Tolong dong bagi yg bisa ntar aku follow dan aku

Berikut ini adalah pertanyaan dari windy2120 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tolong dong bagi yg bisa ntar aku follow dan aku kasih 25 poin

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Sistem Persamaan Linier (SPL) :

$x_1+x_2+2x_3=8$

$-x_1+2x_2+3x_3=1$

$3x_1-7x_2+4x_3=10$

Penyajian SPL dalam bentuk operasi matriks :

$\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=[\left[\begin{array}{ccc}8\\1\\10\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\boxed{\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right]^{-1}}[\left[\begin{array}{ccc}8\\1\\10\end{array}\right]$

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Dimisalkan : $\text{A}=\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right]$

$(~i~)~$ Menghitung determinan (lihat lampiran) :

$det\text{~(A)}$ = (1 × 2 × 4) + (1 × 3 × 3) + (2 × –1 × –7) – (2 × 2 × 3) – (1 × 3 × –7) – (1 × –1 × 4)

$det\text{~(A)}$ = 8 + 9 + 14 – 12 – (–21) – (–4)

$\boxed{\purple{det\text{~(A)}=44}}$

$\\$

$(~ii~)~$ Menentukan matriks Kofaktor :

$\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+\left|\begin{array}{ccc}2&3\\-7&4\end{array}\right|&-\left|\begin{array}{ccc}-1&3\\3&4\end{array}\right|&+\left|\begin{array}{ccc}-1&2\\3&-7\end{array}\right|\\-\left|\begin{array}{ccc}1&2\\-7&4\end{array}\right|&+\left|\begin{array}{ccc}1&2\\3&4\end{array}\right|&-\left|\begin{array}{ccc}1&1\\3&-7\end{array}\right|\\+\left|\begin{array}{ccc}1&2\\2&3\end{array}\right|&-\left|\begin{array}{ccc}1&2\\-1&3\end{array}\right|&+\left|\begin{array}{ccc}1&1\\-1&2\end{array}\right|\end{array}\right]$

$\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+(~(2\times 4)-(3\times -7)~)&-(~(-1\times 4)-(3\times 3)~)&+(~(-1\times -7)-(2\times 3)~)\\-(~(1\times 4)-(2\times -7)~)&+(1\times 4)-(2\times 3)~)&-(~(1\times -7)-(1\times 3)~)\\+(~(1\times 3)-(2\times 2)~)&-(~(1\times 3)-(2\times -1)~)&+(~(1\times 2)-(1\times -1)~)\end{array}\right]$

$\boxed{\purple{\text{Kofaktor~(A)}=\left[\begin{array}{ccc}29&13&1\\-18&-2&10\\-1&-5&3\end{array}\right]}}$

$\\$

$(~iii~)~$ Menentukan matriks $Adj.$ ( Adjoin ) :

$Adj.\text{~(X)}=\left(\text{~Kofaktor~(A)~}\right)^\text{T}$

$Adj.\text{~(A)}=\left[\begin{array}{ccc}29&13&1\\-18&-2&10\\-1&-5&3\end{array}\right]^\text{T}$

$\boxed{\purple{Adj.\text{~(A)}=\left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]}}$

$\\$

$(~iv~)~$ Menentukan matriks invers :

$\text{A}^{-1}=\frac{1}{det~\text{(A)}}\times Adj.\text{~(A)}$

$\boxed{\pink{\text{A}^{-1}=\frac{1}{44}\times \left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]}}$

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Maka :

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]\left[\begin{array}{ccc}8\\1\\10\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}(29\times 8)+(-18\times 1)+(-1\times 10)\\(13\times 8)+(-2\times 1)+(-5\times 10)\\(1\times 8)+(10\times 1)+(3\times 10)\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}204\\52\\48\end{array}\right]$

$\left[\begin{array}{ccc}x_1\\~\\x_2\\~\\x_3\end{array}\right]=\left[\begin{array}{ccc}\frac{204}{44}\\~\\\frac{52}{44}\\~\\\frac{48}{44}\end{array}\right]$

$\red{\huge{\left[\begin{array}{ccc}x_1\\~\\x_2\\~\\x_3\end{array}\right]=\left[\begin{array}{ccc}\frac{51}{11}\\~\\\frac{13}{11}\\~\\\frac{12}{11}\end{array}\right]}}$

