Getal & Ruimte (13e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}p + 3 q = -1 \\ 5 p - 3 q = 4\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 306ms - dynamic variables a Optellen geeft \(6 p = 3 \text{,}\) dus \(p = \frac{1}{2} \text{.}\) 1p ○ \(\begin{rcases}p + 3 q = -1 \\ p = \frac{1}{2}\end{rcases} \begin{matrix}\frac{1}{2} + 3 q = -1 \\ 3 q = -1\frac{1}{2} \\ q = -\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((p , q) = (\frac{1}{2} , -\frac{1}{2}) \text{.}\) 1p 4p b \(\begin{cases}4 x - 5 y = -4 \\ 2 x - 3 y = 3\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 12ms - dynamic variables b \(\begin{cases}4 x - 5 y = -4 \\ 2 x - 3 y = 3\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}4 x - 5 y = -4 \\ 4 x - 6 y = 6\end{cases}\) 1p ○ Aftrekken geeft \(y = -10 \text{.}\) 1p ○ \(\begin{rcases}4 x - 5 y = -4 \\ y = -10\end{rcases} \begin{matrix}4 x - 5 ⋅ -10 = -4 \\ 4 x = -54 \\ x = -13\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x , y) = (-13\frac{1}{2} , -10) \text{.}\) 1p 4p c \(\begin{cases}5 a - 4 b = 4 \\ 4 a - 3 b = -1\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 11ms - dynamic variables c \(\begin{cases}5 a - 4 b = 4 \\ 4 a - 3 b = -1\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}15 a - 12 b = 12 \\ 16 a - 12 b = -4\end{cases}\) 1p ○ Aftrekken geeft \(-a = 16 \text{,}\) dus \(a = -16 \text{.}\) 1p ○ \(\begin{rcases}5 a - 4 b = 4 \\ a = -16\end{rcases} \begin{matrix}5 ⋅ -16 - 4 b = 4 \\ -4 b = 84 \\ b = -21\end{matrix}\) 1p ○ De oplossing is \((a , b) = (-16 , -21) \text{.}\) 1p |