乘法二进制

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Mr. Hwan writes the following matrices on the board and asks his students to multiply them.
::Hwan先生在董事会上写了以下矩阵,要求学生们增加这些矩阵。

$$\begin{align*}\begin{bmatrix} -3 & 1\\ 2 & 0 \end{bmatrix}\cdot\begin{bmatrix} 4\\ -1 \end{bmatrix}\end{align*}$$

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When they are done multiplying, Mr. Hwan asks, "What element is in the first row, second column of your answer?"
::当他们完成乘法时, 许先生问道, “答案的第一行,第二列中是什么元素?”

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Wanda says that because the two matrices do not have the same dimensions they can't be multiplied. Therefore , there is no answer.
::Wanda说,因为这两个矩阵的尺寸不同,所以不能乘以。因此,没有答案。

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Xavier says that the product is a 2 x 1 matrix so there is no element in the first row, second column.
::Xavier说产品是一个 2 x 1 矩阵, 因此第一行第二列中没有元素 。

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Zach says that the element requested is 8.
::Zach说,所要求的要素是8。

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Who is correct?
::谁是正确的?

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Multiplying Two Matrices
::乘法二进制

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To multiply matrices together we will be multiplying each element in each row of the first matrix by each element in each column in the second matrix. Each of these products will be added together to get the result for a particular row and column as shown below.
::为了将矩阵相乘,我们将将第一个矩阵每一行的每个元素乘以第二矩阵每一列的每个元素,其中每种产品将加在一起,以获得以下特定行和列的结果。

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$$\begin{align*}\begin{bmatrix} \color{red}{a} & \color{red}{b}\\ c & d \end{bmatrix} \cdot \begin{bmatrix} \color{red}{e} & f\\ \color{red}{g} & h \end{bmatrix} = \begin{bmatrix} \color{red}{ae}+\color{red}{bg} & af+bh\\ ce+dg & cf+dh \end{bmatrix}\end{align*}$$
::[abcd]=[ae+bgaf+bhce+dgcf+dh] [efgh]=[ae+bgaf+bhce+dgcf+dh]

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Let's multiply the following matrices.
::让我们乘以以下矩阵。

1.     
   $$\begin{align*}\begin{bmatrix} 2 & 3\\ -1 & 5 \end{bmatrix}\cdot\begin{bmatrix} -6 & 0\\ 4 & 1 \end{bmatrix}\end{align*}$$

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By following the rule given above, we get:
::根据上述规则,我们得到:

$$\begin{align*}\begin{bmatrix} 2(-6)+3(4) & 2(0)+3(1)\\ -1(-6)+5(4) & -1(0)+5(1) \end{bmatrix} = \begin{bmatrix} -12+12 & 0+3\\ 6+20 & 0+5 \end{bmatrix} = \begin{bmatrix} 0 & 3\\ 26 & 5 \end{bmatrix}\end{align*}$$

2.     
   $$\begin{align*}\begin{bmatrix} -1 & 3\\ 7 & 0\\ 4 & -2 \end{bmatrix}\cdot\begin{bmatrix} 5 & 8\\ -1 & 2 \end{bmatrix}\end{align*}$$

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In order to multiply these two matrices we need to extend the pattern given in the guidance to apply to a $\begin{align*}3\times2\end{align*}$ matrix and a $\begin{align*}2\times2\end{align*}$ matrix. It is important to note that matrices do not need to have the same dimensions in order to multiply them together. However, there are limitations that will be introduced later. All matrix multiplication problems in this section  are possible.
::为了将这两个矩阵乘以倍增,我们需要扩大指南中给出的模式,以适用于3x2矩阵和2x2矩阵,必须指出,矩阵不必具有相同的维度才能将它们相乘,不过,有些限制将在以后提出,本节中所有矩阵的乘法问题都是可能的。

