# 將可逆矩陣展開為基本矩陣的乘積

This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/).

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from lingeo import random_good_matrix, row_operation_process

## Main idea

Recall that the elementary matrix of a row operation is the resulting matrix of performing the row opertion on the identity matrix.
(See Section 113 for more details.)

Let $A$ be a matrix and $R$ its reduced echelon form.
Then one may perform row operations on $A$ to obtain $R$.
Therefore, there is a sequence of elementary matrices $E_1, \ldots, E_k$ such that
$$
E_k\cdots E_1 A = R.
$$
Since any row operation is revertible,
$$
E_1^{-1}\cdots E_k^{-1} R = A
$$
where $E_i^{-1}$ is the elementary matrix of the reverse row operation corresponding to $E_i$.

When $A$ is a square matrix and is invertible, its reduced echelon form is the identity matrix $I_n$.
Therefore, $A$ can be written as the product of a sequence of elementary matrices
$$
A = F_1 \cdots F_k.
$$

Note that this decomposition is not unique.
In particular, any swapping operation
$$
\rho_i\leftrightarrow\rho_j
$$
can be replaced by
$$
\begin{aligned}
\rho_j&: + \rho_i, \\
\rho_i&: - \rho_i, \\
\rho_j&: + \rho_i,\text{ and} \\
\rho_i&: \times (-1)
\end{aligned}
$$
in order.

## Side stories
- column operation
- congruent
- swap variables

## Experiments

##### Exercise 1

執行以下程式碼。

### code
set_random_seed(0)
print_ans = False
n = 3
A = random_good_matrix(n,n,n)

print("A")
pretty_print(A)

if print_ans:
    elems = row_operation_process(A, inv=True)
    pretty_print(elems)

執行程式碼 `seed = 0` 得
$$A=\begin{bmatrix}
1 & 3 & 5 \\
-5 & -14 & -30\\
-15 & -42 & -89
\end{bmatrix}.
$$
##### Exercise 1(a)

將 $A$ 用列運算化簡為單位矩陣，
並將過程依序記錄下來。

答:
$$\begin{bmatrix}
1 & 3 & 5 \\
-5 & -14 & -30\\
-15 & -42 & -89
\end{bmatrix}\xrightarrow{\rho_2: +5\rho_1}
\begin{bmatrix}
1 & 3 & 5 \\
0 & 1 & -5\\
-15 & -42 & -89
\end{bmatrix}
\xrightarrow{\rho_3:+15\rho_1}
\begin{bmatrix}
1 & 3 & 5 \\
0 & 1 & -5\\
0 & 3 & -14
\end{bmatrix}
\xrightarrow{\rho_1: -3\rho_2}
$$
$$\begin{bmatrix}
1 & 0 & 20 \\
0 & 1 & -5\\
0 & 3 & -14
\end{bmatrix}
\xrightarrow{\rho_3:-3\rho_2}
\begin{bmatrix}
1 & 0 & 20 \\
0 & 1 & -5\\
0 & 0 & 1
\end{bmatrix}
\xrightarrow{\rho_1: -20\rho_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & -5\\
0 & 0 & 1
\end{bmatrix}
\xrightarrow{\rho_2: +5\rho_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
$$

:::success
水~
:::

##### Exercise 1(b)

將 $A$ 寫成基本矩陣的乘積。

答:
因為$$
\begin{bmatrix}
1 & 0 & -20 \\
0 & 1 & 5\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & -3 & 0 \\
0 & 1 & 0\\
0 & -3 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 \\
5 & 1 & 0\\
15 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 3 & 5 \\
-5 & -14 & -30\\
-15 & -42 & -89
\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
$$
所以$$
\begin{bmatrix}
1 & 0 & 0 \\
-5 & 1 & 0\\
-15 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 3 & 0 \\
0 & 1 & 0\\
0 & 3 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 20 \\
0 & 1 & -5\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix} =
\begin{bmatrix}
1 & 3 & 5 \\
-5 & -14 & -30\\
-15 & -42 & -89
\end{bmatrix}=A.
$$
## Exercises

##### Exercise 2

執行以下程式碼。

### code
set_random_seed(0)
print_ans = False
n = 3
A, R, pvts = random_good_matrix(n,n + 1,n, return_answer=True)

print("A")
pretty_print(A)
print("R")
pretty_print(R)

if print_ans:
    elems = row_operation_process(A)
    pretty_print(*(elems[::-1]), A, LatexExpr("="), R)
    elems = row_operation_process(A, inv=True)
    pretty_print(*elems, R, LatexExpr("="), A)

藉由 `seed = 0`得到
$$A=\begin{bmatrix}
1 & -5 & 0 & -22\\
-3 & 16 & 3 & 56\\
-9 & 49 & 13 & 153
\end{bmatrix}、
$$
$$R=\begin{bmatrix}
1 & 0 & 0 & 3\\
0 & 1 & 0 & 5\\
0 & 0 & 1 & -5
\end{bmatrix}.
$$
##### Exercise 2(a)

