Course: 1.4 矩阵和矩阵操作

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 矩阵和矩阵操作

Collapse Expand
矩阵和矩阵操作
You may have worked with matrices before in high school math, but just to reiterate a matrix is simply an n x m array of numbers. Matrices essentially can represent anything that can be stored in a grid. Mathematically we are interested in matrices for reasons ranging from solving linear equations to probability theory to abstract algebra and group theory. For now, we just want to study operations on matrices.
::您可能曾经在高中数学中与矩阵合作过,但只要重申一个矩阵只是一个 n x m 数阵。矩阵基本上可以代表任何可以存储在网格中的东西。从数学角度讲,我们对矩阵感兴趣的原因从解决线性方程式到概率理论到抽象代数和组理。目前,我们只想研究矩阵上的操作。

Sum and difference
::和差额和差额

To take the sum or the difference of any matrix is just to apply the operation to the corresponding entries of the matrix. So for example, $\begin{align*}\begin{bmatrix}1 & -1\\2 & -3\end{bmatrix}-\begin{bmatrix}0 & 5\\-2 & 3\end{bmatrix}=\begin{bmatrix}1 & -6\\4 & -6\end{bmatrix}\end{align*}$ and $\begin{align*}\begin{bmatrix}1 & 2 & -1\end{bmatrix}+\begin{bmatrix}3 & 4 & -9\end{bmatrix}=\begin{bmatrix}4 & 6 & -10\end{bmatrix}\end{align*}$ ; however, these operations are only defined on matrices of the same dimensions , because doing something like $\begin{align*}\begin{bmatrix}2\\4\end{bmatrix}-\begin{bmatrix}3 & 3 & -1\\4 & -9 & 8\end{bmatrix}\end{align*}$ makes no sense because certain entries have no corresponding entries in another matrix.
::任何矩阵的总和或差数只是为了将操作适用于矩阵的相应条目。例如,[1-12-3]-[05-23]=[1-64-6]和[12-1]+[34-9]=[46-10]];然而,这些操作仅以同一维度的矩阵来界定,因为某些条目在另一个矩阵中没有相应的条目,因此,采取类似[24]-[33-14-98]的做法是没有意义的。

Matrix Scalar Multiplication
::矩阵缩数乘法

Another operation defined on matrices is scalar multiplication, so given a matrix and a scalar constant , the product just has all of the entries of the matrix multiplied by the scalar. For example, $\begin{align*}3\begin{bmatrix}-1 & 1\\2 & -5\\4 & 0\end{bmatrix}=\begin{bmatrix}-3 & 3\\6 & -15\\12 & 0\end{bmatrix}\end{align*}$ .
::矩阵上定义的另一个操作是计算乘法,因此,如果给定一个矩阵和一个计算常数,产品只是将矩阵的所有条目乘以弧数。例如,3[-112-540]=[-336-15120]。

Now we can also multiply two matrices, but that is a bit harder to do. First note an imperative fact that is NOT COMMUTATIVE, so order of multiplication matters; however, matrix multiplication is associative, so given three matrices $\begin{align*}A, B\text{ and } C, A\cdot B \neq B\cdot A,\text{ but } A\cdot(B\cdot C) = (A\cdot B)\cdot C\end{align*}$ .
::现在我们也可以将两个矩阵乘以两个矩阵,但这要困难得多。首先指出一个绝对的事实,即非通信性,因此是乘法事项的顺序;然而,矩阵乘法是关联性的,因此考虑到三个矩阵A、B和C,ABBA,但A(BC)=(AB)C。

Also in order for matrix multiplication to work the number of columns in the first matrix must equal the number of rows in the second matrix. An $\begin{align*}(n\times m)\times (m\times k) = (n\times k)\end{align*}$ where that product is a matrix with the dimensions n x m times a matrix with the dimensions m x k yields a matrix with dimensions n x k.
::为了让矩阵乘法工作,第一个矩阵中的列数必须等于第二个矩阵中的行数。如果该产品是尺寸为nxxm乘以尺寸为mxk的矩阵,则该产品为尺寸为nxm乘以尺寸为mxk的矩阵,则该矩阵的尺寸为nxxk。

We can define matrix multiplication as $\begin{align*}(AB)_{ij} = A_{i}\cdot B_{j}\end{align*}$ where the dot is the dot product. So essentially the product matrix's entry of the i-th row and j-th column is the dot product of the first matrices' i-th column and the second matrices' j-th column.
::我们可以将矩阵乘法定义为( AB) ij= AiBj, 点是点产品。因此, 产品矩阵输入 i- 行和 j-th 列就是第一个矩阵 i-th 列和第二个矩阵 j-th 列的点产物。