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Now, let’s multiply each of the rows in the first matrix by each of the columns in the second matrix to get:
::现在,让我们把第一个矩阵中的每行乘以第二个矩阵中的每列,然后得出:

$$\begin{align*}\begin{bmatrix} -1 & 3\\ 7 & 0\\ 4 & -2 \end{bmatrix}\cdot\begin{bmatrix} 5 & 8\\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1(5)+3(-1) & -1(8)+3(2)\\ 7(5)+0(-1) & 7(8)+0(2)\\ 4(5)+-2(-1) & 4(8)+-2(2) \end{bmatrix} = \begin{bmatrix} -5-3 & -8+6\\ 35+0 & 56+0\\ 20+2 & 32-4 \end{bmatrix} = \begin{bmatrix} -8 & -2\\ 35 & 56\\ 22 & 28 \end{bmatrix}\end{align*}$$

3.     
   $$\begin{align*}\begin{bmatrix} 3 & -7 & 1\\ 2 & 8 & -5 \end{bmatrix}\cdot\begin{bmatrix} 2\\ 0\\ 9 \end{bmatrix}\end{align*}$$

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Sometimes, not only are the matrices different dimensions, but the result has dimensions other than either of the original matrices as is the case in this problem . Multiply the matrices row by column to get:
::有时,不仅矩阵的维度不同,结果的维度也不同于最初的矩阵的维度,这个问题中的情况就是如此。

$$\begin{align*}\begin{bmatrix} 3 & -7 & 1\\ 2 & 8 & -5\end{bmatrix}\cdot\begin{bmatrix} 2\\ 0\\ 9\end{bmatrix} = \begin{bmatrix} 3(2)+-7(0)+1(9) & 2(2)+8(0)+-5(9)\end{bmatrix} = \begin{bmatrix} 6+0+9 & 4+0-45\end{bmatrix} = \begin{bmatrix} 15 & -41\end{bmatrix}\end{align*}$$

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked to find the student that had the correct matrix. 
::早些时候,你被要求找到 拥有正确矩阵的学生

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When the two matrices are multiplied, the resulting matrix is:
::当两个矩阵乘以两个矩阵时,由此产生的矩阵如下:

$$\begin{align*}\begin{bmatrix} -7\\ 8 \end{bmatrix}\end{align*}$$

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The matrices can be multiplied, but there is no first row, second column in the resulting matrix. The element 8 is in the second row, first column. Therefore Xavier is right.
::矩阵可以乘以, 但生成的矩阵中没有第一行第二列。 元素 8 在第二行第一列。 因此, Xavier 是对的 。

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Example 2
::例2

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Multiply the following matrices.
::乘以下列矩阵。

$$\begin{align*}\begin{bmatrix} -4 & 1\\ 5 & -3 \end{bmatrix}\cdot\begin{bmatrix} 2\\ -1 \end{bmatrix}\end{align*}$$

$$\begin{align*}\begin{bmatrix} -4 & 1\\ 5 & -3 \end{bmatrix}\cdot\begin{bmatrix} 2\\ -1 \end{bmatrix} = \begin{bmatrix} -4(2)+1(-1)\\ 5(2)+-3(-1) \end{bmatrix} = \begin{bmatrix} -8-1\\ 10+3 \end{bmatrix} = \begin{bmatrix} -9\\ 13 \end{bmatrix}\end{align*}$$

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Example 3
::例3

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Multiply the following matrices.

$$\begin{align*}\begin{bmatrix} -3\\ 2\\ 5 \end{bmatrix}\cdot\begin{bmatrix} -2 & -1 & 4 \end{bmatrix}\end{align*}$$
::乘以下列矩阵。 [-325]-[-2-14]

$$\begin{align*}\begin{bmatrix} -3\\ 2\\ 5 \end{bmatrix}\cdot\begin{bmatrix}-2 & -1 & 4 \end{bmatrix} = \begin{bmatrix} -3(-2) & -3(-1) & -3(4)\\ 2(-2) & 2(-1) & 2(4)\\ 5(-2) & 5(-1) & 5(4) \end{bmatrix} = \begin{bmatrix} 6 & 3 & -12\\ -4 & -2 & 8\\ -10 & -5 & 20 \end{bmatrix}\end{align*}$$

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Example 4
::例4

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Multiply the following matrices.

