PAS MATEMATIKA BLOG

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{ll}\\ 36.&\textrm{Persamaan}\: \: ^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\textrm{mempunyai akar}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2},\: \: \textrm{maka}\\ &\textrm{nilai}\: \: x_{1}+x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Alternatif 1}\\ &^{x}\log 2+\: ^{x}\log (3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 2(3x-4)=2\\ &\Leftrightarrow \: ^{x}\log 6x-8=2\\ &\Leftrightarrow \: \color{black}6x-8=x^{2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0,\: \: \color{purple}\textrm{dengan}\: \begin{cases} a &=1 \\ b &=-6 \\ c &=8 \end{cases}\\ &\Leftrightarrow \: x_{1}+x_{2}=-\displaystyle \frac{b}{a}\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}-\displaystyle \frac{-6}{1}=6\\ &\color{purple}\textrm{Alternatif 2}\\ &\Leftrightarrow \: \color{black}x^{2}-6x+8=0\\ &\Leftrightarrow \: (x-2)(x-4)\\ &\Leftrightarrow \: x_{1}=2\: \: \textrm{atau}\: \: x_{2}=4\\ &\Leftrightarrow \: x_{1}+x_{2}=\color{red}2+4=6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\\ &(\log x)(2\log x-3)=\log 100\\ &\textrm{maka}\: \: x_{1}\times x_{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&100\\ \color{red}\textrm{b}.&10\sqrt{10}\\ \textrm{c}.&\sqrt{10}\\ \textrm{d}.&-\sqrt{10}\\ \textrm{e}.&-10\sqrt{10} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&(\log x)(2\log x-3)=\color{red}\log 100\\ &\Leftrightarrow (\log x)\left ( 2\log x-3 \right )=\color{red}2\\ &\color{black}\Leftrightarrow 2\log ^{2}x-3\log x-2=0\: \color{purple}\begin{cases} a &=2 \\ b &=-3 \\ c &=-2 \end{cases}\\ &\Leftrightarrow \log x_{1}+\log x_{2}=-\displaystyle \frac{-3}{2}=\frac{3}{2}\\ &\Leftrightarrow \log \left ( x_{1}\times x_{2} \right )=1\displaystyle \frac{1}{2}\\ &\Leftrightarrow \color{red}\left ( x_{1}\times x_{2} \right )=10^{1\frac{1}{2}}=10\sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{x}\log (x+12)-3.\: ^{x}\log 4+1=0\\ &^{x}\log (x+12)-\: ^{x}\log 4^{3}=-1\\ &^{x}\log \displaystyle \frac{x+12}{64}=-1\\ &\displaystyle \frac{x+12}{64}=x^{-1}=\frac{1}{x}\\ &x+12=\displaystyle \frac{64}{x}\\ &\color{purple}x^{2}+12x-64=0\\ &\color{purple}(x+16)(x-4)=0\\ &x=-16\: \: \color{black}\textrm{atau}\: \: \color{red}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai}\: \: x\: \: \: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\: \: \textrm{dan}\: \: 6\\ \textrm{b}.&-2\: \: \textrm{dan}\: \: 6\\ \textrm{c}.&-1\\ \textrm{d}.&-2\\ \color{red}\textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{^{2}\log (2x-3)}{^{2}\log x}-\: ^{x}\log (x+6)+\displaystyle \frac{1}{^{(x+2)}\log x}=1\\ &^{x}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1\\ &^{x}\log (2x-3)(x+2)=1+\log (x+6)\\ &^{x}\log \displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=1\\ &\displaystyle \frac{\left ( 2x^{2}+x-6 \right )}{(x+6)}=x^{1}\\ &\left ( 2x^{2}+x-6 \right )=x^{2}+6x\\ &\color{purple}x^{2}-5x-6=0\\ &\color{purple}(x+1)(x-6)=0\\ &x=-1\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

