Course: 4.10 解决母体赤道

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解决矩阵等量
Katel and Juan are in charge of building a stage for their school's upcoming play. They go to the hardware store to buy supplies. Katel buys 10 yards of wood and 2 hammers. Her total comes to $130. Juan buys 8 yards of wood and 4 hammers. His total comes to $116. How could you use a matrix inverse to find the cost of a yard of wood and the cost of a hammer?
::Katel和Juan负责为他们学校即将上演的戏剧建造舞台,他们去五金店购买用品,Katel购买10码的木头和2个锤子,总数达130美元,Juan购买8码的木头和4个锤子,总数达116美元,你怎么用一个矩阵来反向找到木头院的费用和锤子的费用?

Solving Equations with Matrices
::与母体的溶解等式

Solving equations with matrices is very similar to solving an equation with real numbers. Just like real numbers, we can add or subtract the same matrix on both sides of an equation to isolate the variable matrix. The big change is that we cannot divide by a matrix - division by a matrix is not defined. We can, however, multiply by the inverse of a matrix to isolate the variable matrix. Just be careful - matrix multiplication is not commutative so you must “right multiply” or “left multiply” on both sides of the equation. To illustrate, let’s look at the variable matrix, $\begin{align*}x\end{align*}$ , and constant matrices, $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ .
::用矩阵解析方程式与用真实数字解析方程式非常相似。就像实际数字一样, 我们可以在方程式的两侧增减相同的矩阵, 以孤立变量矩阵。巨大的变化是我们不能用矩阵分割 — — 无法定义以分裂为矩阵。但是, 我们可以乘以矩阵反面来孤立变量矩阵。注意 - 矩阵乘法不具有通融性, 因此您必须在方程式的两侧“ 右倍” 或“ 左倍 ” 。为了说明, 让我们看看变量矩阵, x 和恒定矩阵, A 和 B 。

If $\begin{align*}AX=B\end{align*}$ , then to solve we must multiply on the left as shown:
::如果 AX=B,那么要解决问题,我们必须在左边按所示乘以:

$\begin{aligned} A X & = B \\ A^{- 1} A X & = A^{- 1} B \\ I X & = A^{- 1} B \\ X & = A^{- 1} B \end{aligned}$
$\begin{align*}AX &= B\\ A^{-1} AX &= A^{-1} B\\ IX &= A^{-1} B\\ X &= A^{-1} B\end{align*}$
::AX=BA-1AX=A-1BIX=A-1BX=A-1BX=A-1BB

If $\begin{align*}XA=B\end{align*}$ , then to solve we must multiply on the right as shown:
::如果 XA=B,那么要解决问题,我们必须在右边乘以显示的乘法 :

$\begin{aligned} X A & = B \\ X A A^{- 1} & = B A^{- 1} \\ X I & = B A^{- 1} \\ X & = B A^{- 1} \end{aligned}$
$\begin{align*}XA &= B\\ XAA^{-1} &= BA^{-1} \\ XI &= BA^{-1}\\ X &= BA^{-1}\end{align*}$
::XA=BXAA-1=BA-1=BA-1=BA-1XI=BA-1XA=BA-1=BA-1

Let's solve the following equations.
::让我们解决以下的方程式。

$\begin{aligned} [\begin{matrix} 2 & - 4 \\ 5 & 1 \end{matrix}] X = [\begin{matrix} - 14 & - 14 \\ - 13 & 9 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -4\\ 5 & 1 \end{bmatrix}X = \begin{bmatrix} -14 & -14\\ -13 & 9 \end{bmatrix}\end{align*}$
::[2-451]X=[-14-14-139]

To isolate the variable matrix, denoted by $\begin{align*}X\end{align*}$ , we need to get rid of the matrix being multiplied by $\begin{align*}X\end{align*}$ on the left. Find the inverse of
$\begin{aligned} [\begin{matrix} 2 & - 4 \\ 5 & 1 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -4\\ 5 & 1 \end{bmatrix}\end{align*}$ and multiply by it on the left of both sides of the equation as shown below.

