高二（理科）数学 | 1.3 Cramer’s Rule 克兰姆法则 - 独中课程 | 线上补习

高二数学 | 高级数学

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1.3 Cramer’s Rule 克兰姆法则

Solve the following simultaneous equations with Cramer’s Rule
用克兰姆法则解以下各联立方程组

[1]

$\begin{cases}2x - y + 3 = 0 \\ x + 3y - 2 = 0 \end{cases}$

Answer:

(-1, 1)

[2]

$\begin{cases} 2x - 2y + 3z + 1= 0 \\ x + 3y - 2z - 3 = 0 \\ 3x - 4y + z - 2 = 0 \end{cases}$

Answer:

(1, 0, –1)

[3]

$\begin{cases}3x - 2y + 3z = 9 \\ x + 5y - z = 10 \\ 2x - 4y + z = -13 \end{cases}$

Answer:

(–2, 3, 3)

[4]

$\begin{cases}3x = 2y + 3z \\ x - 3y = - z - 3 \\ 2x - 4y + 5z - 6 = 0 \end{cases}$

Answer:

(4, 3, 2)

[5]

$\begin{cases} \frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2 \\ \frac{5}{x}-\frac{3}{y}-\frac{2}{z}=4 \\ \frac{10}{x}-\frac{4}{z}=2\end{cases}$

Answer:

(1, –1, $\frac{1}{2}$ )

[6]

$\begin{cases} \frac{1}{x+1}+\frac{2}{y-2}-\frac{2}{z}=0 \\ \frac{2}{x+1}-\frac{3}{y-2}+\frac{2}{z}=4 \\ \frac{15}{x+1}+\frac{8}{y-2}-\frac{4}{z}=9 \end{cases}$

Answer:

(0, 1, –2)