找出一些基本矩陣 $E_1,\ldots, E_k$ 使得 $E_k\cdots E_1 A = R$。

答:
因為
$$\begin{bmatrix}
1 & -5 & 0 & -22\\
-3 & 16 & 3 & 56\\
-9 & 49 & 13 & 153
\end{bmatrix}\xrightarrow{\rho_2: +3\rho_1}
\begin{bmatrix}
1 & -5 & 0 & -22\\
0 & 1 & 3 & -10\\
-9 & 49 & 13 & 153
\end{bmatrix}\xrightarrow{\rho_3: +9\rho_1}
\begin{bmatrix}
1 & -5 & 0 & -22\\
0 & 1 & 3 & -10\\
0 & 4 & 13 & -45
\end{bmatrix}\xrightarrow{\rho_1:+5\rho_2}
\begin{bmatrix}
1 & 0 & 15 & -72\\
0 & 1 & 3 & -10\\
0 & 4 & 13 & -45
\end{bmatrix}$$
$$\xrightarrow{\rho_3: -4\rho_2}
\begin{bmatrix}
1 & 0 & 15 & -72\\
0 & 1 & 3 & -10\\
0 & 0 & 1 & -5
\end{bmatrix}\xrightarrow{\rho_1: -15\rho_3}
\begin{bmatrix}
1 & 0 & 0 & 3\\
0 & 1 & 3 & -10\\
0 & 0 & 1 & -5
\end{bmatrix}\xrightarrow{\rho_2: -3\rho_3}
\begin{bmatrix}
1 & 0 & 0 & 3\\
0 & 1 & 0 & 5\\
0 & 0 & 1 & -5
\end{bmatrix}
$$

可得

$$E_1=\begin{bmatrix}
1 & 0 & 0 \\
3 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、E_2=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
9 & 0 & 1
\end{bmatrix}、E_3=\begin{bmatrix}
1 & 5 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、$$
$$E_4=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & -4 & 1
\end{bmatrix}、E_5=\begin{bmatrix}
1 & 0 & -15 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、E_6=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & -3\\
0 & 0 & 1
\end{bmatrix}，
$$

使得 $E_6 E_5 E_4 E_3 E_2 E_1 A = R$ 。

:::success
漂亮
:::

##### Exercise 2(b)

找出一些基本矩陣 $F_1,\ldots, F_k$ 使得 $F_1\cdots F_k R = A$。

答:
因為 $E_6 E_5 E_4 E_3 E_2 E_1 A = R$ ，且 $E_1,E_2,E_3,E_4,E_5,E_6$ 皆有反方陣，可知 $A =E_1 ^{-1} E_2 ^{-1} E_3 ^{-1} E_4 ^{-1} E_5 ^{-1} E_6 ^{-1} R$ ，
所以可得
$$F_1=E_1 ^{-1}=\begin{bmatrix}
1 & 0 & 0 \\
-3 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、F_2=E_2 ^{-1}=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
-9 & 0 & 1
\end{bmatrix}、
$$
$$F_3=E_3 ^{-1}=\begin{bmatrix}
1 & -5 & 0 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、F_4=E_4 ^{-1}=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 4 & 1
\end{bmatrix}、$$
$$F_5=E_5 ^{-1}=\begin{bmatrix}
1 & 0 & 15 \\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}、F_6=E_6 ^{-1}=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 3\\
0 & 0 & 1
\end{bmatrix}。
$$

##### Exercise 3

執行以下程式碼。

### code
set_random_seed(0)
print_ans = False
n = 3
A, R, pvts = random_good_matrix(n,n + 1,n - 1, return_answer=True)

print("A")
pretty_print(A)
print("R")
pretty_print(R)

if print_ans:
    elems = row_operation_process(A)
    pretty_print(*(elems[::-1]), A, LatexExpr("="), R)
    elems = row_operation_process(A, inv=True)
    pretty_print(*elems, R, LatexExpr("="), A)

題目給定 `seed=0` ， $A = \begin{bmatrix}
1 & 0 & 3 & 5 \\
3 & 1 & 4 & 10 \\
6 & 3 & 3 & 15
\end{bmatrix} , R = \begin{bmatrix}
1 & 0 & 3 & 5 \\
0 & 1 & -5 & -5 \\
0 & 0 & 0 & 0
\end{bmatrix}$

##### Exercise 3(a)

找出一些基本矩陣 $E_1,\ldots, E_k$ 使得 $E_k\cdots E_1 A = R$。

答:

因為
$$\begin{bmatrix}
1 & 0 & 3 & 5 \\
3 & 1 & 4 & 10 \\
6 & 3 & 3 & 15
\end{bmatrix}\xrightarrow{\rho_2: -3\rho_1}
\begin{bmatrix}
1 & 0 & 3 & 5 \\
0 & 1 & -5 & -5 \\
6 & 3 & 3 & 15
\end{bmatrix}
\xrightarrow{\rho_3:-6\rho_1}
\begin{bmatrix}
1 & 0 & 3 & 5 \\
0 & 1 & -5 & -5 \\
0 & 3 & -15 & -15
\end{bmatrix}
\xrightarrow{\rho_3:-3\rho_2}
\begin{bmatrix}
1 & 0 & 3 & 5 \\
0 & 1 & -5 & -5 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$

所以得 $E_1 = \begin{bmatrix}
1 & 0 & 0 \\
-3 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} , E_2 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
-6 & 0 & 1
\end{bmatrix} , E_3 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & -3 & 1
\end{bmatrix} 。$

##### Exercise 3(b)

找出一些基本矩陣 $F_1,\ldots, F_k$ 使得 $F_1\cdots F_k R = A$。

答:

因為 $E_3 E_2 E_1 A = R$ , 故我們取其反矩陣使得 $A = E_1 ^{-1} E_2 ^{-1} E_3 ^{-1} R$ ，

得 $F_1 = E_1 ^{-1} = \begin{bmatrix}
1 & 0 & 0 \\
3 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} , F_2 = E_2 ^{-1} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
6 & 0 & 1
\end{bmatrix} , F_3 = E_3 ^{-1} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 3 & 1
\end{bmatrix} 。$

##### Exercise 4

將
$$
A = \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
$$
寫成基本矩陣的乘積，
且所用到的基本矩陣沒有對應到「兩列交換」這個動作。

答:
由矩陣列運算
$$\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}\xrightarrow{\rho_1: +\rho_2}
\begin{bmatrix}
1 & 1 \\
0 & 1 \\
\end{bmatrix}
\xrightarrow{\rho_2:+\rho_1}
\begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
\xrightarrow{\rho_1: -\rho_2}
$$
$$\begin{bmatrix}
0 & -1 \\
1 & 2 \\
\end{bmatrix}
\xrightarrow{\rho_2:+2\rho_1}
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\xrightarrow{\rho_1: \times(-1)}
\begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix}
$$
所以
$$\begin{bmatrix}
-1 & 0 \\
0 & 1 \\
\end{bmatrix}\begin{bmatrix}
1 & 0 \\
2 & 1 \\
\end{bmatrix}\begin{bmatrix}
1 & -1 \\
0 & 1 \\
\end{bmatrix}\begin{bmatrix}
1 & 0 \\
1 & 1 \\
\end{bmatrix}\begin{bmatrix}
1 & 1 \\
0 & 1 \\
\end{bmatrix}\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}=
\begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix}=A.
$$

:::warning
- [x] 經由列運算... 的最後一個矩陣寫錯
:::

##### Exercise 5

列運算所對應到的基本矩陣會乘在左邊，
而行運算所對應到的基本矩陣會乘在右邊。
（參考 113-6。）
令
$$
A = \begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4
\end{bmatrix}.
$$
找出一些基本矩陣 $E_1,\ldots, E_k$ 使得 $E_k\cdots E_1 A E_1\trans\cdots E_k\trans = I_4$。

$Ans:$

:::warning

- [ ] 加一些說明文字。標點、句子要完整。
- [x] $K$ --> $k$
:::

做矩陣的行列運算，使其接近單位矩陣。

$E_1A E_1\trans=\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 2 & 2 & 0 \\
1 & 2 & 3 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}，$
$E_1=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1
\end{bmatrix}$.

$E_2E_1A E_1\trans E_2\trans=\begin{bmatrix}
1 & 1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}，$
$E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$.

$E_3E_2E_1A E_1\trans E_2\trans E_3\trans =\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}，$
$E_3=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$.

$k=3$。

$$E_1=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1
\end{bmatrix},
E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},E_3=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
$$

**[由李昌諺同學提供]**

##### Exercise 5

列運算所對應到的基本矩陣會乘在左邊，
而行運算所對應到的基本矩陣會乘在右邊。
（參考 113-6。）
令
$$
A = \begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4
\end{bmatrix}.
$$
找出一些基本矩陣 $E_1,\ldots, E_k$ 使得 $E_k\cdots E_1 A E_1\trans\cdots E_k\trans = I_4$。