$\begin{aligned} {[\begin{matrix} 2 & - 4 \\ 5 & 1 \end{matrix}]}^{- 1} = \frac{1}{2 - (- 20)} [\begin{matrix} 1 & 4 \\ - 5 & 2 \end{matrix}] = \frac{1}{22} [\begin{matrix} 1 & 4 \\ - 5 & 2 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -4\\ 5 & 1 \end{bmatrix}^{-1} = \frac{1}{2-(-20)} \begin{bmatrix} 1 & 4\\ -5 & 2 \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 1 & 4\\ -5 & 2 \end{bmatrix}\end{align*}$

Now, multiply by this inverse on the left on both sides of the equation:
::现在,以方程两侧左边的倒数乘法乘以:

$\begin{aligned} \frac{1}{22} [\begin{matrix} 1 & 4 \\ - 5 & 2 \end{matrix}] [\begin{matrix} 2 & - 4 \\ 5 & 1 \end{matrix}] X & = \frac{1}{22} [\begin{matrix} 1 & 4 \\ - 5 & 2 \end{matrix}] [\begin{matrix} - 14 & - 14 \\ - 13 & 9 \end{matrix}] \\ \frac{1}{22} [\begin{matrix} 22 & 0 \\ 0 & 22 \end{matrix}] X & = \frac{1}{22} [\begin{matrix} - 66 & 22 \\ 44 & 88 \end{matrix}] \\ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] X & = [\begin{matrix} - 3 & 1 \\ 2 & 4 \end{matrix}] \\ X & = [\begin{matrix} - 3 & 1 \\ 2 & 4 \end{matrix}] \end{aligned}$
$\begin{align*}\frac{1}{22} \begin{bmatrix} 1 & 4\\ -5 & 2 \end{bmatrix} \begin{bmatrix} 2 & -4\\ 5 & 1 \end{bmatrix} X &= \frac{1}{22} \begin{bmatrix} 1 & 4\\ -5 & 2 \end{bmatrix} \begin{bmatrix} -14 & -14\\ -13 & 9 \end{bmatrix}\\ \frac{1}{22} \begin{bmatrix} 22 & 0\\ 0 & 22 \end{bmatrix} X &= \frac{1}{22} \begin{bmatrix} -66 & 22\\ 44 & 88 \end{bmatrix}\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} X &= \begin{bmatrix} -3 & 1\\ 2 & 4 \end{bmatrix}\\ X &= \begin{bmatrix} -3 & 1\\ 2 & 4 \end{bmatrix}\end{align*}$
::122[14-52][2-451]X=122[14-52][-14-14-14-139]122[220022]X=122[-66224488][1001]X=[-3124]X=[-3124]

$\begin{aligned} X [\begin{matrix} - 8 & 0 \\ 7 & 13 \end{matrix}] = [\begin{matrix} 96 & 104 \\ - 60 & 52 \end{matrix}] \end{aligned}$
$\begin{align*}X \begin{bmatrix} -8 & 0\\ 7 & 13 \end{bmatrix} = \begin{bmatrix} 96 & 104\\ -60 & 52 \end{bmatrix}\end{align*}$
::X[-80713]=[96104-6052]

This time the variable matrix, $\begin{align*}X\end{align*}$ , is being multiplied by another matrix on the right. This time we will need to find the inverse of
$\begin{aligned} [\begin{matrix} - 8 & 0 \\ 7 & 13 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -8 & 0\\ 7 & 13 \end{bmatrix}\end{align*}$ and multiply by it on the right side as shown below.

$\begin{aligned} {[\begin{matrix} - 8 & 0 \\ 7 & 13 \end{matrix}]}^{- 1} = \frac{1}{- 104} [\begin{matrix} 13 & 0 \\ - 7 & - 8 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -8 & 0\\ 7 & 13 \end{bmatrix}^{-1} = \frac{1}{-104} \begin{bmatrix} 13 & 0\\ -7 & -8 \end{bmatrix}\end{align*}$

Now, multiply by this inverse on the right on both sides of the equation:
::乘以方程两侧右侧的逆向

$\begin{aligned} X [\begin{matrix} - 8 & 0 \\ 7 & 13 \end{matrix}] \frac{1}{- 104} [\begin{matrix} 13 & 0 \\ - 7 & - 8 \end{matrix}] = [\begin{matrix} 96 & 104 \\ - 60 & 52 \end{matrix}] \frac{1}{- 104} [\begin{matrix} 13 & 0 \\ - 7 & - 8 \end{matrix}] \end{aligned}$
$\begin{align*}X \begin{bmatrix} -8 & 0\\ 7 & 13 \end{bmatrix} \frac{1}{-104} \begin{bmatrix} 13 & 0\\ -7 & -8 \end{bmatrix} = \begin{bmatrix} 96 & 104\\ -60 & 52 \end{bmatrix} \frac{1}{-104} \begin{bmatrix} 13 & 0\\ -7 & -8 \end{bmatrix} \end{align*}$
::X[-80713]1-104[130-7-8]=[96104-6052]1-104[130-7-8]

