Course: 4.9 寻找矩阵的反面 | The Way To Learn

The Way To Learn

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 查找矩阵的反向

Collapse Expand
查找矩阵的反向
At your school book fair, you buy 3 paperback and 2 hardcovers. Your best friend buys 2 paperbacks and 4 hardcovers. What is the inverse of the matrix represented by this situation?
::在你的学校书展上,你买了3张纸背和2张硬封面。你最好的朋友买了2张纸背和4张硬封面。这种情况代表的矩阵的反面是什么?

Inverse of a Matrix
::矩阵的逆数

Recall that the multiplicative inverse of a real number is the reciprocal of the number and that the product of a number and its multiplicative inverse is the multiplicative identity , or 1. For example: $\begin{align*}\frac{3}{7} \times \frac{7}{3}=1\end{align*}$ . Now we need to define a multiplicative identity and a multiplicative inverse for a square matrix . For real numbers, 1 is considered the identity because we can multiply any number, $\begin{align*}a\end{align*}$ , by 1 and the result is $\begin{align*}a\end{align*}$ . In other words, the value of the number does not change. For matrices, the multiplicative inverse of a square matrix will be a square matrix in which the values of the main diagonal are 1 and the remaining values are all zero. The following are examples of identity matrices.
::回顾一个实际数字的倍数反比是数字的对等值,数字的产物及其倍数反比是多复制性身份,或1. 例如:37x73=1. 现在我们需要定义一个多复制性身份和一个方形矩阵的倍复制性反比。对于真实数字,1被视为身份,因为我们可以乘以任何数字,a乘以1,结果就是a。换句话说,数字的值不会改变。对于矩阵,一个平方矩阵的倍复制性反比是一个方矩阵,其中主对角矩阵的值为1,其余的值均为零。以下是身份矩阵的例子。

$\begin{align*}I= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\end{align*}$
::I=[1001]I=[1001001001]I=[10010010010010001]I=[100100100010010001]

The products below illustrate how we can multiply a matrix by the identity and the result will be the original matrix.
::以下产品说明我们如何能够根据特征乘以一个矩阵,结果将是原始矩阵。

Given:
::参照:

$\begin{align*}A= \begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix}, \ AI=\begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 2(1)+(-1)(0) & 2(0)+(-1)(1)\\ (-3)(1)+5(0) & (-3)(0)+5(1) \end{bmatrix} = \begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix}\end{align*}$
::A=[2-1-1-35],AI=[2-1-35][1001] =[2(1)+(-1)(0)(0)(0)+(-1)(1)(1)(1)(3)(3)(1)+5(0)(3)(0)+5(1)] =[2-1—35]

In fact, it does not matter which order we multiply by the identity matrix . In other words, $\begin{align*}AI=IA=A\end{align*}$ .
::事实上,用身份矩阵乘以哪个顺序并不重要,换句话说,是AI=IA=A。

Now that we have defined an identity matrix, we can determine an inverse matrix such that $\begin{align*}A(A^{-1})=(A^{-1})A=I\end{align*}$ .
::既然我们已经确定了身份矩阵,我们可以确定一个反向矩阵,如A(A-1)=(A-1)A=I。

The formula for finding the Inverse of a $\begin{align*}2 \times 2\end{align*}$ matrix is:
::查找 2x2 矩阵反向值的公式是:

Given:
::参照:

$\begin{align*}A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \ A^{-1}= \frac{1}{det[A]}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix},\end{align*}$
::A=[abcd],A-1=1det[A][d-b-ca]=1ad-bc[d-b-ca],

where $\begin{align*}ad-bc \ne 0\end{align*}$
:bc)0 和(bc)%0

$\begin{align*}^*\end{align*}$ Note: If $\begin{align*}ad-bc = 0\end{align*}$ or $\begin{align*}det[A]=0\end{align*}$ , the matrix $\begin{align*}A\end{align*}$ is called singular . The inverse of a singular matrix cannot be determined.
::* 注:如果 ad-bc=0或det[A]=0,则矩阵A称为单数。单数矩阵的反面无法确定。

