课程： 3.4 Cramer规则 | The Way To Learn

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选择节Cramer 规则

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Cramer 规则
Given any matrix $\begin{align*}A\end{align*}$ , the matrix $\begin{align*}A_{i}(\vec{b})\end{align*}$ is the matrix constructed by matrix $\begin{align*}A\end{align*}$ and replacing $\begin{align*}i\end{align*}$ with $\begin{align*}\vec{b}\end{align*}$ so if $\begin{align*}A = [\vec{a_{1}}\cdots \vec{a_{i}}\cdots \vec{a_{n}}] \to A_{i}(\vec{b}) = [\vec{a_{1}}\cdots \vec{b}\cdots \vec{a_{n}}]\end{align*}$ .
::就任何矩阵A而言,矩阵Ai(b____)是由矩阵A构建的矩阵,如果A=[a1][a1]-ai-an] i(b__)=[a1]-bban__],则以 b__取代 i。

Now, let's go to a rule you've probably seen in high school, Cramer's rule. This rule states that given a system of linear equations represented by the matrix-vector product $\begin{align*}A\vec{x} = \vec{b}\end{align*}$ we get that the solution vector $\begin{align*}\vec{x}\end{align*}$ has the form $\begin{align*}\vec{x} = A^{-1}\vec{b}\end{align*}$ the vector $\begin{align*}\vec{x}\end{align*}$ has the representation
::现在,让我们来谈谈你可能在高中时看到的一条规则, Cramer 的规则。该规则指出, 给以矩阵- 矢量产品Ax\\\ {b} 代表的线性方程系统, 我们得到的答案矢量 x_ 具有 xA- 1b} 矢量 x_ 具有代表的窗体

$\begin{align*}\vec{x} = \begin{bmatrix}x_{1}\\\cdot\\\cdot\\\cdot\\x_{n}\end{bmatrix} \\ x_{i} = \frac{\det(A_{i}(\vec{b}))}{\det(A)}\end{align*}$
::x[x1]xn]xI=det(Ai(b))det(A)

Let's look at the proof for this theorem :
::让我们看看这个理论的证明:

$\begin{align*}I = \begin{bmatrix}\vec{e_{1}}&\cdots&\vec{e_{i}}&\cdots&\vec{e_{n}}\end{bmatrix}\end{align*}$ where $\begin{align*}I\end{align*}$ is the identity matrix and the column vectors $\begin{align*}\vec{e_{i}}\end{align*}$ are the standard basis vectors for $\begin{align*}\mathbb{R}^{n}\end{align*}$
::I=[e1]=[e1]=身份矩阵和矢量列为Rn标准基矢量的I=[e1]=身份矩阵和矢量列为Rn标准基矢量的I=Rn

Now, by definition we can have that $\begin{align*}I_{i}(\vec{x}) = \begin{bmatrix}\vec{e_{1}}\cdots\vec{e_{i-1}}&\vec{x}\cdots\vec{e_{n}}\end{bmatrix}\end{align*}$ and $\begin{align*}A\cdot I_{i}(\vec{x}) = A[\vec{e_{1}}\cdots\vec{x}\cdots\vec{e_{n}}] \\ A\cdot I_{i}(\vec{x}) = [A\vec{e_{1}}\cdots A\vec{x}\cdots A\vec{e_{n}}] \\ A\cdot I_{i}(\vec{x}) = [\vec{a_{1}}\cdots A\vec{x}\cdots \vec{a_{n}}] \\ A\cdot I_{i}(\vec{x}) = [\vec{a_{1}}\cdots\vec{b}\cdots\vec{a_{n}}] \\ A\cdot I_{i}(\vec{x}) = A_{i}(\vec{b}) \\ \det(A\cdot I_{i}(\vec{x})) = \det(A_{i}(\vec{b})) \\ \det(A)\cdot\det(I_{i}(\vec{x})) = \det(A_{i}(\vec{b})) \\ \det(I_{i}(\vec{x})) = \det([\vec{e_{1}}\cdots\vec{x}\cdots\vec{e_{n}}]) \\ \det(I_{i}(\vec{x})) = x_{i} \\ \det(A)\cdot x_{i} = \det(A_{i}(\vec{b})) \\ x_{i} = \frac{\det(A_{i}(\vec{b}))}{\det(A)}\end{align*}$
::现在,根据定义,我们可以有I(x) = [e1-1-xen] 和 AI(x) = A[e1x = A) AI(x) = AI(x) = [a1-Ax ] AIi(x) = [a1_B an} AI(x) = AI(b) diet (AI(b) ) =det(A) diet(A) = deit(Ai(b) ) diet(A) id(A) = dett([e1x )) deti(I(x) ) ) det(i(i(x) ) =x(A) xx(A) =det(A) exi=de(A)) ex=de(A)) (A)))) dede(A) (A))))))))