$$\begin{align*}\begin{bmatrix} 5 & -2 \end{bmatrix} \cdot \begin{bmatrix} 1 & 4\\ -3 & -7 \end{bmatrix}\end{align*}$$
::[5-2]][14-3-7]

$$\begin{align*}\begin{bmatrix} 5 & -2 \end{bmatrix}\cdot\begin{bmatrix} 1 & 4\\ -3 & -7 \end{bmatrix} = \begin{bmatrix} 5(1)+-2(-3) & 5(4)+-2(-7) \end{bmatrix} = \begin{bmatrix} 5+6 & 20+14 \end{bmatrix} = \begin{bmatrix} 11 & 34 \end{bmatrix}\end{align*}$$

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Review
::回顾

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Multiply the matrices together.
::将矩阵乘在一起。

1. .

$$\begin{align*}\begin{bmatrix} 2 & -1\\ 0 & 3 \end{bmatrix}\cdot\begin{bmatrix} -4 & 3\\ 2 & 5 \end{bmatrix}\end{align*}$$

2. .

$$\begin{align*}\begin{bmatrix} -2 & 7 \end{bmatrix}\cdot\begin{bmatrix} 1 & -4\\ 5 & 3 \end{bmatrix}\end{align*}$$

3. .

$$\begin{align*}\begin{bmatrix} -3 & 2\\ 5 & -4\\ -1 & 6 \end{bmatrix}\cdot\begin{bmatrix} -1 & 3\\ 4 & -2 \end{bmatrix}\end{align*}$$

4. .

$$\begin{align*}\begin{bmatrix} -8 & 1\\ 3 & 5 \end{bmatrix}\cdot\begin{bmatrix} -4\\ 7 \end{bmatrix}\end{align*}$$

5. .

$$\begin{align*}\begin{bmatrix} -1\\ 4\\ 8 \end{bmatrix}\cdot\begin{bmatrix} 2 & -3 & 6 \end{bmatrix}\end{align*}$$

6. .

$$\begin{align*}\begin{bmatrix} -5 & 1 \end{bmatrix}\cdot\begin{bmatrix} -9 & 3 & 0\\ 2 & -1 & 6 \end{bmatrix}\end{align*}$$

7. .

$$\begin{align*}\begin{bmatrix} 4 & -1\\ 5 & 3 \end{bmatrix}\cdot\begin{bmatrix} 2\\ 6 \end{bmatrix}\end{align*}$$

8. .

$$\begin{align*}\begin{bmatrix} -2 & 4 & 7 \end{bmatrix}\cdot\begin{bmatrix} 3\\ -1\\ 5 \end{bmatrix}\end{align*}$$

9. .

$$\begin{align*}\begin{bmatrix} -1 & 2 & -4\\ 5 & 3 & 1\\ -5 & 2 & -1 \end{bmatrix}\cdot\begin{bmatrix} 2 & -3 & 5\\ 6 & 2 & 1\\ -4 & 1 & 0 \end{bmatrix}\end{align*}$$

10. .

$$\begin{align*}\begin{bmatrix} 1 & -2 & 4 \end{bmatrix}\cdot\begin{bmatrix} -3 & 1 & 2\\ 5 & -1 & 6\\ 1 & 2 & -7 \end{bmatrix}\end{align*}$$

11. .

$$\begin{align*}\begin{bmatrix} 6 & 11\\ 1 & 2 \end{bmatrix}\cdot\begin{bmatrix} 2 & -11\\ -1 & 6 \end{bmatrix}\end{align*}$$

12. .

$$\begin{align*}\begin{bmatrix} -4 & 5 & 7\\ 1 & 2 & -3\\ 9 & -6 & 8 \end{bmatrix}\cdot\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix}\end{align*}$$

13. .

$$\begin{align*}\begin{bmatrix} -2 & -1 & 3\end{bmatrix}\cdot\begin{bmatrix} -4\\ 5\\ -8 \end{bmatrix}\end{align*}$$

14. .

$$\begin{align*}\begin{bmatrix} -10 \end{bmatrix}\cdot\begin{bmatrix} 15\end{bmatrix}\end{align*}$$

15. .

$$\begin{align*}\begin{bmatrix} 2 & 4 \\ -1 & -2 \\ 7 & -12 \end{bmatrix}\cdot\begin{bmatrix} 5\\ -3\end{bmatrix}\end{align*}$$

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Review (Answers)
::回顾(答复)

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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 。