Let's find the inverse of the following matrices and verify our solution is correct (if the inverse exists).
::让我们找出以下矩阵的逆差, 并验证我们的解决方案是正确的( 如果逆差存在的话 ) 。

$\begin{align*}\begin{bmatrix} 1 & 2\\ -3 & 7 \end{bmatrix}\end{align*}$

First, use the formula above to find the inverse.
::首先,使用上面的公式查找反向公式。

$\begin{align*}\begin{bmatrix} 1 & 2\\ -3 & 7 \end{bmatrix}^{-1} = \frac{1}{(1)(7)-(2)(-3)} \begin{bmatrix} 7 & -2\\ 3 & 1 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 7 & -2\\ 3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{7}{13} & -\frac{2}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}\end{align*}$

To verify that this is indeed the inverse, we must show that the product of the inverse and the original matrix is the identity matrix for a $\begin{align*}2 \times 2\end{align*}$ matrix. It will be easier to find this product using the form of the inverse where the reciprocal of the determinant has not been distributed inside the matrix as shown below:
::为了证实这确实是反向的,我们必须证明反向和原始矩阵的产物是2x2矩阵的身份矩阵。

$\begin{align*}\frac{1}{13} \begin{bmatrix} 7 & 2\\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2\\ 3 & 7 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} (7)(1)+(-2)(-3) & (7)(-2)+(2)(7)\\ (-3)(1)+(1)(3) & (-3)(-2)+(1)(7) \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 13 & 0\\ 0 & 13 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}$

Are

$\begin{align*}\begin{bmatrix} 4 & -3\\ -3 & 2 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} -2 & -3\\ -3 & -4 \end{bmatrix}\end{align*}$ inverses?
::[4-3-3-32]和[-2-3-3-3-4]是否相反?

If the matrices are inverses then their product will be the identity matrix.
::如果矩阵反转,其产品将是身份矩阵。

$\begin{align*}\begin{bmatrix} 4 & -3\\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 & -3\\ -3 & -4 \end{bmatrix}= \begin{bmatrix} (4)(-2)+(-3)(-3) & (4)(-3)+(-3)(-4)\\ (-3)(-2)+(2)(-3) & (-3)(-3)+(2)(-4) \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}$

Since the product is the identity matrix, the matrices are inverses of one another.
::由于产品是身份矩阵,这些矩阵是相互反向的。

$\begin{align*}\begin{bmatrix} 4 & 6\\ -2 & -3 \end{bmatrix}.\end{align*}$

Use the formula above to find the inverse.
::使用上面的公式查找反向值。

$\begin{align*}\begin{bmatrix} 4 & 6\\ -2 & -3 \end{bmatrix}^{-1} = \frac{1}{(4)(-3)-(6)(-2)} \begin{bmatrix} -3 & -6\\ 2 & 4 \end{bmatrix} = \frac{1}{0} \begin{bmatrix} -3 & -6\\ 2 & 4 \end{bmatrix} \Rightarrow \ & \text{the inverse does not exist.}\\ & \text{This matrix is singular.}\end{align*}$
::[46-2-3]-1=1(4)(-3)-(6)(-2)[-3-624]=10[-3-624]=10[-3-624]}* 反之则不存在。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the inverse of the matrix that represents the book fair purchases.
::早些时候,你被要求找到代表书集购买的矩阵的反面。

The matrix that represents this situation is:

$\begin{align*}\begin{bmatrix} 3 & 2\\ 2 & 4 \end{bmatrix}\end{align*}$
::反映这种情况的矩阵表是:[3224]

Use the formula you learned in this lesson to find the inverse.
::使用您在此课中学习的公式查找反向。

$\begin{align*}\begin{bmatrix} 3 & 2\\ 2 & 4 \end{bmatrix}^{-1} = \frac{1}{[(3)(4)-(2)(2)]} \begin{bmatrix} 4 & -2\\ -2 & 3 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} 4 & -2\\ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{4}\\ -\frac{1}{4} & \frac{3}{8} \end{bmatrix}\end{align*}$

Example 2
::例2

Are matrices

$\begin{align*}\begin{bmatrix} -\frac{1}{3} & 1\\ -\frac{1}{2} & 2 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} -12 & 6\\ -3 & 2 \end{bmatrix}\end{align*}$ inverses of each other?