Let's try an example of this
::让我们试一试这个例子

$\begin{align*}3x_{1} - 2x_{2} = 4 \\ -x_{1} + 5x_{2} = 7 \\ A = \begin{bmatrix}3&-2\\-1&5\end{bmatrix} \\ \vec{x} = \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} \\ \vec{b} = \begin{bmatrix}4\\7\end{bmatrix} \\ A\vec{x} = \vec{b} \\ \begin{bmatrix}3&-2\\-1&5\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = \begin{bmatrix}4\\7\end{bmatrix} \\ \det(A) = \det\bigg(\begin{bmatrix}3&-2\\-1&5\end{bmatrix}\bigg) \\ \det(A) = 3(5) - (-1)(-2) \\ \det(A) = 15 - 2 = 13 \\ A_{1}(\vec{b}) = \begin{bmatrix}4&-2\\7&5\end{bmatrix} \\ \det(A_{1}(\vec{b})) = 4(5) - (-2)(7) \\ \det(A_{1}(\vec{b})) = 34 \\ x_{1} = \frac{34}{13} \\ A_{2}(\vec{b}) = \begin{bmatrix}3&4\\-1&7\end{bmatrix} \\ \det(A_{2}\vec{b}) = 3(7) - (4)(-1) \\ \det(A_{2}(\vec{b})) = 25 \\ x_{2} = \frac{25}{13} \\ \vec{x} = \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = \begin{bmatrix}\frac{34}{13}\\\frac{25}{13}\end{bmatrix} \end{align*}$
::3x1-2x2=4-x1+5x2=7A=[3-2-15]x[x1x2]{[47]Ax{{{{{{{{{{{{{}}{[3-2--2-15}}{[x1x2]=[47]det([3-2-2--2-15])det([3-A)=3-(1)(1)(1)(1)(1)(2)(2)(2)(2)(2)det(A)=152=13A1(1)(2-275})d(A1(b)())))=4(5)-(2(7)(7(7)(A1)(2(b))=34x1(3-13(A2)(17)det(A2(2)(2-))=25x2=2513x*[x2][x1x2]=[34132-5133]

Now, that we've seen an example using a 2x2 matrix, let's look at some higher dimensional examples and the we can use Cramer's rule to again derive the formula for the inverse of a matrix A.
::现在,我们已经看到一个使用 2x2 矩阵的例子, 让我们看看一些更高维的示例, 我们可以使用 Cramer 的规则再次得出矩阵 A 的反向公式。

$\begin{align*}3x_{1} - 2x_{2} + 4x_{3} = 7 \\ -1x_{1} + 4x_{2} - x_{3} = 2 \\ 2x_{1} + 5x_{2} - 3x_{3} = -1 \\ A = \begin{bmatrix}3 & -2&4\\-1&4&-1\\2&5&-3\end{bmatrix} \\ \vec{x} = \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix} \\ \vec{b} = \begin{bmatrix}7\\2\\-1\end{bmatrix} \\ A\vec{x} = \vec{b} \\ \begin{bmatrix}3&-2&4\\-1&4&-1\\2&5&-3\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix} = \begin{bmatrix}7\\2\\-1\end{bmatrix} \\ \det(A) = \det\Bigg(\begin{bmatrix}3&-2&4\\-1&4&-1\\2&5&-3\end{bmatrix}\Bigg) \\ \det(A) = 3\det\bigg(\begin{bmatrix}4&-1\\5&-3\end{bmatrix}\bigg) + 2\det\bigg(\begin{bmatrix}-1&-1\\2&-3\end{bmatrix}\bigg) + 4\det\bigg(\begin{bmatrix}-1&4\\2&5\end{bmatrix}\bigg) \\ \det(A) = 3(4(-3) - 5(-1)) + 2(-3(-1) - (-1(2))) + 4((-1)(5) - (4)(2)) \\ \det(A) = 3(-7) + 2(5) + 4(-13) \\ \det(A) = -21 + 10 - 52 \\ \det(A) = -63 \\ A_{1}(\vec{b}) = \begin{bmatrix}7&-2&4\\2&4&-1\\-1&5&-3\end{bmatrix} \\ \det(A_{1}(\vec{b})) = \det\Bigg(\begin{bmatrix}7&-2&4\\2&4&-1\\-1&5&-3\end{bmatrix}\Bigg) \\ \det(A_{1}(\vec{b})) = 7(4(-3) - 5(-1)) - (-2)(2(-3) - (-1)(-1)) + 4(2(5) - 4(-1)) \\ \det(A_{1}(\vec{b})) = -7 \\ A_{2}(\vec{b}) = \begin{bmatrix}3&7&4\\-1&2&-1\\2&-1&-3\end{bmatrix} \\ \det(A_{2}(\vec{b})) = \det\Bigg(\begin{bmatrix}3&7&4\\-1&2&-1\\2&-1&-3\end{bmatrix}\Bigg) \\ \det(A_{2}(\vec{b})) = 3(2(-3) - (-1)(-1)) - 7(-1(-3) - (-1)(2)) + 4((-1)(-3) - (-1)(2)) \\ \det(A_{2}(\vec{b})) = -68 \\ A_{3}(\vec{b}) = \begin{bmatrix}3&-2&7\\-1&4&2\\2&5&-1\end{bmatrix} \\ \det(A_{3}(\vec{b})) = \det\Bigg(\begin{bmatrix}3&-2&7\\-1&4&2\\2&5&-1\end{bmatrix}\Bigg) \\ \det(A_{3}(\vec{b})) = 3(4(-1) - (2)(5)) - (-2)(-1(-1) - (2)(2)) + 7((-1)(5) - (4)(2)) \\ \det(A_{3}(\vec{b})) = -139 \\ x_{1} = \frac{7}{63} \\ x_{2} = \frac{68}{63} \\ x_{3} = \frac{139}{63} \\ \vec{x} = \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix} = \begin{bmatrix}\frac{7}{63}\\\frac{68}{63}\\\frac{139}{63}\end{bmatrix}\end{align*}$
::3x1-2x2-2x2+4-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1x3-3-3-3-3-2-3-2-3xx1A=[3-24-14-125-3,3xx3,3x[7-2-1,1x2x3,b[3-24—4—4][[4-14-14—4][[A)=(3-24-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2--2-2-2-2--2-2-2-2-2-2--2-2-2-2--2-2-2-2-2--2--2-2--2-2--2-2--2-2--2--2--2--2--2--2--2--2--2--2--2--2--2--2---2--2---2------2-----------2-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Cramer 规则