$\begin{array}{ll}\\ 32.&\textrm{Agar}\: \: \log \left ( x^{2}-1 \right )<0\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<1\\ \textrm{b}.&-\sqrt{2}<x<\sqrt{2}\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>1\\ \textrm{d}.&x<-\sqrt{2}\: \: \textrm{atau}\: \: x>\sqrt{2}\\ \color{red}\textrm{e}.&-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\log \left ( x^{2}-1 \right )<0\\ &\textrm{Diketahui}\: \: \color{red}\log f(x)<0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-1>0\\ &\Leftrightarrow x<-1\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: \log \left ( x^{2}-1 \right )<0\\ &\log \left ( x^{2}-1 \right )<\log 1\\ &\Leftrightarrow x^{2}-1<1\\ &\Leftrightarrow x^{2}-2<0\\ &\Leftrightarrow x^{2}-\left ( \sqrt{2} \right )^{2}<0\\ &\Leftrightarrow -\sqrt{2}<x<\sqrt{2}\\ &\textrm{Jadi},\: \: \color{red}-\sqrt{2}<x<-1\: \: \textrm{atau}\: \: 1<x<\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Himpunan penyelesaian dari}\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{b}.&\left \{ x|-\sqrt{3}<x<-1\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \right \}\\ \textrm{c}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{d}.&\left \{ x|-2<x<-\sqrt{3} \right \}\\ \textrm{e}.&\left \{ x|\sqrt{3}<x<2 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &\textrm{Diketahui}\: \: \color{red}^{.^{\frac{1}{2}}}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-3>0\\ &\Leftrightarrow x<-\sqrt{3}\: \: \textrm{atau}\: \: x>\sqrt{3}\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>0\\ &^{.^{\frac{1}{2}}}\log \left ( x^{2}-3 \right )>\: ^{.^{\frac{1}{2}}}\log 1\\ &\Leftrightarrow x^{2}-3<1\quad \left (\color{black}\textrm{karena basisnya}\: \: \displaystyle \frac{1}{2}<1 \right )\\ &\Leftrightarrow x^{2}-4<0\\ &\Leftrightarrow x^{2}-2^{2}>0\\ &\Leftrightarrow -2<x<2\\ &\textrm{Jadi},\: \: \color{red}-2<x<-\sqrt{3}\: \: \textrm{atau}\: \: \sqrt{3}<x<2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &^{2}\log \left ( x^{2}-x \right )\leq 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-1\leq x\leq 2,\: x\neq 1\: \: \textrm{atau}\: \: x\neq 0\\ \color{red}\textrm{c}.&-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2\\ \textrm{d}.&-1< x\leq 0\: \: \textrm{atau}\: \: 1\leq x< 2\\ \textrm{e}.&-1\leq x\leq 0\: \: \textrm{atau}\: \: 1\leq x\leq 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{2}\log \left ( x^{2}-x \right )\leq 1\\ &\textrm{Diketahui}\: \: \color{red}^{2}\log f(x)>0,\: \: \color{blue}\textrm{maka}\\ &\color{purple}\textrm{Syarat (1)},\: \: \color{red}f(x)>0\\ &\Leftrightarrow x^{2}-x>0\Leftrightarrow x(x-1)>0\\ &\Leftrightarrow x<0\: \: \textrm{atau}\: \: x>1\\ &\color{purple}\textrm{Syarat (2)},\: \: ^{2}\log \left ( x^{2}-x \right )\leq 1\\ &^{2}\log \left ( x^{2}-x \right )\leq \: ^{2}\log 2\\ &\Leftrightarrow x^{2}-x\leq 2\\ &\Leftrightarrow x^{2}-x-2\leq 0\\ &\Leftrightarrow (x+1)(x-2)\leq 0\\ &\Leftrightarrow -1\leq x\leq 2\\ &\textrm{Jadi},\: \: \color{red}-1\leq x< 0\: \: \textrm{atau}\: \: 1<x\leq 2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left | \log (x+1) \right |> 1 \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{b}.&x<-9\: \: \textrm{atau}\: \: x>9\\ \color{red}\textrm{c}.&-1<x<-0,9\: \: \textrm{atau}\: \: x>9\\ \textrm{d}.&-9< x<0,9\\ \textrm{e}.&-0,9<x<9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Ingat bahwa}\\ &\left | x \right |>A\Leftrightarrow \color{black}x<-A\: \: \textrm{atau}\: \: x>A,\: \: \color{red}A>0\\ &\Leftrightarrow \log (x+1)<-1\: \: \textrm{atau}\: \: \log (x+1)>1\\ &\color{red}\textrm{Syarat (1) buat keduanya},\: \: \color{red}f(x)>0\\ &(x+1)>0\Leftrightarrow x>-1\\ &\color{red}\textrm{Syarat (2)},\: \: \log \left ( x+1 \right )<-1\\ &\log (x+1)<\log 10^{-1}\\ &x+1<\displaystyle \frac{1}{10}\Leftrightarrow x<-\frac{9}{10}\\ &\color{red}\textrm{Syarat (3)},\: \: \log \left ( x+1 \right )> 1\\ &\log (x+1)>\log 10^{1}\\ &(x+1)>10\Leftrightarrow x>9\\ &\textrm{Jadi},\: \: \color{red}-1<x<-0,9\: \: \textrm{atau}\: \: x>9 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