Because scalar multiplication is commutative, we can move this factor to the end and do the matrix multiplication first to avoid fractions.
::由于天平乘法具有通量性,我们可以将这个系数移到终点,首先进行矩阵乘法的乘法,以避免分数。

$\begin{aligned} X [\begin{matrix} - 8 & 0 \\ 7 & 13 \end{matrix}] [\begin{matrix} 13 & 0 \\ - 7 & - 8 \end{matrix}] \frac{1}{- 104} & = [\begin{matrix} 96 & 104 \\ - 60 & 52 \end{matrix}] [\begin{matrix} 13 & 0 \\ - 7 & - 8 \end{matrix}] \frac{1}{- 104} \\ X [\begin{matrix} - 104 & 0 \\ 0 & - 104 \end{matrix}] \frac{1}{- 104} & = [\begin{matrix} 520 & - 832 \\ - 1144 & - 416 \end{matrix}] \frac{1}{- 104} \\ X [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] & = [\begin{matrix} - 5 & 8 \\ 11 & 4 \end{matrix}] \\ X & = [\begin{matrix} - 5 & 8 \\ 11 & 4 \end{matrix}] \end{aligned}$
$\begin{align*}X \begin{bmatrix} -8 & 0\\ 7 & 13 \end{bmatrix} \begin{bmatrix} 13 & 0\\ -7 & -8 \end{bmatrix} \frac{1}{-104} &= \begin{bmatrix} 96 & 104\\ -60 & 52 \end{bmatrix} \begin{bmatrix} 13 & 0\\ -7 & -8 \end{bmatrix} \frac{1}{-104}\\ X \begin{bmatrix} -104 & 0\\ 0 & -104 \end{bmatrix} \frac{1}{-104} &= \begin{bmatrix} 520 & -832\\ -1144 & -416 \end{bmatrix} \frac{1}{-104}\\ X \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} &= \begin{bmatrix} -5 & 8\\ 11 & 4 \end{bmatrix}\\ X &= \begin{bmatrix} -5 & 8\\ 11 & 4 \end{bmatrix}\end{align*}$
::[-80713][130-7-7-8]1-104=[96104-6052][130-7-8]1-104X[-1400-104]1-104=[520-832-1144-416]1-104X[1001]=[58114]X=[58114]

$\begin{aligned} [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] X + [\begin{matrix} 15 \\ - 13 \end{matrix}] = [\begin{matrix} 10 \\ 13 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix} X + \begin{bmatrix} 15 \\ -13 \end{bmatrix} = \begin{bmatrix} 10 \\ 13 \end{bmatrix}\end{align*}$
::[112-57]X+[15-13]=[1013]

This equation is a little different. First, we have a matrix that we must subtract on both sides before we can multiply by the inverse of
$\begin{aligned} [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] . \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix}.\end{align*}$
::这个等式有点不同。首先,我们有一个矩阵,我们必须在两边减去这个矩阵,然后才能乘以[112-57]的反数。

$\begin{aligned} [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] X + [\begin{matrix} 15 \\ - 13 \end{matrix}] - [\begin{matrix} 15 \\ - 13 \end{matrix}] & = [\begin{matrix} 10 \\ 13 \end{matrix}] - [\begin{matrix} 15 \\ - 13 \end{matrix}] \\ [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] X & = [\begin{matrix} - 5 \\ 26 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix}X + \begin{bmatrix} 15 \\ -13 \end{bmatrix} - \begin{bmatrix} 15 \\ -13 \end{bmatrix} &= \begin{bmatrix} 10 \\ 13 \end{bmatrix} - \begin{bmatrix} 15 \\ -13 \end{bmatrix} \\ \begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix}X &= \begin{bmatrix} -5 \\ 26\end{bmatrix}\end{align*}$
::[112-57]X+[15-13]-[15-13]-[15-13]=[1013]-[15-13][112-57]X=[-526]

