CONTOH SOAL MATRIKS DAN SPL

 $\begin{array}{ll}\\ 1.&\textrm{Diketahui persamaan}\begin{cases} x-y & =2 \\ kx+y & =3 \end{cases}\\ &\textrm{memiliki solusi}\: \: (x,y)\: \: \textrm{di kuadran I}\\ &\textrm{Jika dan hanya jika nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle k=-1\\ \textrm{b}.&k>-1\\ \textrm{c}.&k<\displaystyle \frac{3}{2}\\ \textrm{d}.&0<k<\displaystyle \frac{3}{2}\\ \textrm{e}.&-1<k<\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x-y=2\: \: \: \quad....(1)\\ kx+y=3\quad\: ....(2)\end{matrix}\right.\\ &\textrm{Dengan metode matriks didapatkan}\\ &x=\displaystyle \frac{\begin{vmatrix} 2 & -1\\ 3& 1 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}\\ &\textrm{Dengan cara yang sama pula}\\ &y=\displaystyle \frac{\begin{vmatrix} 1 & 2\\ k & 3 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{3-2k}{k+1}\\ &\textrm{Supaya memiliki solusi di kwadran I},\\ &\textrm{maka baik}\: \: x\: \: \textrm{maupun}\: \: y\\ &\textrm{haruslah positif, akibatnya}:\\ & k+1>0\Rightarrow k>-1\\ &\textrm{Sebagai akibat yang lain adalah}:\\ &3-2k>0\Rightarrow k<\displaystyle \frac{3}{2}\\ &\textrm{Jadi},\: \: -1<k<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{tidak berpenyelesaian, maka}\\ &\textrm{ingat penyelesaian metode matrik}\\ &\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Semunya dikerjakan dengan metode}\\ &\textrm{matriks}\: (\textbf{Cara Cramer})\\ &\begin{aligned} x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \textrm{Cara di atas}$  full matriks-Cramer