Course: 8.9 逆进数 | The Way To Learn

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逆矩阵

Two numbers are multiplicative inverses if their product is 1. Every number besides the number 0 has a multiplicative inverse . For matrices, two matrices are inverses of each other if they multiply to be the identity matrix .

What kinds of matrices do not have inverses?
::哪些类型的矩阵没有反言?

Inverses of Matrices
::儿科逆数

Multiplicative inverses are two numbers or matrices whose product is one or the identity matrix. Consider a matrix $\begin{align*}A\end{align*}$ that has inverse $\begin{align*}A^{-1}\end{align*}$ . How do you find matrix $\begin{align*}A^{-1}\end{align*}$ if you just have matrix $\begin{align*}A\end{align*}$ ?
::多重反向是两个数字或矩阵,其产品为一个或身份矩阵。考虑一个矩阵A,其反向为A-1。如果只有矩阵A,你如何找到矩阵A-1?

\begin{aligned} A = [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}], A^{- 1} = ? \end{aligned}

$\begin{align*}A=\begin{bmatrix} 1 & 2 & 3\\ 1 & 0 & 1\\ 0 & 2 & -1 \end{bmatrix}, \ A^{-1}=?\end{align*}$
::A=[12310102-1],A-1=?

The answer is that you augment matrix $\begin{align*}A\end{align*}$ with the identity matrix and row reduce.
::答案是,用身份矩阵和减排来补充矩阵A。

\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 2 \\ 0 & 2 & - 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] \\ R_{2} + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 2 \\ 0 & 0 & - 3 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 1 & 1 \end{matrix} \end{matrix}] \\ R_{2} \div - 2 \to [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & - 3 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ - 1 & 1 & 1 \end{matrix} \end{matrix}] \\ R_{3} \div - 3 \to [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ R_{3} \cdot - 3 + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 0 & 1 & 1 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ R_{3} \cdot - 1 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 0 & 1 & 1 \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ R_{2} \cdot - 2 + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \end{aligned}

$\begin{align*}\begin{bmatrix} \begin{matrix} 1 & 2 & 3\\ 1 & 0 & 1\\ 0 & 2 & -1\\ \end{matrix} \left| \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{1} \cdot -1 + {R}_{2} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 3\\ 0 & -2 & -2\\ 0 & 2 & -1\\ \end{matrix} \left| \begin{matrix} 1 & 0 & 0\\ -1 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{2}+{R}_{3} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 3\\ 0 & -2 & -2\\ 0 & 0 & -3\\ \end{matrix} \left| \begin{matrix} 1 & 0 & 0\\ -1 & 1 & 0\\ -1 & 1 & 1\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{2} \div -2 \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 3\\ 0 & 1 & 1\\ 0 & 0 & -3\\ \end{matrix} \left| \begin{matrix} 1 & 0 & 0\\ \frac {1}{2} & -\frac{1}{2} & 0\\ -1 & 1 & 1\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{3} \div -3 \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 3\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{matrix} \left| \begin{matrix} 1 & 0 & 0\\ \frac {1}{2} & -\frac{1}{2} & 0\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{3} \cdot -3 + {R}_{1} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{matrix} \left| \begin{matrix} 0 & 1 & 1\\ \frac {1}{2} & -\frac{1}{2} & 0\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{3} \cdot -1 + {R}_{2} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \left| \begin{matrix} 0 & 1 & 1\\ \frac {1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{2} \cdot -2 + {R}_{1} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{matrix} \left| \begin{matrix} -\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\ \frac {1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\\ \end{matrix} \right. \end{bmatrix}\\\end{align*}$
::[1231002-11-1100100001]R1__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

The matrix on the right is the inverse matrix $\begin{align*}A^{-1}\end{align*}$ .
::右边的矩阵是逆矩阵A-1。

\begin{aligned} A^{- 1} = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \end{aligned}

$\begin{align*}A^{-1}=\begin{bmatrix} -\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\ \frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \end{bmatrix}\end{align*}$
::A-1=[-13431316-1613-1313-13-13]

Fractions are usually unavoidable when computing inverses.
::当计算反向时,分数通常是不可避免的。

One reason why inverses are so powerful is because they allow you to solve systems of equations with the same logic as you would solve a single linear equation . Consider the following system based on the coefficients of matrix $\begin{align*}A\end{align*}$ from above.
::反向如此强大的原因之一是,它们允许你用与您相同的逻辑来解决单一线性方程式的等式系统。根据以上矩阵A的系数来考虑下面的系统。

\begin{aligned} x + 2 y + 3 z & = 96 \\ x + 0 y + z & = 36 \\ 0 x + 2 y - z & = - 12 \end{aligned}