Second, $\begin{align*}X\end{align*}$ , is not a $\begin{align*}2 \times 2\end{align*}$ matrix. What are the dimensions of $\begin{align*}X\end{align*}$ ? If we multiply a $\begin{align*}2 \times 2\end{align*}$ matrix by a $\begin{align*}2 \times 1\end{align*}$ matrix then we will get a $\begin{align*}2 \times 1\end{align*}$ matrix, so $\begin{align*}X\end{align*}$ is a $\begin{align*}2 \times 1\end{align*}$ matrix. Let’s find the inverse of
::第二, X, 不是一个 2x2 矩阵。 X 的尺寸是什么? 如果我们将 2x2 矩阵乘以 2x2 矩阵乘以 2x1 矩阵, 那么我们就会得到 2x1 矩阵, 所以X 是一个 2x1 矩阵。让我们找到相反的

$\begin{aligned} [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] . \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix}.\end{align*}$

$\begin{aligned} {[\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}]}^{- 1} = \frac{1}{87} [\begin{matrix} 7 & - 2 \\ 5 & 11 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix}^{-1} = \frac{1}{87} \begin{bmatrix} 7 & -2\\ 5 & 11 \end{bmatrix}\end{align*}$

Now we can “left multiply” on both sides of the equation and solve for $\begin{align*}X\end{align*}$ .
::现在,我们可以在方程的两侧“左倍数”解决 X。

$\begin{aligned} \frac{1}{87} [\begin{matrix} 7 & - 2 \\ 5 & 11 \end{matrix}] [\begin{matrix} 11 & 2 \\ - 5 & 7 \end{matrix}] & = \frac{1}{87} [\begin{matrix} 7 & - 2 \\ 5 & 11 \end{matrix}] [\begin{matrix} - 5 \\ 26 \end{matrix}] \\ \frac{1}{87} [\begin{matrix} 87 & 0 \\ 0 & 87 \end{matrix}] X & = \frac{1}{87} [\begin{matrix} - 87 \\ 261 \end{matrix}] \\ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] X & = [\begin{matrix} - 1 \\ 3 \end{matrix}] \\ X & = [\begin{matrix} - 1 \\ 3 \end{matrix}] \end{aligned}$
$\begin{align*}\frac{1}{87}\begin{bmatrix} 7 & -2\\ 5 & 11 \end{bmatrix} \begin{bmatrix} 11 & 2\\ -5 & 7 \end{bmatrix} &= \frac{1}{87} \begin{bmatrix} 7 & -2\\ 5 & 11 \end{bmatrix} \begin{bmatrix} -5 \\ 26 \end{bmatrix} \\ \frac{1}{87}\begin{bmatrix} 87 & 0\\ 0 & 87 \end{bmatrix} X &= \frac{1}{87} \begin{bmatrix} -87 \\ 261 \end{bmatrix}\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}X &= \begin{bmatrix} -1\\ 3 \end{bmatrix}\\ X &= \begin{bmatrix} -1 \\ 3 \end{bmatrix}\end{align*}$
::187[7-2511][112-57]=187[7-2511][-526]187[870087]X=187[87261][1001]X=[-13]X=[-13]

Examples
::实例

Example 1
::例1

Earlier, you were asked how you could use a matrix inverse to find the cost of a yard of wood and the cost of a hammer.
::早些时候,有人问你们如何使用一个反向矩阵来寻找一个木场的成本和锤子的成本。

Solve the equation:
$\begin{aligned} [\begin{matrix} 10 & 2 \\ 8 & 4 \end{matrix}] X = [\begin{matrix} 130 \\ 116 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 10 & 2\\ 8 & 4 \end{bmatrix}X = \begin{bmatrix} 130 \\ 116 \end{bmatrix}\end{align*}$
::解决方程: [10284]X=[130116]

To isolate the variable matrix, denoted by $\begin{align*}X\end{align*}$ , we need to get rid of the matrix being multiplied by $\begin{align*}X\end{align*}$ on the left. Find the inverse of
$\begin{aligned} [\begin{matrix} 10 & 2 \\ 8 & 4 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 10 & 2\\ 8 & 4 \end{bmatrix}\end{align*}$ and multiply by it on the left of both sides of the equation as shown below.

