Lanjutan Contoh Soal Matriks 2

\begin{array}{ll}\\ \fbox{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{C}=\begin{pmatrix} ^{a}\log 2 & -\displaystyle \frac{1}{2}\\ -2(b+c) & 3 \end{pmatrix},\\ &\textrm{dan}\: \: \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{d}.&11\\ \textrm{b}.&5&\textrm{c}.&7&\textrm{e}.&13 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\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} 2.\: ^{a}\log 6-2.\: ^{2}\log 2 & 2.2-5-2\left ( -\displaystyle \frac{1}{2} \right )\\ 2.1 -6-2(-2(b+c))& 2(a+3b)+3a-5b-2.3 \end{pmatrix}&=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\\ \left.\begin{matrix} \textcircled{1}...............\quad 2.\: ^{a}\log 6-2.\: ^{2}\log 2&=2\\ \textcircled{2}...............\quad 2.1 -6-2(-2(b+c))&=0\\ \textcircled{3}...............\quad2(a+3b)+3a-5b-2.3&=2 \end{matrix}\right\}&\\\\ \begin{array}{|c|c|}\hline \textrm{dari persamaan}\: \: \textcircled{1}&\textrm{dari persamaan}\: \: \textcircled{2}\\\hline \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^{2}\\ 3&=a\\ 12&=4a \end{aligned}&\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 \end{aligned}\\\hline \end{array}& \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{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}\\ \textrm{a}.&-6&&&\textrm{d}.&3\\ \textrm{b}.&-3&\textrm{c}.&2&\textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{array}{|c|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} &\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}&\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}\\ \fbox{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}\\ \textrm{a}.&2&&&\textrm{d}.&5\\ \textrm{b}.&3&\textrm{c}.&4&\textrm{e}.&6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\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 \begin{aligned}-2p+q&=1\\ -p+q&=-2\quad _{-}\\ ----&---\\ -p\qquad&=3\\ p&=-3\\ &\\ & \end{aligned}&\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}\\ \fbox{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&&&\textrm{d}.&12\\ \textrm{b}.&7&\textrm{c}.&9&\textrm{e}.&15 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(\textrm{Z})&=\begin{pmatrix} -3p-8q & 12\\ -6 & -2(p+q) \end{pmatrix}\\ \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}\\ \textrm{Sehingga}&\\ -12&=-3p-8q\quad.................\textcircled{1}\\ -1&=p+q\quad.......................\textcircled{2}\\ \textrm{persamaan}&\: \textcircled{2}\: \: \textrm{ke persamaan}\: \: \textcircled{1}\\ -12&=-3p-3q-5q=-3(p+q)-5q\\ -12&=-3(-1)-5q\\ -12&=3-5q\\ 5q&=3+12\\ q&=3\quad..............................\textcircled{3}\\ \textrm{persamaan}&\: \: \textcircled{3}\: \: \textrm{ke persamaan}\: \: \textcircled{2}\\ p+q&=-1\\ p&=-1-q=-1-3=-4\\ p^{2}-q^{2}&=(-4)^{2}-3^{2}=16-9\\ &=7 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -5\\ 3 & -1 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-17&&&\textrm{d}.&13\\ \textrm{b}.&-13&\textrm{c}.&11&\textrm{e}.&17 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\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}\\ \fbox{16}.&\textrm{Determinan untuk matriks}\: \: \begin{pmatrix} 2 & -1&-1\\ 1 & 4&-1\\ 1&-2&3 \end{pmatrix}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&10&&&\textrm{d}.&30\\ \textrm{b}.&18&\textrm{c}.&22&\textrm{e}.&36 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\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)-(1.4.-1)\\ &\quad -(-2.-1.2)-(3.1.-1)\\ &=24+1+2+4-24+3\\ &=10 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika diketahu matriks}\: \: \textrm{A}=\begin{pmatrix} x+3&-2\\ -16&2x-6 \end{pmatrix},\: \: \textrm{maka nilai dari}\: \: \: x\\ &\textrm{supaya matriks A tidak memiliki invers adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&4\\ \textrm{b}.&2&\textrm{c}.&3&\textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Invers}&\: \textrm{dari matriks A adalah}\: \: \: \textrm{A}^{-1}.\\ \textrm{A}^{-1}&=\displaystyle \frac{1}{det\: \textrm{A}}\times Adjoin\: \textrm{A}.\\ \textrm{Karen}&\textrm{a}\: \textrm{tidak memiliki invers, 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\qquad (\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 \: \: \: \, \, x=-5\quad \textrm{atau}\quad x=5 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Diketahu matriks}\: \: \textrm{H}\: \: \textrm{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{d}.&1\\ \textrm{b}.&-2&\textrm{c}.&-1&\textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|c|}\hline \textbf{Alternatif 1}&\textbf{Alternatif 2}\\\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=1 \end{aligned}&\begin{aligned}\textrm{H.A}&=\textrm{B}\begin{cases} det\: \textrm{H} &=\left | \textrm{H} \right | \\ det\: \textrm{A} &=\left | \textrm{A} \right |=\begin{vmatrix} 3 & 4\\ 2 & 1 \end{vmatrix}=12-2=10 \\ det\: \textrm{B} &=\left | \textrm{B} \right |=\begin{vmatrix} 7 & 8\\ 4 & 6 \end{vmatrix}=42-32=10 \end{cases}\\ \left | \textrm{H} \right |.\left | \textrm{A} \right |&=\left | \textrm{B} \right |\\ \left | \textrm{H} \right |&=\displaystyle \frac{\left | \textrm{B} \right |}{\left | \textrm{A} \right |}\\ &=\displaystyle \frac{10}{10}\\ &=1\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{19}.&\textrm{Diketahu determinan suatu 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 yang memenuhipersamaantersebut},\\ &\textrm{maka nilai dari}\: \: \: p+q\: \: \textrm{adalah}....\\ &\begin{array}{llllllll}\\ \textrm{a}.&-3&&&\textrm{d}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&-\displaystyle \frac{1}{3}&\textrm{c}.&-1&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\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}\begin{cases} a &=3 \\ b &=-1 \\ c &=-10 \end{cases}.\\ \textrm{maka}\: \: \: p+q\: \: &=-\displaystyle \frac{b}{a}=-\displaystyle \frac{-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{20}.&\textrm{Diketahu matriks}\: \: \textrm{Z}=\begin{pmatrix} -1 & 2 & 1\\ 2 & -4 & 1\\ 1 & 2 & 2 \end{pmatrix}\\ &\textrm{Invers dari matriks}\: \: \: \textrm{Z}\: \: \textrm{tersebut adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{12}\begin{pmatrix} -10 & -2 & 6\\ -3 & -4 & 3\\ 1 & 3 & 0 \end{pmatrix}&\textrm{d}.&\displaystyle \frac{1}{12}\begin{pmatrix} -5 & -2 & 6\\ -4 & -4 & 1\\ 2 & -1 & 0 \end{pmatrix}\\ \textrm{b}.&\displaystyle \frac{1}{12}\begin{pmatrix} -10 & -2 & 6\\ -4 & 4 & 3\\ 8 & 4 & 0 \end{pmatrix}&\textrm{e}.&\displaystyle \frac{1}{12}\begin{pmatrix} -5 & -6 & 2\\ -3 & -3 & 3\\ 2 & 1 & 0 \end{pmatrix}\\ \textrm{c}.&\displaystyle \frac{1}{12}\begin{pmatrix} -10 & -2 & 6\\ -3 & -3 & 3\\ 8 & 4 & 0 \end{pmatrix} \end{array}\\\\  \end{array}.