Example 1: Evaluate the matrix product $\begin{align*}\begin{bmatrix}1&2\\-1&3\end{bmatrix}\cdot\begin{bmatrix}3&4\\8&-2\end{bmatrix}\end{align*}$
::实例1:评估矩阵产品[12-13]和[348-2]

Recall by definition that the i - th row and j - th column of the product matrix is the dot product of row i of matrix one and j - th column of matrix two. So $\begin{align*}(A\cdot B)_{(1,1)} = \begin{bmatrix}1\\2\end{bmatrix} \cdot \begin{bmatrix}3\\8\end{bmatrix} = 3 + 16 = 19.\end{align*}$
::顾名思义地回顾,产品矩阵的一行和j-j-th列是矩阵一一一一行和j-j-表二第列一行的点产物。因此(AB(1,1,1)=[12][38]=3+16=19)。

Then $\begin{align*}(A\cdot B)_{(1,2)} = \begin{bmatrix}1\\2\end{bmatrix}\cdot\begin{bmatrix}4\\-2\end{bmatrix} = 4\cdot 1 + 2\cdot(-2) = 4 - 4 = 0\end{align*}$
::然后(AB1,2) =[12] =[4-2] =[4-2] =[4-2] =[4-2] 1+2(2) (-2) =4-4=0

$\begin{align*}(A\cdot B)_{(2,1)} = \begin{bmatrix}-1\\3\end{bmatrix}\cdot\begin{bmatrix}3\\8\end{bmatrix} = 3\cdot(-1) + 8\cdot 3 = -3 + 24 = 21\end{align*}$
:AB2,1) =[-13] =[38] =3(-1) +83] 3+24=21

$\begin{align*}(A\cdot B)_{(2,2)} = \begin{bmatrix}-1\\3\end{bmatrix}\cdot\begin{bmatrix}4\\-2\end{bmatrix} = -1(4) + 3(-2) = -10\end{align*}$
:A_B(2,2) =[-13] [4-2] 1(4)+3(-2) 10

So the matrix product is $\begin{align*}A\cdot B = \begin{bmatrix}19 & 0 \\ 21 & -10\end{bmatrix}\end{align*}$
::所以矩阵产品是AB=[19021-10]

Matrix Inverses
::矩阵逆数

The identity matrix is the matrix such that any matrix times the identity matrix is equal to that matrix. This matrix is denoted as $\begin{align*}I = \begin{bmatrix}1 & 0 &\cdot &\cdot &\cdot & 0 \\ 0 & 1 & \cdot &\cdot & \cdot\ & 0\\ & & \cdot \\ & & & \cdot \\ & & & &\cdot \\ & & & & & 1\end{bmatrix}\end{align*}$ or it is the matrix with 1's down the diagonal and 0's everywhere else.
::身份矩阵是矩阵,因此,身份矩阵的任何时间与该矩阵相同,该矩阵以I=[1000101]表示,或为1的矩阵,在对角下方为0,在其他地方为0。

To prove this matrix is the identity we see that (looking specifically in the two by two case),
::为了证明这个矩阵就是我们所看到的身份(具体地看在两个案例中,两个案例一个案例),

Now let
::现在让我们

$\begin{align*}A=\begin{bmatrix}a & b \\ c & d\end{bmatrix} \to\end{align*}$ $\begin{align*}A\cdot I =\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot\begin{bmatrix}1&0\\0&1\end{bmatrix}= \begin{bmatrix}A_{i=1}\cdot I_{j=1} & A_{i=1}\cdot I_{j=2} \\ A_{i=2}\cdot I_{j=1} & A_{i=2}\cdot j_{2}\end{bmatrix} = \begin{bmatrix}a(1) + 0(b) & a(0) + b(1) \\ c(1) + d(0) & c(0) + d(1)\end{bmatrix} = \begin{bmatrix}a & b \\ c& d\end{bmatrix}\end{align*}$
::[1001]=[Ai=1Ij=1Ai=1Ij=1Ai=1Ij=2Ai=2Ij=1Ai=2Ij=1Ai=2j2]=[a(1)+0+0(b)a(0)+b(1)c(1)+d(0)c(0)+d(1)]=[abd]

Proving it for the general case is similar, because you are taking the dot product of the i-th row and the j - th column which will end up equal to the entry (i,j) in the first matrix because everything is 0 except the j - th entry of that row which yields (i,j) and then the matrix is equal. You can verify this yourself with other examples as well as proving that multiplication by the identity matrix is commutative, i.e., $\begin{align*}A\cdot I = I\cdot A\end{align*}$ .