$\begin{aligned} {[\begin{matrix} 10 & 2 \\ 8 & 4 \end{matrix}]}^{- 1} = \frac{1}{(10) (4) - (2) (8)} [\begin{matrix} 4 & - 2 \\ - 8 & 10 \end{matrix}] = \frac{1}{24} [\begin{matrix} 4 & - 2 \\ - 8 & 10 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 10 & 2\\ 8 & 4 \end{bmatrix}^{-1} = \frac{1}{(10)(4)-(2)(8)} \begin{bmatrix} 4 & -2\\ -8 & 10 \end{bmatrix} = \frac{1}{24} \begin{bmatrix} 4 & -2\\ -8 & 10 \end{bmatrix}\end{align*}$

Now, multiply by this inverse on the left on both sides of the equation:
::现在,以方程两侧左边的倒数乘法乘以:

$\begin{aligned} \frac{1}{24} [\begin{matrix} 4 & - 2 \\ - 8 & 10 \end{matrix}] [\begin{matrix} 10 & 2 \\ 8 & 4 \end{matrix}] X & = \frac{1}{24} [\begin{matrix} 4 & - 2 \\ - 8 & 10 \end{matrix}] [\begin{matrix} 130 \\ 116 \end{matrix}] \\ \frac{1}{24} [\begin{matrix} 24 & 0 \\ 0 & 24 \end{matrix}] X & = \frac{1}{24} [\begin{matrix} 288 \\ 120 \end{matrix}] \\ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] X & = [\begin{matrix} 12 \\ 5 \end{matrix}] \\ X & = [\begin{matrix} 12 \\ 5 \end{matrix}] \end{aligned}$
$\begin{align*}\frac{1}{24} \begin{bmatrix} 4 & -2\\ -8 & 10 \end{bmatrix} \begin{bmatrix} 10 & 2\\ 8 & 4 \end{bmatrix} X &= \frac{1}{24} \begin{bmatrix} 4 & -2\\ -8 & 10 \end{bmatrix} \begin{bmatrix} 130\\ 116\end{bmatrix}\\ \frac{1}{24} \begin{bmatrix} 24 & 0\\ 0 & 24 \end{bmatrix} X &= \frac{1}{24} \begin{bmatrix} 288\\ 120 \end{bmatrix}\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} X &= \begin{bmatrix} 12\\ 5 \end{bmatrix}\\ X &= \begin{bmatrix} 12\\ 5 \end{bmatrix}\end{align*}$
::124[240024]X=124[288120][1001]X=[125]X=[125]X=[125]

Therefore , wood costs $12 per yard and hammers cost $5 apiece.
::因此,每院子里的木柴费用为12美元,锤子费用为每人5美元。

Example 2
::例2

Solve the following matrix equation .
::解决以下矩阵等式。

$\begin{aligned} [\begin{matrix} 2 & - 5 \\ 6 & 1 \end{matrix}] X = [\begin{matrix} 39 \\ 37 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -5\\ 6 & 1 \end{bmatrix} X = \begin{bmatrix} 39\\ 37 \end{bmatrix}\end{align*}$
::[2-561]X=[3937]

“Left multiply” by the inverse on both sides.
::双方的反向“向左乘”。

$\begin{aligned} \frac{1}{32} [\begin{matrix} 1 & 5 \\ - 6 & 2 \end{matrix}] [\begin{matrix} 2 & - 5 \\ 6 & 1 \end{matrix}] X & = \frac{1}{32} [\begin{matrix} 1 & 5 \\ - 6 & 2 \end{matrix}] [\begin{matrix} 39 \\ 37 \end{matrix}] \\ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] X & = \frac{1}{32} [\begin{matrix} 224 \\ - 160 \end{matrix}] \\ X & = [\begin{matrix} 7 \\ - 5 \end{matrix}] \end{aligned}$
$\begin{align*}\frac{1}{32} \begin{bmatrix} 1 & 5\\ -6 & 2 \end{bmatrix} \begin{bmatrix} 2 & -5\\ 6 & 1 \end{bmatrix} X &= \frac{1}{32} \begin{bmatrix} 1 & 5\\ -6 & 2 \end{bmatrix} \begin{bmatrix} 39\\ 37 \end{bmatrix}\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} X &= \frac{1}{32} \begin{bmatrix} 224\\ -160 \end{bmatrix}\\ X &= \begin{bmatrix} 7\\ -5 \end{bmatrix}\end{align*}$
::132[15-62][2-561]X=132[15-62][3937][1001]X=132[224-160]X=[7-5]

Example 3
::例3

Solve the following matrix equation.

