Getal & Ruimte (13e editie) - vwo wiskunde C

'Stelsels oplossen'.

vwo wiskunde A k.1 Stelsels van lineaire vergelijkingen

Stelsels oplossen (3)

opgave 1

Los 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

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