$\begin{align*}x+2y+3z &= 96\\ x+0y+z &= 36\\ 0x+2y-z &= -12\end{align*}$
::x+2y+3z=96x+0y+z=360x+2y-z%12

By writing this system as a matrix equation you get:
::将这个系统写成矩阵方程式后,您可以得到:

\begin{aligned} [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}] \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}

$\begin{align*}\begin{bmatrix} 1 & 2 & 3\\ 1 & 0 & 1\\ 0 & 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\\ A \cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\end{align*}$
::[123102-1]-[xyz]=[9636-12]A[xyz]=[9636-12]

If this were a normal linear equation where you had a constant times the variable equals a constant, you would multiply both sides by the multiplicative inverse of the coefficient . Do the same in this case.
::如果这是正常的线性方程式, 变量的常数乘以常数, 变量等于常数, 则以乘法乘以乘以系数的倍数反差。在此情况下, 也这样做。

\begin{aligned} A^{- 1} \cdot A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}

$\begin{align*}A^{-1} \cdot A \cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= A^{-1} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\\ \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= A^{-1} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\end{align*}$
::A-1__A__[xyz]=A-1__[9636-12][xyz]=A-1__[9636-12][9636-12][xyz]=A-1__[9636-12]

All that is left is for you to substitute in and to perform the matrix multiplication to get the solution.
::剩下的只能由你来替代并进行矩阵乘法以获得解决方案

\begin{aligned} [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} - \frac{1}{3} \cdot 96 + \frac{4}{3} \cdot 36 + \frac{1}{3} \cdot (- 12) \\ \frac{1}{6} \cdot 96 - \frac{1}{6} \cdot 36 + \frac{1}{3} \cdot (- 12) \\ \frac{1}{3} \cdot 96 - \frac{1}{3} \cdot 36 - \frac{1}{3} \cdot (- 12) \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 12 \\ 6 \\ 24 \end{matrix}] \end{aligned}

$\begin{align*}\begin{bmatrix} x\\ y\\ z \end{bmatrix} &= A^{-1} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\\ \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} -\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\ \frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \end{bmatrix} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\\ \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} -\frac{1}{3} \cdot 96+\frac{4}{3} \cdot 36+\frac{1}{3} \cdot (-12)\\ \frac{1}{6} \cdot 96-\frac{1}{6} \cdot 36+\frac{1}{3} \cdot (-12)\\ \frac{1}{3} \cdot 96-\frac{1}{3} \cdot 36-\frac{1}{3} \cdot (-12) \end{bmatrix}\\ \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} 12\\ 6\\ 24 \end{bmatrix}\end{align*}$
::[xyz]=A-1____[9636-12][xyz]=[-13431316-16131313-13-13-13]__[9636-12][xyz]=[-131996+43136+13__(-12)[1619-96-161936+13__(-12)-(-12)131996-1313-36-13__(-12)][xyz]=[12624]

Examples
::实例

Example 1
::例1

Earlier, you were asked what types of matrices do not have inverses. Non- square matrices do not generally have inverses. Square matrices that have equal to zero do not have inverses.
::早些时候,有人问到哪些类型的矩阵没有反向。非平方矩阵通常没有反向。等于零的方矩阵没有反向。

Example 2
::例2

Find the inverse of the following matrix.
::查找以下矩阵的反向。

$\begin{aligned} [\begin{matrix} 1 & 6 \\ 4 & 24 \end{matrix}] \end{aligned}$

$\begin{align*}\begin{bmatrix} 1 & 6\\ 4 & 24 \end{bmatrix}\end{align*}$

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 6 \\ 4 & 24 \end{matrix} | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 4 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 6 \\ 0 & 0 \end{matrix} | \begin{matrix} 1 & 0 \\ - 4 & 1 \end{matrix} \end{matrix}] \end{aligned}$