From this you'll end up getting that the identity matrix is also the matrix yielded when you take the product of a matrix and its inverse . So now we can define the inverse of a matrix as the matrix that multiplies a given matrix to yield the identity.

Let's now try to derive this matrix in the 2 by 2 case:

Let $\begin{align*}A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\end{align*}$ and the inverse obviously has to have dimensions of 2x2 because the product yields the two by two identity matrix. So $\begin{align*}A^{-1} = \begin{bmatrix}e & f \\ g & h\end{bmatrix}\end{align*}$ and now we can try to solve for the entries of the inverse matrix in terms of the entries of the regular matrix such that the product is equal to the identity matrix.

So $\begin{align*}A\cdot A^{-1} = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\cdot\begin{bmatrix} e & f \\ g & h\end{bmatrix} = \begin{bmatrix}ae + bg & af + bh \\ ce + gd & cf + dh\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\end{align*}$

From this we get the system of equations

$\begin{align*}ae + bg = 1 \\ af + bh = 0 \\ ce + gd = 0 \\ cf + dh = 1\end{align*}$

Solving in terms of the entries of the matrix we get

%20%2B%20%5Cfrac%7B1%7D%7Bc%7D%20%5C%5C%20%0Ag%20%3D%20%5Cfrac%7B1%7D%7Bb%7D%20-%20%5Cfrac%7Ba%7D%7Bb%7D(e)%20%5C%5C%0Ah%20%3D%20-%5Cfrac%7Ba%7D%7Bb%7Df"> $\begin{align*}e = (-d)\frac{g}{c} \\ f = -\frac{d}{c}(h) + \frac{1}{c} \\ g = \frac{1}{b} - \frac{a}{b}(e) \\ h = -\frac{a}{b}f\end{align*}$

Solving comes out to

$\begin{align*}g = \frac{1}{b} - \frac{a}{b}(-\frac{d}{c}g) = \frac{1}{b} + \frac{ad}{bc}(g) \\ g - \frac{ad}{bc}g = \frac{1}{b} \\ g(1-\frac{ad}{bc}) = \frac{1}{b} \\ g = \frac{-bc}{ad-bc}\cdot\frac{1}{b} \\ g = \frac{-c}{ad-bc}\end{align*}$

Then $\begin{align*}e = -\frac{d}{c}(\frac{-c}{ad-bc}) = \frac{d}{ad-bc}\end{align*}$

Now solving for the other two entries we get

$\begin{align*}h = -\frac{a}{b}f \\ f = -\frac{d}{c}h + \frac{1}{c} \\ f = \frac{ad}{bc}f + \frac{1}{c} \\ f(1-\frac{ad}{bc}) = \frac{1}{c} \\ f = -\frac{b}{ad-bc}\end{align*}$

Then

$\begin{align*}h = -\frac{a}{b}(-\frac{b}{ad-bc}) = \frac{a}{ad-bc}\end{align*}$ .

Now we have the matrix

$\begin{align*}A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}\end{align*}$ .

As you see we have this quantity out in front $\begin{align*}ad-bc\end{align*}$ which is known as the determinant of the matrix A. The determinant has many mathematical uses and people define it in many different ways. The determinant tells us many different facts about matrices as well as linear transformations represented by matrices and can also be used to tell us about span and linear independence. It also represents the area of the object (parallelogram in 2 dimensions) created by the two vectors formed by the columns of the matrix.

If the determinant of a matrix is 0 then obviously the matrix has no inverse. Also, if the matrix has no inverse then the columns are linearly dependent vectors. Looking at the two by two case you get that two vectors are linearly dependent if they are scalar multiples of each other and they are only scalar multiples of each other if $\begin{align*}\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix}, \begin{bmatrix}x_{3} \\ x_{4}\end{bmatrix} \text{ has } \begin{bmatrix}x_{3} \\ x_{4}\end{bmatrix} = c\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix},\text{ then } \text{det}\begin{bmatrix}x_{1} & x_{3} \\ x_{2} & x_{4}\end{bmatrix} = x_{1}(cx_{2}) - x_{2}(cx_{1}) = 0.\end{align*}$

Review Problems

Evaluate   $\begin{align*}\begin{bmatrix}1 & 1\\ 2 & -1\\ 3 & 1\end{bmatrix} + \begin{bmatrix} 3 & -1\\2 & 1\\-5 & 4\end{bmatrix}\end{align*}$
::评价[112-131]+[3-112-154]

