Getal & Ruimte (13e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}3 x + 5 y = -1 \\ 5 x - 5 y = 5\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 318ms - dynamic variables a Optellen geeft \(8 x = 4 \text{,}\) dus \(x = \frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}3 x + 5 y = -1 \\ x = \frac{1}{2}\end{rcases} \begin{matrix}3 ⋅ \frac{1}{2} + 5 y = -1 \\ 5 y = -2\frac{1}{2} \\ y = -\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (\frac{1}{2} , -\frac{1}{2}) \text{.}\) 1p 4p b \(\begin{cases}4 p + 2 q = 4 \\ 5 p + q = -4\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 10ms - dynamic variables b \(\begin{cases}4 p + 2 q = 4 \\ 5 p + q = -4\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 p + 2 q = 4 \\ 10 p + 2 q = -8\end{cases}\) 1p ○ Aftrekken geeft \(-6 p = 12 \text{,}\) dus \(p = -2 \text{.}\) 1p ○ \(\begin{rcases}4 p + 2 q = 4 \\ p = -2\end{rcases} \begin{matrix}4 ⋅ -2 + 2 q = 4 \\ 2 q = 12 \\ q = 6\end{matrix}\) 1p ○ De oplossing is \((p , q) = (-2 , 6) \text{.}\) 1p 4p c \(\begin{cases}4 a - 6 b = -3 \\ 3 a - 5 b = 4\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 11ms - dynamic variables c \(\begin{cases}4 a - 6 b = -3 \\ 3 a - 5 b = 4\end{cases}\) \(\begin{vmatrix}5 \\ 6\end{vmatrix}\) geeft \(\begin{cases}20 a - 30 b = -15 \\ 18 a - 30 b = 24\end{cases}\) 1p ○ Aftrekken geeft \(2 a = -39 \text{,}\) dus \(a = -19\frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}4 a - 6 b = -3 \\ a = -19\frac{1}{2}\end{rcases} \begin{matrix}4 ⋅ -19\frac{1}{2} - 6 b = -3 \\ -6 b = 75 \\ b = -12\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-19\frac{1}{2} , -12\frac{1}{2}) \text{.}\) 1p |