$\begin{align*}\begin{bmatrix} \begin{matrix} 1 & 6\\ 4 & 24\\ \end{matrix} \left| \begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix} \right. \end{bmatrix}\\ {R}_{1} \cdot -4 + {R}_{2} \rightarrow \begin{bmatrix} \begin{matrix} 1 & 6\\ 0 & 0\\ \end{matrix} \left| \begin{matrix} 1 & 0\\ -4 & 1\\ \end{matrix} \right. \end{bmatrix} \end{align*}$
::[164241001]R14+R2}[160010-41]

This matrix is not invertible because its rows are not linearly independent . To test to see if a square matrix is invertible, check whether or not the determinant is zero. If the determinant is zero then the matrix is not invertible because the rows are not linearly independent.
::此矩阵不是不可倒置的, 因为它的行并非线性独立的。要测试一个平方矩阵是否是不可倒置的, 请检查决定数是否为零。如果决定因素为零, 则该矩阵不可倒置, 因为行并非线性独立的。

Example 3
::例3

Confirm matrix $\begin{align*}A\end{align*}$ and $\begin{align*}A^{-1}\end{align*}$ are inverses by computing $\begin{align*}A^{-1} \cdot A\end{align*}$ and $\begin{align*}A \cdot A^{-1}\end{align*}$ .
::通过计算A-1-A和A-A-1,确认的矩阵A和A-1是反向的。

$\begin{align*}A=\begin{bmatrix} 1 & 2 & 3\\ 1 & 0 & 1\\ 0 & 2 & -1 \end{bmatrix}, \ A^{-1}=\begin{bmatrix} -\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\ \frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \end{bmatrix}\end{align*}$
::A=[12310102-1-1],A-1=[-13431316-161313-13-13]

\begin{aligned} A^{- 1} \cdot A & = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \cdot [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}] \\ a_{11} & = - \frac{1}{3} \cdot 1 + \frac{4}{3} \cdot 1 + \frac{1}{3} \cdot 0 = 1 \\ a_{22} & = \frac{1}{6} \cdot 2 - \frac{1}{6} \cdot 0 + \frac{1}{3} \cdot 2 = 1 \\ a_{33} & = \frac{1}{3} \cdot 3 - \frac{1}{3} \cdot 1 - \frac{1}{3} (- 1) = 1 \end{aligned}

$\begin{align*}A^{-1} \cdot A &= \begin{bmatrix} -\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\ \frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3\\ 1 & 0 & 1\\ 0 & 2 & -1 \end{bmatrix}\\ a_{11} &= -\frac{1}{3} \cdot 1+\frac{4}{3} \cdot 1+\frac{1}{3} \cdot 0=1\\ a_{22} &= \frac{1}{6} \cdot 2-\frac{1}{6} \cdot 0+\frac{1}{3} \cdot 2=1\\ a_{33} &= \frac{1}{3} \cdot 3-\frac{1}{3} \cdot 1-\frac{1}{3} (-1)=1\end{align*}$
::A-1-A=[-13431316-161313-13-13-13]-[1231010-11-1]-a11-1-1+431-1+131-1+131-1+130=1a22=16_2-1-60+13_2=1a33=13_3-131-1-13-1-1-1-1-1=1

Note that the rest of the entries turn out to be zero. This is left for you to confirm.
::请注意, 其余的条目结果为零。这留待您确认。

Example 4
::例4

Use a calculator to compute $\begin{align*}A^{-1}\end{align*}$ , compute $\begin{align*}A^{-1} \cdot A\end{align*}$ , compute $\begin{align*}A \cdot A^{-1}\end{align*}$ and compute $\begin{align*}A^{-1} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\end{align*}$ .
::使用计算器计算 A-1, 计算 A-1-A, 计算 A-1A, 计算 A-1A, 计算 A-1A, [9636- 12] 。

Start by entering just matrix $\begin{align*}A\end{align*}$ into the calculator.
::从输入矩阵A进入计算器开始。

To compute matrix $\begin{align*}A^{-1}\end{align*}$ use the inverse button programmed into the calculator. Do not try to raise the matrix to the negative one exponent . This will not work.
::要计算矩阵 A-1 , 请使用计算器中的反键。不要试图将矩阵提升到负指数。这行不通。