Calculate $\begin{align*}\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix} + \begin{bmatrix}1 & 1 & -1 & -1 \\ 1 & 1 & 0 & -5 \\ 0 & 0 & 4 & -3\end{bmatrix}\end{align*}$
::计算 [111111]+[11-1-1-1110-5004-3]

Evaluate the sum $\begin{align*}\begin{bmatrix}1 & -1 & 2 \\ 3 & -1 & 1 \\ -1 & 2 & 73\end{bmatrix} + \begin{bmatrix}1 & -1 & \frac{1}{2} \\ -1 & -1 & 0 \\ 0 & 0 & 5\end{bmatrix}\end{align*}$
::评价[1-123-11-11-11273]+[1-112-1-1-1-1-1-10005]

Calculate the difference between the two matrices $\begin{align*}\begin{bmatrix}1&1\\-1&5\end{bmatrix} - \begin{bmatrix}0&-1\\3&-5\end{bmatrix}\end{align*}$
::计算两个矩阵表[11-15]-[0-13-5]之间的差异

Multiply the matrix by the scalar constant $\begin{align*}\frac{7}{2}\begin{bmatrix}\frac{1}{2} & -\frac{3}{2} \\ \frac{2}{3} & -1\end{bmatrix}\end{align*}$
::用 scalar 常量72 乘以矩阵[12-3223-1]

Evaluate the "linear combination" of the matrices $\begin{align*}-1\cdot\begin{bmatrix}3&2\\-1&1\end{bmatrix} + 4\begin{bmatrix}2 & -1\\1 & \frac{1}{2}\end{bmatrix}\end{align*}$
::评价矩阵表-1[32-11]+4[2-1112]的“线性组合”

Calculate the product of the two matrices $\begin{align*}\begin{bmatrix}1 & 1 \\ 2 & 2\end{bmatrix}\cdot\begin{bmatrix}0 & 0 \\ -1 & \frac{1}{2}\end{bmatrix}\end{align*}$
::计算两个矩阵的产值 [1122] [00-112]

Calculate the product of the 2x3 and the 3x2 matrices $\begin{align*}\begin{bmatrix}3 & 1 & 3 \\ -1 & 1 & -2\end{bmatrix}\cdot\begin{bmatrix}3 & -1 \\ 1 & 3 \\ 2 & 5\end{bmatrix}\end{align*}$
::计算 2x3 和 3x2 矩阵的产值 [313- 11-2][3- 11325]

Evaluate $\begin{align*}\begin{bmatrix}1&1\\-1&1\end{bmatrix}\cdot\begin{bmatrix}-1&\frac{1}{2} \\ 2 & 4 \\ 3 & -2\end{bmatrix}\end{align*}$
::评价[11-11] 和[-112243-2]

Find the product of $\begin{align*}\begin{bmatrix} -1 & 1 \\ 2 & 5 \end{bmatrix} \text{ and } \begin{bmatrix}1 & 4 & 5 & -6 \\ 2 & 3 & -7 & \frac{1}{3}\end{bmatrix}\end{align*}$
::查找[-1125]和[145-623-713]的产物。

Calculate the determinant of the matrix $\begin{align*}\begin{bmatrix} 1 & 1 \\ -2 & 5\end{bmatrix}\end{align*}$
::计算矩阵的决定因素[11-25]

Evaluate $\begin{align*}\text{det}\Bigg(\begin{bmatrix}1 & 0 \\ 0 &1\end{bmatrix}\Bigg)\end{align*}$
::评价 det([[1001])

Generalize the last result to calculate $\begin{align*}\text{det}(D)\end{align*}$ where D is a diagonal matrix where the only nonzero entries are on the diagonal
::将最后一个结果普遍化, 计算 d( D) 的 det( D) 时, D 是一个对角矩阵, 其中唯一的非零条目位于对角

Calculate $\begin{align*}\text{det}\Bigg(\begin{bmatrix}3 & -1 \\ -9 & 3\end{bmatrix}\Bigg)\end{align*}$
::计算 det( [3- 1- 93])

Find the inverse $\begin{align*}\begin{bmatrix}7 & -2 \\ -14 & 4\end{bmatrix}\end{align*}$
::查找反向 [7-2--2-144]

Calculate $\begin{align*}\Bigg ( \begin{bmatrix} -2 & 1 \\ \frac{1}{2}& \frac{14}{3} \end{bmatrix} \Bigg )^{-1}\end{align*}$
::计算 ([- 2112143])- 1

Lastly, evaluate $\begin{align*}\Bigg(\begin{bmatrix}14 & -7 \\ -2 & 1\end{bmatrix}\Bigg)^{-1}\end{align*}$
::最后,评价([14-7-21]-1

Section outline

General

矩阵和矩阵操作