$\begin{aligned} X [\begin{matrix} - 3 & 8 \\ - 2 & 15 \end{matrix}] = [\begin{matrix} 0 & 87 \\ 33 & - 88 \end{matrix}] \end{aligned}$
$\begin{align*} X \begin{bmatrix} -3 & 8\\ -2 & 15 \end{bmatrix} = \begin{bmatrix} 0 & 87\\ 33 & -88 \end{bmatrix}\end{align*}$
::解决下列矩阵方程式。 X[-38-215]=[08733-88]

“Right multiply” by the inverse on both sides.
::双方的“右乘”反反。

$\begin{aligned} X [\begin{matrix} - 3 & 8 \\ - 2 & 15 \end{matrix}] \frac{1}{- 29} [\begin{matrix} 15 & - 8 \\ 2 & - 3 \end{matrix}] & = [\begin{matrix} 0 & 87 \\ 33 & - 88 \end{matrix}] \frac{1}{- 29} [\begin{matrix} 15 & - 8 \\ 2 & - 3 \end{matrix}] \\ X [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] & = \frac{1}{- 29} [\begin{matrix} 174 & - 261 \\ 319 & 0 \end{matrix}] \\ X & = [\begin{matrix} - 6 & 9 \\ - 11 & 0 \end{matrix}] \end{aligned}$
$\begin{align*}X \begin{bmatrix} -3 & 8\\ -2 & 15 \end{bmatrix} \frac{1}{-29} \begin{bmatrix} 15 & -8\\ 2 & -3 \end{bmatrix} &= \begin{bmatrix} 0 & 87\\ 33 & -88 \end{bmatrix} \frac{1}{-29} \begin{bmatrix} 15 & -8\\ 2 & -3 \end{bmatrix}\\ X \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} &= \frac{1}{-29} \begin{bmatrix} 174 & -261\\ 319 & 0 \end{bmatrix}\\ X &= \begin{bmatrix} -6 & 9\\ -11 & 0 \end{bmatrix}\end{align*}$
::X[-38-215]1-29[15-82-33]=[08733-88]1-29[15-82-3]X[1001]=1-29[174-2613190]X=[69-110]

Example 4
::例4

Solve the following matrix equation.
::解决以下矩阵等式。

$\begin{aligned} [\begin{matrix} - 1 & 3 \\ 0 & 5 \end{matrix}] X - [\begin{matrix} 11 & 7 \\ 13 & 21 \end{matrix}] = [\begin{matrix} - 7 & 15 \\ 7 & 14 \end{matrix}] \end{aligned}$
$\begin{align*} \begin{bmatrix} -1 & 3\\ 0 & 5 \end{bmatrix}X - \begin{bmatrix} 11 & 7\\ 13 & 21 \end{bmatrix} = \begin{bmatrix} -7 & 15 \\ 7 & 14 \end{bmatrix}\end{align*}$
::[-1305]X-[1171321]=[-715714]

Add the matrix
$\begin{aligned} [\begin{matrix} 11 & 7 \\ 13 & 21 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 11 & 7\\ 13 & 21 \end{bmatrix}\end{align*}$ to both sides and then “left multiply” by the inverse on both sides.
::将汇总表[1171321] 添加到双方,然后将“左倍数”乘以双方的反数。

