Tentukan dengan cara invers, cramer dan dauss yordan :

1) 2x – y = 0
x + 3y = 7
2) x + 2y – z = – 2
3x + 4z = 8
4x + 4y + z = 5

Tentukan dengan cara invers, cramer dan dauss yordan :

Jawaban Terkonfirmasi

Dengan metode invers:
(Sengaja disatukan antar metode untuk tidak menyulitkan)
Nomor 1.
$$begin{align}left[begin{array}{cc}2&-1 \ 1&3end{array}right]left[begin{array}{c}x\yend{array}right]&=left[begin{array}{c}0\7end{array}right] \ left[begin{array}{c}x\yend{array}right]&=left[begin{array}{cc}2&-1\1&3end{array}right]^{-1}left[begin{array}{c}0\7end{array}right] \ left[begin{array}{c}x\yend{array}right]&=frac{1}{6-(-1)}left[begin{array}{cc}3&1\-1&2end{array}right]left[begin{array}{c}0\7end{array}right] \ end{align}$
Sehingga, berlaku:
$left[begin{array}{c}x\yend{array}right]&=displaystylefrac{1}{7}left[begin{array}{cc}0+7 \ 0+14 end{array}right] \\ left[begin{array}{c}x\yend{array}right]=left[begin{array}{cc}1 \ 2 end{array}right]$

Nomor 2.
$$begin{align}left[begin{array}{ccc}1&2&-1 \ 3&0&4 \ 4&4&1end{array}right]left[begin{array}{c}x\y \ zend{array}right]&=left[begin{array}{c}-2\8 \ 5end{array}right] \ left[begin{array}{c}x\y \ zend{array}right]&=left[begin{array}{ccc}1&2&-1 \ 3&0&4 \ 4&4&1end{array}right]^{-1}left[begin{array}{c}-2\8\5end{array}right] \ &=-frac12left[begin{array}{ccc}-16&-6&8 \ 13&5&-7 \ 12&4&-6end{array}right]left[begin{array}{c}-2\8\5end{array}right] end{align}$
Note : Invers matriks 3 x 3 itu penjelasan ada di lampiran 1.
Sehingga, diperoleh:
$displaystyleleft[begin{array}{c}x\y \ zend{array}right]=-frac12left[begin{array}{ccc}32-48+40\ -26+40-35 \ -24+32-30end{array}right] \\ left[begin{array}{c}x\y \ zend{array}right]=-frac12left[begin{array}{ccc}24\ -21 \ -22end{array}right] \\ left[begin{array}{c}x\y \ zend{array}right]=left[begin{array}{ccc}-12\ 21/2 \ 11end{array}right]$

Dengan cramer.
Nomor 1.
$$begin{align}x&=frac{left|begin{array}{ccc}0&-1\7&3end{array}right|}{left|begin{array}{ccc}2&-1\1&3end{array}right|}=frac{0-(-7)}{6-(-1)}=frac77=1end{align}$
Serta:
$$begin{align}y&=frac{left|begin{array}{ccc}2&0\1&7end{array}right|}{left|begin{array}{ccc}2&-1\1&3end{array}right|}=frac{14-0}{6-(-1)}=frac{14}7=2end{align}$

Nomor 2.
$$begin{align}x&=frac{left|begin{array}{ccc}-2&2&-1\8&0&4 \ 5&4&1end{array}right|}{left|begin{array}{ccc}1&2&-1\3&0&4 \ 4&4&1end{array}right|}=frac{24}{-2}=-12end{align}$
Dan:
$$begin{align}y&=frac{left|begin{array}{ccc}1&-2&-1\3&8&4 \ 4&5&1end{array}right|}{left|begin{array}{ccc}1&2&-1\3&0&4 \ 4&4&1end{array}right|}=frac{-21}{-2}=frac{21}2end{align}$
Lalu:
$$begin{align}z&=frac{left|begin{array}{ccc}1&2&-2\3&0&8 \ 4&4&5end{array}right|}{left|begin{array}{ccc}1&2&-1\3&0&4 \ 4&4&1end{array}right|}=frac{-22}{-2}=11end{align}$

Untuk metode terakhir diserahkan pada 2 lampiran akhir yang tersedia.
#TulisanKecilDanKacauHiraukanSaja

Gambar Jawaban

Jawaban Terkonfirmasi

2x – y = 0
x + 3y = 7
metode invers
$left[begin{array}{ccc}2 &- 1\1&3\end{array}right]$ $left[begin{array}{ccc}x\y\end{array}right]$ = $left[begin{array}{ccc}0\7\end{array}right]$
$frac{1}{6 - (- 1)}$ $left[begin{array}{ccc}3&1\-1&2\end{array}right]$ $left[begin{array}{ccc}2&-1\1&3\end{array}right]$ . $left[begin{array}{ccc}x\y\end{array}right]$ = $frac{1}{6 - (-1)}$ . $left[begin{array}{ccc}3&1\-1&2\end{array}right]$ . $left[begin{array}{ccc}0\7\end{array}right]$
$left[begin{array}{ccc}1&0\0&1\end{array}right]$ . $left[begin{array}{ccc}x\y\end{array}right]$ = $frac{1}{7}$ . $left[begin{array}{ccc}7\14\end{array}right]$
$left[begin{array}{ccc}x\y\end{array}right]$ = $left[begin{array}{ccc}1\2\end{array}right]$
x = 1 ; y = 2
metode Cramer
x = $frac{ left[begin{array}{ccc}0&-1\7&3\end{array}right] }{ left[begin{array}{ccc}2&-1\1&3\end{array}right] }$
x = $frac{7}{7}$
x = 1
y = $frac{ left[begin{array}{ccc}2&0\1&7\end{array}right] }{ left[begin{array}{ccc}2&-1\1&3\end{array}right] }$
y = $frac{14}{7}$
y = 2