答:
將 $A$ 列運算 $$\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4
\end{bmatrix}\xrightarrow{\rho_2: -1\rho_1}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4
\end{bmatrix}
\xrightarrow{\rho_3:-1\rho_1}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 1 & 2 & 2 \\
1 & 2 & 3 & 4
\end{bmatrix}
\xrightarrow{\rho_4: -1\rho_1}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 1 & 2 & 2 \\
0 & 1 & 2 & 3
\end{bmatrix}
$$
$$\xrightarrow{\rho_3:-1\rho_2}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 1 & 2 & 3
\end{bmatrix}
\xrightarrow{\rho_4: -1\rho_2}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
\xrightarrow{\rho_4: -1\rho_3}
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}.
$$
再用行運算 $$\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}\xrightarrow{\kappa_2: -1\kappa_1}
\begin{bmatrix}
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{\kappa_3: -1\kappa_1}
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{\kappa_4: -1\kappa_1}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
$$\xrightarrow{\kappa_3: -1\kappa_2}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{\kappa_4: -1\kappa_2}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{\kappa_4: -1\kappa_3}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
$$
可求出 $E_1,\ldots, E_k$ 為
$$E_1=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\end{bmatrix},
E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\end{bmatrix},
E_3=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-1 & 0 & 0 & 1\end{bmatrix}.$$
$$E_4=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 1\end{bmatrix},
E_5=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 1\end{bmatrix},
E_6=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1\end{bmatrix}.
$$

**[由廖緯程同學提供]**

令 $R_i$ 為第 $i$ 列，$C_i$ 為第 $i$ 行。
先做列運算
$$
R_2:=+(-1)R_1\\R_3:=+(-1)R_1\\R_4:=+(-1)R_1
$$
，以及對應的行運算
$$
C_2:=+(-1)C_1\\
C_3:=+(-1)C_1\\
C_4:=+(-1)C_1
$$
，他們對應的基本矩陣分別為
$$
\begin{aligned}
E_1 &=
\begin{bmatrix}
1 & 0 & 0 & 0\\
-1 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_2 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-1 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_3 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
-1 & 0 & 0 & 1
\end{bmatrix},\\
E_1\trans &=
\begin{bmatrix}
1 & -1 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_2\trans =
\begin{bmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_3\trans =
\begin{bmatrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
\end{aligned}
$$
$A$ 做完這六個運算後為
$$
E_3E_2E_1AE_1\trans E_2\trans E_3\trans =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 1 & 1\\
0 & 1 & 2 & 2\\
0 & 1 & 2 & 3
\end{bmatrix}.
$$
接下來做列運算
$$
R_3:=+(-1)R_2\\R_4:=+(-1)R_2
$$
以及他們對應到的行運算
$$
C_3:=+(-1)C_2\\C_4:=+(-1)C_2
$$
這四個運算對應的基本矩陣分別為
$$
E_4 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_5 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & -1 & 0 & 1
\end{bmatrix},\\
E_4\trans =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
E_5\trans =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix},
$$
$A$ 做完十個運算後為
$$
E_5E_4E_3E_2E_1AE_1\trans E_2\trans E_3\trans E_4\trans E_5\trans =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 0 & 1 & 2
\end{bmatrix}.
$$
最後做列運算
$$
R_4:=+(-1)R_3
$$
及對應的行運算
$$
C_4:=+(-1)C_3
$$
他們對應到的基本矩陣分別為
$$
E_6 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & -1 & 1
\end{bmatrix},
E_6\trans =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 1
\end{bmatrix}
$$
$A$ 做完十二個運算後為
$$
E_6E_5E_4E_3E_2E_1AE_1\trans E_2\trans E_3\trans E_4\trans E_5\trans E_6\trans = I_4 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}.
$$
因為這十二個運算都是將某列(行)乘上常數加到另一列(行)上，所以 $\det(A) = \det(I_4) = 1$

**[由汪駿佑同學提供]**

我們先找到

$$
E_1=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1
\end{bmatrix}，
E_1\trans=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$

這個基本矩陣。

當我們把它擺成如 $E_1 A E_1\trans$ 時，$E_1$ 的意思是對 $A$ 的列作列運算：$\rho_4:+(-1)\rho_3$，
而 $E_1\trans$ 的意思是對 $A$ 的行作行運算：$\kappa_4:+(-1)\kappa_3$。

當我們做完運算後會得到：

$$
E_1A E_1\trans=\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 2 & 2 & 0 \\
1 & 2 & 3 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
$$

同理我們也可以找到

$$
E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}，
E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$

使得

$$
E_2 (E_1 A E_1\trans) E_2\trans=\begin{bmatrix}
1 & 1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$

以及

$$
E_3=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}，
E_3\trans=\begin{bmatrix}
1 & -1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$

使得

$$
E_3 (E_2 E_1 A E_1\trans E_2\trans) E_3\trans =\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
$$

因此我們找出了

$$E_1=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1 & 1
\end{bmatrix},
E_2=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},E_3=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.$$

使得

$$
E_3 E_2 E_1 A E_1\trans E_2\trans E_3\trans =\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} = I_4.
$$

:::info
分數 = 6 ± 檢討 = 6.5

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