$\begin{aligned} [\begin{matrix} - 1 & 3 \\ 0 & 5 \end{matrix}] X - [\begin{matrix} 11 & 7 \\ 13 & 21 \end{matrix}] + [\begin{matrix} 11 & 7 \\ 13 & 21 \end{matrix}] & = [\begin{matrix} - 7 & 15 \\ 7 & 14 \end{matrix}] + [\begin{matrix} 11 & 7 \\ 13 & 21 \end{matrix}] \\ [\begin{matrix} - 1 & 3 \\ 0 & 5 \end{matrix}] X & = [\begin{matrix} 4 & 22 \\ 20 & 35 \end{matrix}] \\ \frac{1}{- 5} [\begin{matrix} 5 & - 3 \\ 0 & - 1 \end{matrix}] [\begin{matrix} - 1 & 3 \\ 0 & 5 \end{matrix}] X & = \frac{1}{- 5} [\begin{matrix} 5 & - 3 \\ 0 & - 1 \end{matrix}] [\begin{matrix} 4 & 22 \\ 20 & 35 \end{matrix}] \\ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] X & = \frac{1}{- 5} [\begin{matrix} - 40 & 5 \\ - 20 & - 35 \end{matrix}] \\ X & = [\begin{matrix} 8 & - 1 \\ 4 & 7 \end{matrix}] \end{aligned}$
$\begin{align*} \begin{bmatrix} -1 & 3\\ 0 & 5 \end{bmatrix} X - \begin{bmatrix} 11 & 7\\ 13 & 21 \end{bmatrix} + \begin{bmatrix} 11 & 7\\ 13 & 21 \end{bmatrix} &= \begin{bmatrix} -7 & 15\\ 7 & 14 \end{bmatrix} + \begin{bmatrix} 11 & 7\\ 13 & 21 \end{bmatrix}\\ \begin{bmatrix} -1 & 3\\ 0 & 5 \end{bmatrix} X &= \begin{bmatrix} 4 & 22\\ 20 & 35 \end{bmatrix}\\ \frac{1}{-5} \begin{bmatrix} 5 & -3\\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 3\\ 0 & 5 \end{bmatrix} X &= \frac{1}{-5} \begin{bmatrix} 5 & -3\\ 0 & -1 \end{bmatrix} \begin{bmatrix} 4 & 22\\ 20 & 35 \end{bmatrix}\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}X &= \frac{1}{-5} \begin{bmatrix} -40 & 5 \\ -20 & -35 \end{bmatrix}\\ X &= \begin{bmatrix} 8 & -1\\ 4 & 7 \end{bmatrix}\end{align*}$
::[-1305]X-[1171321]+[117321][-1305]+[171321][-1305]X=[42220355]1-5[5-30-1][-1-1305]X=1-5[5-330-1][4222035][1001]X=1-5[405-203-35]X=[8-447]

Review
::回顾

Answer the following questions to best of your ability.
::尽你最大能力回答下列问题。

Explain the steps used to solve for matrix X in the equation $\begin{align*}AX=B\end{align*}$ if A and B are 2x2 matrices.
::如果 A 和 B 是 2x2 矩阵, 请解释用于解析公式 AX=B 中矩阵X 所使用的步骤。

How is solving a matrix equation like solving a linear equation? How is it different?
::如何解决矩阵方程式和解决线性方程式一样?这有什么不同呢?

In the matrix equation $\begin{align*}XA=B\end{align*}$ , can the equation be solved for matrix X if there is no inverse of A ?
::在矩阵方程式XA=B中,如果没有A的反义,能否解决矩阵X的方程式?

Solve for the unknown matrix in each equation below.
::解决以下方程式中未知的矩阵。

$\begin{aligned} [\begin{matrix} 2 & - 1 \\ 3 & 5 \end{matrix}] X = [\begin{matrix} - 10 & 4 \\ 11 & 6 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -1\\ 3 & 5 \end{bmatrix} X = \begin{bmatrix} -10 & 4\\ 11 & 6 \end{bmatrix}\end{align*}$
::[2-135]X=[-104116]

$\begin{aligned} X [\begin{matrix} 6 & 7 \\ 11 & - 3 \end{matrix}] = [\begin{matrix} 47 & - 56 \\ 81 & 47 \end{matrix}] \end{aligned}$
$\begin{align*} X \begin{bmatrix} 6 & 7\\ 11 & -3 \end{bmatrix} = \begin{bmatrix} 47 & -56\\ 81 & 47 \end{bmatrix}\end{align*}$
::X[6711-3]=[47-568147]

$\begin{aligned} [\begin{matrix} 5 & - 10 \\ 1 & 9 \end{matrix}] X = [\begin{matrix} 50 \\ - 23 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 5 & -10\\ 1 & 9 \end{bmatrix} X = \begin{bmatrix} 50\\ -23 \end{bmatrix}\end{align*}$
::[50-23] [5-1019]X=[50-23]

$\begin{aligned} [\begin{matrix} 2 & 8 \\ 12 & - 7 \end{matrix}] X = [\begin{matrix} - 2 & - 76 \\ - 67 & 204 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & 8\\ 12 & -7 \end{bmatrix} X = \begin{bmatrix} -2 & -76\\ -67 & 204 \end{bmatrix}\end{align*}$
::[2812-7]X=[-2-76-67204]

$\begin{aligned} X [\begin{matrix} 2 & 9 \\ 5 & - 1 \end{matrix}] = [\begin{matrix} 10 & - 2 \\ 22 & - 42 \end{matrix}] \end{aligned}$
$\begin{align*} X \begin{bmatrix} 2 & 9\\ 5 & -1 \end{bmatrix} = \begin{bmatrix} 10 & -2\\ 22 & -42 \end{bmatrix}\end{align*}$
::[X295-1]=[10-222-42]