MENGINGAT KEMBALI

$\color{blue}\textrm{E. Sifat-Sifat Eksponen dan Logaritma}$

$\color{blue}\textrm{E.1 Persamaan Eksponen dan Logaritma}$

$\color{blue}\begin{array}{|l|l|}\hline \color{red}\textrm{Eksponens}&\color{black}\textrm{Logaritma}\\\hline \displaystyle a^{n}=\underset{n\: \: faktor}{\underbrace{a\times a\times a\times \cdots \times a}}&\color{black}^{a}\log b=c\: \Rightarrow \: a^{c}=b\\\hline \bullet \quad a^{p}\times a^{q}=a^{p+q}&\bullet \quad \color{black}^{a}\log x+\: ^{a}\log y=\: ^{a}\log xy\\\hline \bullet \quad a^{p}: a^{q}=a^{p-q}&\bullet \quad \color{black}^{a}\log x-\: ^{a}\log y=\: ^{a}\log \displaystyle \frac{x}{y}\\\hline \bullet \quad \left ( a^{p} \right )^{q}=a^{p.q}&\bullet \quad ^{a}\log x=\: \displaystyle \frac{^{m}\log x}{^{m}\log a}\\\hline \bullet \quad \displaystyle \sqrt[q]{a^{p}}=\displaystyle a^{ \left (\frac{p}{q} \right )}&\bullet \quad ^{a}\log b\: \times \: ^{b}\log c=\: ^{a}\log c\\\hline \bullet \quad \left ( a\times b \right )^{p}=a^{p}\: \times \: b^{p}&\bullet \quad ^{a^{m}}\log b^{n}=\displaystyle \frac{n}{m}\times \: ^{a}\log b\\\hline \bullet \quad \left ( \displaystyle \frac{a}{b} \right )^{p}=\displaystyle \frac{\displaystyle a^{p}}{\displaystyle b^{p}}&\bullet \quad \displaystyle a^{\: {^{a}}\log b}=b\\\hline \bullet \quad a^{-p}=\displaystyle \frac{1}{\displaystyle a^{p}}&\bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{b}\log a}\\\hline \bullet \quad a^{0}=1,\: \: \: \: \: a\neq 0&\bullet \quad ^{a}\log 1=0\\\hline \bullet \quad a^{1}=1&\bullet \quad \color{black}^a\log a=1\\\hline \begin{cases} a,b\: \in \mathbb{R} \\ p,q\: \in \mathbb{Q} \end{cases}&\begin{cases} a\neq 0 &\\ a>0&(\textrm{bilangan pokok}) \\ x,y>0 & (\textrm{numerus}) \end{cases}\\\hline \end{array}$

