Getal & Ruimte (12e editie) - vwo wiskunde C
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}4x-y=3 \\ 5x-y=-2\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - 361ms - dynamic variables a Aftrekken geeft \(-x=5\text{,}\) dus \(x=-5\text{.}\) 1p ○ \(\begin{rcases}4x-y=3 \\ x=-5\end{rcases}\begin{matrix}4⋅-5-y=3 \\ -y=23 \\ y=-23\end{matrix}\) 1p ○ De oplossing is \((x, y)=(-5, -23)\text{.}\) 1p 4p b \(\begin{cases}2a+b=2 \\ a-2b=6\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - 15ms - dynamic variables b \(\begin{cases}2a+b=2 \\ a-2b=6\end{cases}\) \(\begin{vmatrix}2 \\ 1\end{vmatrix}\) geeft \(\begin{cases}4a+2b=4 \\ a-2b=6\end{cases}\) 1p ○ Optellen geeft \(5a=10\text{,}\) dus \(a=2\text{.}\) 1p ○ \(\begin{rcases}2a+b=2 \\ a=2\end{rcases}\begin{matrix}2⋅2+b=2 \\ b=-2\end{matrix}\) 1p ○ De oplossing is \((a, b)=(2, -2)\text{.}\) 1p 4p c \(\begin{cases}4a+4b=2 \\ 6a+3b=-3\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - 12ms - dynamic variables c \(\begin{cases}4a+4b=2 \\ 6a+3b=-3\end{cases}\) \(\begin{vmatrix}3 \\ 4\end{vmatrix}\) geeft \(\begin{cases}12a+12b=6 \\ 24a+12b=-12\end{cases}\) 1p ○ Aftrekken geeft \(-12a=18\text{,}\) dus \(a=-1\frac{1}{2}\text{.}\) 1p ○ \(\begin{rcases}4a+4b=2 \\ a=-1\frac{1}{2}\end{rcases}\begin{matrix}4⋅-1\frac{1}{2}+4b=2 \\ 4b=8 \\ b=2\end{matrix}\) 1p ○ De oplossing is \((a, b)=(-1\frac{1}{2}, 2)\text{.}\) 1p |