$Sistem Persamaan Linier (SPL) :[tex]x_1+x_2+2x_3=8[/tex][tex]-x_1+2x_2+3x_3=1[/tex][tex]3x_1-7x_2+4x_3=10[/tex]Penyajian SPL dalam bentuk operasi matriks :[tex]\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=[\left[\begin{array}{ccc}8\\1\\10\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\boxed{\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right]^{-1}}[\left[\begin{array}{ccc}8\\1\\10\end{array}\right][/tex]•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Dimisalkan : [tex]\text{A}=\left[\begin{array}{ccc}1&1&2\\-1&2&3\\3&-7&4\end{array}\right][/tex][tex](~i~)~[/tex]Menghitung determinan (lihat lampiran) :[tex]det\text{~(A)}[/tex] = (1 × 2 × 4) + (1 × 3 × 3) + (2 × –1 × –7) – (2 × 2 × 3) – (1 × 3 × –7) – (1 × –1 × 4)[tex]det\text{~(A)}[/tex] = 8 + 9 + 14 – 12 – (–21) – (–4)[tex]\boxed{\purple{det\text{~(A)}=44}}[/tex][tex]\\[/tex][tex](~ii~)~[/tex]Menentukan matriks Kofaktor :[tex]\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+\left|\begin{array}{ccc}2&3\\-7&4\end{array}\right|&-\left|\begin{array}{ccc}-1&3\\3&4\end{array}\right|&+\left|\begin{array}{ccc}-1&2\\3&-7\end{array}\right|\\-\left|\begin{array}{ccc}1&2\\-7&4\end{array}\right|&+\left|\begin{array}{ccc}1&2\\3&4\end{array}\right|&-\left|\begin{array}{ccc}1&1\\3&-7\end{array}\right|\\+\left|\begin{array}{ccc}1&2\\2&3\end{array}\right|&-\left|\begin{array}{ccc}1&2\\-1&3\end{array}\right|&+\left|\begin{array}{ccc}1&1\\-1&2\end{array}\right|\end{array}\right][/tex][tex]\text{Kofaktor~(A)}=\left[\begin{array}{ccc}+(~(2\times 4)-(3\times -7)~)&-(~(-1\times 4)-(3\times 3)~)&+(~(-1\times -7)-(2\times 3)~)\\-(~(1\times 4)-(2\times -7)~)&+(1\times 4)-(2\times 3)~)&-(~(1\times -7)-(1\times 3)~)\\+(~(1\times 3)-(2\times 2)~)&-(~(1\times 3)-(2\times -1)~)&+(~(1\times 2)-(1\times -1)~)\end{array}\right][/tex][tex]\boxed{\purple{\text{Kofaktor~(A)}=\left[\begin{array}{ccc}29&13&1\\-18&-2&10\\-1&-5&3\end{array}\right]}}[/tex][tex]\\[/tex][tex](~iii~)~[/tex]Menentukan matriks [tex]Adj.[/tex] ( Adjoin ) :[tex]Adj.\text{~(X)}=\left(\text{~Kofaktor~(A)~}\right)^\text{T}[/tex][tex]Adj.\text{~(A)}=\left[\begin{array}{ccc}29&13&1\\-18&-2&10\\-1&-5&3\end{array}\right]^\text{T}[/tex][tex]\boxed{\purple{Adj.\text{~(A)}=\left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]}}[/tex][tex]\\[/tex][tex](~iv~)~[/tex]Menentukan matriks invers :[tex]\text{A}^{-1}=\frac{1}{det~\text{(A)}}\times Adj.\text{~(A)}[/tex][tex]\boxed{\pink{\text{A}^{-1}=\frac{1}{44}\times \left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]}}[/tex]•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Maka :[tex]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}29&-18&-1\\13&-2&-5\\1&10&3\end{array}\right]\left[\begin{array}{ccc}8\\1\\10\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}(29\times 8)+(-18\times 1)+(-1\times 10)\\(13\times 8)+(-2\times 1)+(-5\times 10)\\(1\times 8)+(10\times 1)+(3\times 10)\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=\frac{1}{44}\times \left[\begin{array}{ccc}204\\52\\48\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x_1\\~\\x_2\\~\\x_3\end{array}\right]=\left[\begin{array}{ccc}\frac{204}{44}\\~\\\frac{52}{44}\\~\\\frac{48}{44}\end{array}\right][/tex][tex]\red{\huge{\left[\begin{array}{ccc}x_1\\~\\x_2\\~\\x_3\end{array}\right]=\left[\begin{array}{ccc}\frac{51}{11}\\~\\\frac{13}{11}\\~\\\frac{12}{11}\end{array}\right]}}[/tex]$

Semoga dengan pertanyaan yang sudah terjawab oleh WillyJember dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

Apabila terdapat kesalahan dalam mengerjakan soal, silahkan koreksi jawaban dengan mengirimkan email ke yomemimo.com melalui halaman Contact

Last Update: Wed, 21 Jul 21

<< Previous Post

Next Post >>

Kerjakan soal, tugas, & PR sekolah semakin mudah

Tolong dong bagi yg bisa ntar aku follow dan aku

Jawaban dan Penjelasan

Video Tutorial Android dan Smartphone

Pertanyaan Lain Mata pelajaran Matematika