Selanjutnya

$\color{blue}\begin{array}{|l|l|l|}\hline \textrm{No}&\qquad\textrm{Bentuk}&\qquad\qquad\color{red}\textrm{Syarat}\\\hline 1.&a^{f(x)}=1&a\neq 0,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 2.&a^{f(x)}=a^{p}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=p\\\hline 3.&a^{f(x)}=a^{g(x)}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: \color{red}f(x)=g(x)\\\hline 4.&a^{f(x)}=b^{f(x)}&a\neq 0,\: b\neq 0\: ,\quad \textrm{maka}\: \: \color{red}f(x)=0\\\hline 5.&f(x)^{g(x)}=1&\begin{cases} f(x)=1 & \\ g(x)=0, & \textrm{jika}\: \: f(x)\neq 0 \\ f(x)=-1, & \textrm{jika}\: \: g(x)=\: \textrm{genap} \end{cases}\\\hline 6.&\color{black}f(x)^{g(x)}=f(x)^{h(x)}&\color{red}\begin{cases} (i).\quad g(x)=h(x)& \\ (ii).\quad f(x)=1& \\ (iii).\quad f(x)=0,&g(x)>0,\: \: h(x)>0 \\ (iv).\quad f(x)=-1,&g(x)\: \textrm{dan}\: h(x)\: \: \\ &\color{black}\textrm{keduanya ganjil}\\ &\color{black}\textrm{atau genap} \end{cases}\\\hline 7.&g(x)^{f(x)}=h(x)^{f(x)}&\begin{cases} (i).\quad g(x) =h(x)& \\ (ii).\quad f(x)=0, & g(x)\neq 0,\: h(x)\neq 0 \end{cases}\\\hline 8.&\begin{aligned}&\color{black}A\left ( a^{f(x)} \right )^{2}\\ &\quad \color{black}+B\left ( a^{f(x)} \right )\\ &\qquad \color{black}+C=0 \end{aligned}&\color{red}a>0,\: \: a\neq 1\\\hline \end{array}$

$\color{blue}\textrm{E.2 Pertidaksamaan Eksponen dan Logaritma}$

Berikut sifat pertidaksamaan Eksponen

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad\qquad \color{black}0<a<1\\\hline a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\leq g(x)&a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\geq g(x)\\\hline a^{f(x)}< a^{g(x)}\Rightarrow f(x)< g(x)&a^{f(x)}< a^{g(x)}\Rightarrow f(x)> g(x)\\\hline a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\geq g(x)&a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\leq g(x)\\\hline a^{f(x)}> a^{g(x)}\Rightarrow f(x)> g(x)&a^{f(x)}> a^{g(x)}\Rightarrow f(x)< g(x)\\\hline \end{array}$

Untuk pertidaksamaan logaritma (dengan syarat $\left (f(x)>0\: \: \textrm{dan}\: \: g(x)>0 \right )$ ) adalah sebagai berikut:

$\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad \color{red}a>1&\qquad \color{black}0<a<1\\\hline \begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\leq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}&\begin{aligned}&^a\log f(x)< \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\geq g(x) \end{aligned}&\begin{aligned}&^a\log f(x)\geq \: ^a\log g(x)\\ &\Rightarrow f(x)\leq g(x) \end{aligned}\\\hline \begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)> g(x) \end{aligned}&\begin{aligned}&^a\log f(x)> \: ^a\log g(x)\\ &\Rightarrow f(x)< g(x) \end{aligned}\\\hline \end{array}$

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

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

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

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

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

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

Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA
Kanginan, M., Terzalgi, Z. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU.
Sharma, S. N. 2017. Jelajah Matematika 2 SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
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)

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

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

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

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

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