--- title: Cramer's rule id: 59713da0a428c1a62d7db430 challengeType: 5 videoUrl: '' localeTitle: 克莱默的统治 --- ## Description

在线性代数中Cramer规则是一个线性方程组解的显式公式,其中包含与未知数一样多的方程,只要系统具有唯一解,就有效。它通过用方程右边的矢量替换一列来表示(方形)系数矩阵的决定因素和从它获得的矩阵的解决方案。

特定

$ \ left \ {\ begin {matrix} a_1x + b_1y + c_1z&= {\ color {red} d_1} \\ a_2x + b_2y + c_2z&= {\ color {red} d_2} \\ a_3x + b_3y + c_3z&= {\颜色{红} D_3} \ {结束矩阵} \权。$

以矩阵格式表示

$ \ begin {bmatrix} a_1&b_1&c_1 \\ a_2&b_2&c_2 \\ a_3&b_3&c_3 \ end {bmatrix} \ begin {bmatrix} x \\ y \\ z \ end {bmatrix} = \ begin {bmatrix} {\ color {red} d_1} \\ {\ color {red} d_2} \\ {\ color {red} d_3} \ end {bmatrix}。$

然后可以找到$ x,y $和$ z $的值,如下所示:

$ x = \ frac {\ begin {vmatrix} {\ color {red} d_1}&b_1&c_1 \\ {\ color {red} d_2}&b_2&c_2 \\ {\ color {red} d_3}&b_3& c_3 \ end {vmatrix}} {\ begin {vmatrix} a_1&b_1&c_1 \\ a_2&b_2&c_2 \\ a_3&b_3&c_3 \ end {vmatrix}},\ quad y = \ frac {\ begin {vmatrix } a_1&{\ color {red} d_1}&c_1 \\ a_2&{\ color {red} d_2}&c_2 \\ a_3&{\ color {red} d_3}&c_3 \ end {vmatrix}} {\ begin {vmatrix} a_1&b_1&c_1 \\ a_2&b_2&c_2 \\ a_3&b_3&c_3 \ end {vmatrix}},\ text {和} z = \ frac {\ begin {vmatrix} a_1&b_1&{\ color {red} d_1} \\ a_2&b_2&{\ color {red} d_2} \\ a_3&b_3&{\ color {red} d_3} \ end {vmatrix}} {\ begin {vmatrix} a_1&b_1& c_1 \\ a_2&b_2&c_2 \\ a_3&b_3&c_3 \ end {vmatrix}}。$

任务

给定以下方程组:

$ \ begin {例} 2w-x + 5y + z = -3 \\ 3w + 2x + 2y-6z = -32 \\ w + 3x + 3y-z = -47 \\ 5w-2x-3y + 3z = 49 \\ \ end {cases} $

使用Cramer的规则解决$ w $,$ x $,$ y $$ z $

## Instructions
## Tests
```yml tests: - text: cramersRule是一个函数。 testString: 'assert(typeof cramersRule === "function", "cramersRule is a function.");' - text: 'cramersRule([[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3]], [-3, -32, -47, 49])应返回[2, -12, -4, 1] 。' testString: 'assert.deepEqual(cramersRule(matrices[0], freeTerms[0]), answers[0], "cramersRule([[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3]], [-3, -32, -47, 49]) should return [2, -12, -4, 1].");' - text: 'cramersRule([[3, 1, 1], [2, 2, 5], [1, -3, -4]], [3, -1, 2])应返回[1, 1, -1] 。' testString: 'assert.deepEqual(cramersRule(matrices[1], freeTerms[1]), answers[1], "cramersRule([[3, 1, 1], [2, 2, 5], [1, -3, -4]], [3, -1, 2]) should return [1, 1, -1].");' ```
## Challenge Seed
```js function cramersRule (matrix, freeTerms) { // Good luck! return true; } ```
### After Test
```js console.info('after the test'); ```
## Solution
```js // solution required ```