Jawabc (Silahkan lengkapi sendiri untuk isiian yang belum lengkap)

Perhatikanlah, langkah awalnya kita tentukan determinan matriks Z dulu, yaitu:

\begin{aligned}det\: \textrm{Z}&=\left | \textrm{Z} \right |=\begin{vmatrix} -1 & 2 & 1\\ 2 & -4 & 1\\ 1 & 2 & 2 \end{vmatrix}\\ &=+(8)+2+4-(-4)-(-2)-8\\ &=2+4+4+2\\ &=12 \end{aligned}.

Selanjutanya kita tentukan kofaktor-kofaktornya.

\begin{array}{ll}\\ &\textrm{kofaktor-kofaktornya adalah}\\ &\begin{array}{lll}\\ K_{11}=(-1)^{1+1}.\begin{vmatrix} M_{11} \end{vmatrix}=....,&K_{12}=(-1)^{1+2}.\begin{vmatrix} M_{12} \end{vmatrix}=....,&K_{13}=(-1)^{1+3}.\begin{vmatrix} M_{13} \end{vmatrix}=....\\ K_{21}=(-1)^{2+1}.\begin{vmatrix} M_{21} \end{vmatrix}=....,&K_{22}=(-1)^{2+2}.\begin{vmatrix} M_{22} \end{vmatrix}=-3,&K_{23}=(-1)^{2+3}.\begin{vmatrix} M_{23} \end{vmatrix}=4\\ K_{31}=(-1)^{3+1}.\begin{vmatrix} M_{31} \end{vmatrix}=....,&K_{32}=(-1)^{3+2}.\begin{vmatrix} M_{32} \end{vmatrix}=....,&K_{33}=(-1)^{3+3}.\begin{vmatrix} M_{33} \end{vmatrix}=.... \end{array}\\ &\\ &\textrm{Jadi matriks kofaktornya adalah} \: \: \textrm{C}=\begin{pmatrix} K_{11} & K_{12} & K_{13}\\ K_{21} & K_{22} & K_{23}\\ K_{31} & K_{32} & K_{33} \end{pmatrix}=\begin{pmatrix} ... & ... & ...\\ ... & -3 & 4\\ ... & ... & ... \end{pmatrix} \end{array}..

\begin{aligned}\textbf{Contoh}&\: \textbf{hitungannya adalah sebagai berikut}:\\K_{22}&=+\begin{vmatrix} -1 & 1\\ 1 & 2 \end{vmatrix}=+\left ( (-2)-1 \right )=-3\\ K_{23}&=-\begin{vmatrix} -1 & 2\\ 1 & 2 \end{vmatrix}=-\left ( (-2)-2 \right )=4 \end{aligned}\\ \begin{array}{ll}\\ &\textrm{dan adjoin matriks Z-nya adalah}:\\ &\\ &\textrm{Adj}\: \: A=\begin{pmatrix} K_{11} & K_{21} & K_{31}\\ K_{12} & K_{22} & K_{32}\\ K_{13} & K_{23} & K_{33} \end{pmatrix}=\begin{pmatrix} ... & ... & ...\\ ... & -3 & ...\\ ... & 4 & ... \end{pmatrix} \end{array}..

\begin{aligned}\textrm{Sehingga}&\: \textrm{invers dari matriks Z tersebut adalah}:\\ Z^{-1}&=\displaystyle \frac{1}{det\: \textrm{Z}}\times \textrm{adjoin matriks}\: \textrm{Z}\\ &=\displaystyle \frac{1}{...}\times \begin{pmatrix} ... & ... & ...\\ ... & -3 & ...\\ ... & 4 & ... \end{pmatrix} \end{aligned}.

Sumber Referensi

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

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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