$\begin{aligned} [\begin{matrix} - 9 \\ - 77 \end{matrix}] = [\begin{matrix} 3 & 2 \\ - 1 & - 6 \end{matrix}] X \end{aligned}$
$\begin{align*}\begin{bmatrix} -9 \\ -77 \end{bmatrix} = \begin{bmatrix} 3 & 2\\ -1 & -6 \end{bmatrix} X \end{align*}$
::[-9-77]=[32-1-6]X

$\begin{aligned} [\begin{matrix} - 1 & 0 \\ 7 & 2 \end{matrix}] X + [\begin{matrix} 2 & 6 \\ 18 & - 12 \end{matrix}] = [\begin{matrix} 5 & 5 \\ 5 & 5 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -1 & 0\\ 7 & 2 \end{bmatrix} X + \begin{bmatrix} 2 & 6\\ 18 & -12 \end{bmatrix} = \begin{bmatrix} 5 & 5\\ 5 & 5 \end{bmatrix}\end{align*}$
::[-1072]X+[2618-12]=[5555]

$\begin{aligned} [\begin{matrix} 2 & - 8 \\ 11 & - 5 \end{matrix}] X - [\begin{matrix} - 14 \\ 0 \end{matrix}] = [\begin{matrix} - 30 \\ 31 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -8\\ 11 & -5 \end{bmatrix} X - \begin{bmatrix} -14\\ 0 \end{bmatrix} = \begin{bmatrix} -30\\ 31 \end{bmatrix}\end{align*}$
::[2-811-5]X-[-140]=[-3031]

$\begin{aligned} [\begin{matrix} - 3 & - 10 \\ 0 & 1 \end{matrix}] + X [\begin{matrix} 2 & - 3 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -3 & -10\\ 0 & 1 \end{bmatrix} + X \begin{bmatrix} 2 & -3\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}\end{align*}$
::[-3-1001]+X[2-310]=[1234]

$\begin{aligned} [\begin{matrix} - 2 \\ 10 \end{matrix}] = [\begin{matrix} - 6 & 7 \\ 1 & - 2 \end{matrix}] X - [\begin{matrix} 12 \\ - 15 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -2 \\ 10 \end{bmatrix} = \begin{bmatrix} -6 & 7\\ 1 & -2 \end{bmatrix} X - \begin{bmatrix} 12\\ -15 \end{bmatrix}\end{align*}$
::[-210]=[-671-2]X-[12-15]

$\begin{aligned} ([\begin{matrix} 2 & - 3 \\ 7 & 5 \end{matrix}] + [\begin{matrix} 3 & 1 \\ - 3 & - 2 \end{matrix}]) X + [\begin{matrix} 3 & - 1 \\ 2 & 5 \end{matrix}] = [\begin{matrix} - 66 & 43 \\ - 21 & 54 \end{matrix}] \end{aligned}$
$\begin{align*}\left(\begin{bmatrix} 2 & -3\\ 7 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 1\\ -3 & -2 \end{bmatrix}\right) X + \begin{bmatrix} 3 & -1\\ 2 & 5 \end{bmatrix} = \begin{bmatrix} -66 & 43\\ -21 & 54 \end{bmatrix}\end{align*}$
::[2-375]+[31-3-3-2])X+[3-125]=[-6643-2154]

$\begin{aligned} X ([\begin{matrix} 5 & - 1 \\ 2 & 4 \end{matrix}] - [\begin{matrix} 3 & - 1 \\ 2 & 2 \end{matrix}]) + [\begin{matrix} 5 & - 2 \\ 9 & 7 \end{matrix}] = [\begin{matrix} 9 & - 16 \\ 11 & 13 \end{matrix}] \end{aligned}$
$\begin{align*}X \left(\begin{bmatrix} 5 & -1\\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 3 & -1\\ 2 & 2 \end{bmatrix}\right) + \begin{bmatrix} 5 & -2\\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 9 & -16\\ 11 & 13 \end{bmatrix}\end{align*}$
::X[5-124]-[3-1122]+[5-297]=[9-161113]

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::回顾(答复)

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::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。

Section outline

General

解决矩阵等量

Solving Equations with Matrices ::与母体的溶解等式

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Solving Equations with Matrices
::与母体的溶解等式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)