SBMPTN Zone : Sistem Persamaan Linear Tiga Variabel

Berikut ini adalah kumpulan soal mengenai Sistem Persamaan Linear Tiga Variabel tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN

No. 1

Jika 3x+y+z=k, x+3y+z=k, x+y+3z=k, k\neq0, maka x^2+y^2+z^2 jika dinyatakan dalam k adalah ....

1. \dfrac{k^2}{25}
2. \dfrac{3k^2}{25}
3. \dfrac{k^2}5

4. \dfrac{9^2}{16}k^2
5. \dfrac3{15}k^2

Ganesha Operation

$\begin{array}{rl}3x+y+z& =x+3y+z\\ x& =y\end{array}$

$\begin{array}{rl}x+3y+z& =x+y+3z\\ y& =z\end{array}$

$\begin{array}{rl}3x+y+z& =k\\ 5x& =k\\ x& ={\displaystyle \frac{k}{5}}\end{array}$
x=y=z=\dfrac{k}5

$\begin{array}{rl}{x}^2+y^2+z^2& ={\left({\displaystyle \frac{k}{5}}\right)}^2+{\left({\displaystyle \frac{k}{5}}\right)}^2+{\left({\displaystyle \frac{k}{5}}\right)}^2\\ & ={\displaystyle \frac{k^2}{25}}+{\displaystyle \frac{k^2}{25}}+{\displaystyle \frac{k^2}{25}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3k^2}{25}}}}}}\end{array}$

Jadi, x^2+y^2+z^2=dfrac{3k^2}{25}.
JAWAB: B

No. 2

Jika (a+b,a,b) memenuhi sistem persamaan:
3x-y+2z=-1
-2x+y+3z=-3
maka a+b=

1. 0
2. 1
3. 2

4. 3
5. 4

$\begin{array}{rl}3x-y+2z& =-1\\ 3(a+b)-a+2b& =-1\\ 3a+3b-a+2b& =-1\\ 2a+5b& =-1\end{array}$

$\begin{array}{rl}-2x+y+3z& =-3\\ -2(a+b)+a+3b& =-3\\ -2a-2b+a+3b& =-3\\ -a+b& =-3\\ -2a+2b& =-6\end{array}$

$\begin{array}{rl}2a+5b& =-1\\ -2a+2b& =-6& +\\ 7b& =-7\\ b& =-1\end{array}$

$\begin{array}{rl}-a+b& =-3\\ -a+(-1)& =-3\\ -a-1& =-3\\ -a& =-3+1\\ & =-2\\ a& =2\end{array}$

$\begin{array}{rl}a+b& =2+(-1)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, a+b=1.
JAWAB: B