Getal & Ruimte (12e editie) - vwo wiskunde A
'Stelsels oplossen'.
| vwo wiskunde A | k.1 Stelsels van lineaire vergelijkingen |
opgave 1Los exact op. 3p a \(\begin{cases}6a-4b=-5 \\ 6a-2b=5\end{cases}\) Eliminatie (1) 003f - Stelsels oplossen - basis - dynamic variables a Aftrekken geeft \(-2b=-10\text{,}\) dus \(b=5\text{.}\) 1p ○ \(\begin{rcases}6a-4b=-5 \\ b=5\end{rcases}\begin{matrix}6a-4⋅5=-5 \\ 6a=15 \\ a=2\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((a, b)=(2\frac{1}{2}, 5)\text{.}\) 1p 4p b \(\begin{cases}3a-2b=-4 \\ 2a-b=-5\end{cases}\) Eliminatie (2) 003g - Stelsels oplossen - basis - dynamic variables b \(\begin{cases}3a-2b=-4 \\ 2a-b=-5\end{cases}\) \(\begin{vmatrix}1 \\ 2\end{vmatrix}\) geeft \(\begin{cases}3a-2b=-4 \\ 4a-2b=-10\end{cases}\) 1p ○ Aftrekken geeft \(-a=6\text{,}\) dus \(a=-6\text{.}\) 1p ○ \(\begin{rcases}3a-2b=-4 \\ a=-6\end{rcases}\begin{matrix}3⋅-6-2b=-4 \\ -2b=14 \\ b=-7\end{matrix}\) 1p ○ De oplossing is \((a, b)=(-6, -7)\text{.}\) 1p 4p c \(\begin{cases}4x-2y=3 \\ 5x+3y=1\end{cases}\) Eliminatie (3) 003h - Stelsels oplossen - basis - dynamic variables c \(\begin{cases}4x-2y=3 \\ 5x+3y=1\end{cases}\) \(\begin{vmatrix}3 \\ 2\end{vmatrix}\) geeft \(\begin{cases}12x-6y=9 \\ 10x+6y=2\end{cases}\) 1p ○ Optellen geeft \(22x=11\text{,}\) dus \(x=\frac{1}{2}\text{.}\) 1p ○ \(\begin{rcases}4x-2y=3 \\ x=\frac{1}{2}\end{rcases}\begin{matrix}4⋅\frac{1}{2}-2y=3 \\ -2y=1 \\ y=-\frac{1}{2}\end{matrix}\) 1p ○ De oplossing is \((x, y)=(\frac{1}{2}, -\frac{1}{2})\text{.}\) 1p |