali keempat datum tersebut}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1.864\\ \textrm{b}.&1.932\\ \color{red}\textrm{c}.&1.960\\ \textrm{d}.&1.976\\ \textrm{e}.&1.983 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui bahwa}\: :x_{1},x_{2},x_{3},\: \: \textrm{dan}\: \: x_{4}\: \: \textrm{adalah}\\ &\textrm{asli dengan}\: \: x_{4}-x_{1}=6\: ........\color{red}(1)\\ &\&\: \textrm{modusnya adalah 7, maka data dapat dituliskan}\\ &:\: x_{1},\color{red}7,7\color{blue},x_{4}.\: \: \textrm{Jawaban ini sesuai karena median} = \color{red}7.\\ &\textrm{Karena mean}\: =7,\: \: \textrm{maka dapat dituliskan}\\ &\displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=7\Leftrightarrow x_{1}+x_{2}+x_{3}+x_{4}=28\\ &\Leftrightarrow x_{1}+14+x_{4}=28\Leftrightarrow x_{4}+x_{1}=14\: ........\color{red}(2)\\ &\color{black}\textrm{Selanjutnya kita eliminasi}\: \: \color{blue}(1)\&(2)\\ &\color{blue}\begin{array}{llll}\\ x_{4}-x_{1}&=6&\\ x_{4}+x_{1}&=14\: \: \: +\\\hline 2x_{4}\: \: \qquad&=20&\\ x_{4}&=10\: .......\color{red}(3)\\ \textrm{maka}\: \: x_{1}&=4\: .........\color{red}(4) \end{array}\\ &\textrm{Jadi},\: \: x_{1}\times x_{2}\times x_{3}\times x_{4}=(4).(7).(7).(10)=\color{red}1960 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Desil ke-8}\: \: \left ( D_{8} \right )\: \: \textrm{dari data berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Nilai}&f\\\hline 41-45&7\\ 46-50&12\\ 51-55&9\\ 56-60&8\\ 61-65&4\\\hline \end{array}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&58\\ \textrm{b}.&57,5\\ \textrm{c}.&57\\ \textrm{d}.&56,75\\ \textrm{e}.&56,25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\: \textrm{desil ke}-8=\color{black}D_{8},\: \: \textrm{dengan}\: \: n=\sum f=40\\ D_{i}&=t_{b}+p\left ( \displaystyle \frac{\displaystyle \frac{i\times n}{10}-F}{f} \right )\\ D_{8}&= \textrm{datum ke}-\left ( \displaystyle \frac{8n}{10} \right )=x_{\frac{8\times 40}{10}}=\color{red}x_{32}\\ \textrm{Dan}\: \: \color{red}x_{32}\: \: &\textrm{terletak di kelas interval}\: :\: \color{red}56-60 \\ D_{8}&=55,5+5\left ( \displaystyle \frac{32-28}{8} \right )\\ &=55,5+2,5\\ &=\color{red}58 \end{aligned} \end{array}$