$\begin{align*}\begin{bmatrix} -\frac{1}{3} & 1\\ -\frac{1}{2} & 2 \end{bmatrix} \begin{bmatrix} -12 & 6\\ -3 & 2 \end{bmatrix} = \begin{bmatrix} \left(-\frac{1}{3}\right)(-12)+(1)(-3) & \left(-\frac{1}{3}\right)(6)+(1)(2) \\ \left(-\frac{1}{2}\right)(-12)+(2)(-3) & \left(-\frac{1}{2}\right)(6)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\end{align*}$

Yes, they are inverses.
::是的,他们是反悔的。

Example 3
::例3

Find the inverse of
::查找反向

$\begin{align*}\begin{bmatrix} 4 & 2\\ 10 & 5 \end{bmatrix}.\end{align*}$

$\begin{align*} \begin{bmatrix} 4 & 2\\ 10 & 5 \end{bmatrix}^{-1} = \frac{1}{(4)(5)-(2)(10)} \begin{bmatrix} 5 & -2\\ -10 & 4 \end{bmatrix} = \frac{1}{0} \begin{bmatrix} 5 & -2\\ -10 & 4 \end{bmatrix} \Rightarrow & \ \text{inverse does not exist.}\\ & \text{This matrix is singular.}\end{align*}$
::[4-2105]-1=1(4)(5)-(2)(10)[5-2-104]=10[5-2-104]- 反之亦然。

Example 4
::例4

Find the inverse of
::查找反向

$\begin{align*}\begin{bmatrix} 3 & -4\\ 6 & -7 \end{bmatrix}.\end{align*}$

$\begin{align*}\begin{bmatrix} 3 & -4\\ 6 & -7 \end{bmatrix}^{-1} = \frac{1}{(3)(-7)-(-4)(6)} \begin{bmatrix} -7 & 4\\ -6 & 3 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -7 & 4\\ -6 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{7}{3} & \frac{4}{3}\\ -2 & 1 \end{bmatrix}\end{align*}$

Review
::回顾

Determine whether the following pairs of matrices are inverses of one another.
::确定以下几对矩阵是否相互反向。

.

$\begin{align*}\begin{bmatrix} 5 & -15\\ 3 & -10 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} 2 & -3\\ \frac{3}{5} & -1 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 3 & 7\\ 1 & 2 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} -2 & 7\\ 1 & -3 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -5 & 8\\ 1 & 3 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} 3 & -8\\ -1 & -5 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -5 & 4\\ -9 & 7 \end{bmatrix}\end{align*}$ and

$\begin{align*}\begin{bmatrix} 7 & -4\\ 9 & -5 \end{bmatrix}\end{align*}$

Find the inverse of each matrix below, if it exists.
::如果存在,请在下面查找每个矩阵的反向。

.

$\begin{align*}\begin{bmatrix} -11 & 7\\ -3 & 2 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 6 & -3\\ 8 & -5 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 1 & 2\\ 4 & 9 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -2 & -3\\ 6 & 9 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -2 & 4\\ 5 & 3 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 0 & 1\\ 3 & 2 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -4 & 7\\ 0 & 2 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 2 & -6\\ -6 & 18 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} 7 & 5\\ 14 & 10 \end{bmatrix}\end{align*}$

.

$\begin{align*}\begin{bmatrix} -2 & 5\\ -2 & 6 \end{bmatrix}\end{align*}$

For two 2x2 matrices, A and B, to be inverses of each other, what must be true of AB and BA ?
::两个2x2矩阵A和B相互对立,AB和BA必须属于什么情况?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

版权所有 © 2024 由瓜瓜科技thewaytolearn.cn设计。保留所有权利。