Note that the calculator may return decimal versions of the fractions and will not show the entire matrix on its limited display. You will have to scroll to the right to confirm that $\begin{align*}A^{-1}\end{align*}$ matches what you have already found. Once you have found $\begin{align*}A^{-1}\end{align*}$ go ahead and store it as matrix $\begin{align*}B\end{align*}$ so you do not need to type in the entries.
::请注意计算器可以返回分数的十进制版本, 并且不会在其有限的显示中显示整个矩阵。您需要滚动到右侧以确认 A- 1 符合您已经找到的内容。一旦您找到了 A- 1 , 请继续将其存储为矩阵 B , 因此您不需要在条目中输入。

$\begin{align*}A^{-1} \cdot A=B \cdot A\end{align*}$
::A-1A=BA

$\begin{align*}A \cdot A^{-1}=A \cdot B\end{align*}$
::A-A-1=A-B

$\begin{align*}A^{-1} \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}=B \cdot \begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}=B \cdot C\end{align*}$
::A-1[9636-12]=B[9636-12]=BC

You need to create matrix $\begin{align*}C=\begin{bmatrix} 96\\ 36\\ -12 \end{bmatrix}\end{align*}$
::您需要创建矩阵 C= [9636- 12]

Being able to effectively use a calculator should improve your understanding of matrices and allow you to check all the work you do by hand.
::能够有效地使用计算器应该能提高您对矩阵的理解,并允许您亲手检查您所做的所有工作。

Example 5
::例5

The identity matrix happens to be its own inverse. Find another matrix that is its own inverse.
::身份矩阵恰好是其自身的反向。另找一个自身反向的矩阵。

Helmert came up with a very clever matrix whose transpose happens to be its own inverse. Here are the $\begin{align*}2 \times 2\end{align*}$ and the $\begin{align*}3 \times 3\end{align*}$ versions.
::Helmert 提出了一个非常聪明的矩阵, 它的转换恰好是它自己的反向。这是 2x2 和 3x3 版本。

$\begin{align*}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}, \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} &-\frac{2}{\sqrt{6}} \end{bmatrix}\end{align*}$

Summary

Two matrices are multiplicative inverses of each other if they multiply to be the identity matrix.
::两个矩阵如果成倍成为身份矩阵,是相互重复的反差。

To find the inverse of a matrix, augment it with the identity matrix and row reduce.
::为了找到矩阵的反面,用身份矩阵和减排来补充矩阵。

Review
::回顾

Find the inverse of each of the following matrices, if possible. Make sure to do some by hand and some with your calculator.
::如果可能的话, 查找以下矩阵的反向。请确定用手做一些或用计算器做一些。

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

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

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

4. $\begin{align*}\begin{bmatrix} 1 & 6\\ 0 & 1 \end{bmatrix}\end{align*}$

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

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

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

8. $\begin{align*}\begin{bmatrix} 4 & 5 & 8\\ 9 & 0 & 1\\ 0 & 3 & -2 \end{bmatrix}\end{align*}$

9. $\begin{align*}\begin{bmatrix} 0 & 7 & -1\\ 2 & -3 & 1\\ 6 & 8 & 0 \end{bmatrix}\end{align*}$

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

11. $\begin{align*}\begin{bmatrix} -2 & -6 & -12\\ -1 & -5 & -2\\ 2 & 3 & 4 \end{bmatrix}\end{align*}$

12. $\begin{align*}\begin{bmatrix} -2 & 6 & 3\\ 2 & 4 & 0\\ -8 & 2 & 1 \end{bmatrix}\end{align*}$

13. Show that transpose Helmert’s $\begin{align*}2 \times 2\end{align*}$ matrix is its own inverse: $\begin{align*}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\end{align*}$
::13. 显示转体赫尔默特的 2x2 矩阵是其本身的反向 : [121212-12]

14. Show that the transpose Helmert’s $\begin{align*}3 \times 3\end{align*}$ matrix is its own inverse: $\begin{align*}\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \end{bmatrix}\end{align*}$ .
::14. 显示转体赫尔默特的3x3矩阵是其本身的反方向:[13131312-1201616-26]。

15. Non-square matrices sometimes have left inverses, where $\begin{align*}A^{-1} \cdot A=I\end{align*}$ , or right inverses, where $\begin{align*}A \cdot A^{-1}=I\end{align*}$ . Why can’t non-square matrices have “regular” inverses?
::15. 非平方矩阵有时会左转,A-1-A=I或右转,A-A-1=I。为什么非平方矩阵不能有 " 常规 " 反转?

Review (Answers)
::回顾(答复)

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Section outline

General

逆矩阵

Inverses of Matrices ::儿科逆数

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)