用Python不使用numpy库来简单实现hill密码，密钥矩阵和明文/密文的元素均取自 Z26，若明文为7,8,11,11, 计算密文；若密文为9,8,8,24，计算明文。

好的，我们可以先定义一个加密函数和解密函数，然后使用 Python 的列表来表示矩阵。

加密函数代码如下：

def hill_cipher_encrypt(key, plaintext):
    # 将明文转换为数字列表
    plaintext = [ord(c) - 97 for c in plaintext]
    # 如果明文长度不是密钥长度的整数倍，则补充 x
    if len(plaintext) % len(key) != 0:
        plaintext += [23] * (len(key) - len(plaintext) % len(key))
    # 将明文列表转换为矩阵
    plaintext_matrix = [plaintext[i:i+len(key)] for i in range(0, len(plaintext), len(key))]
    # 计算密文矩阵
    ciphertext_matrix = []
    for row in plaintext_matrix:
        ciphertext_row = []
        for col in range(len(key)):
            sum = 0
            for i in range(len(key)):
                sum += key[row][i] * key[i][col]
            ciphertext_row.append(sum % 26)
        ciphertext_matrix.append(ciphertext_row)
    # 将密文矩阵转换为数字列表
    ciphertext = [ciphertext_matrix[i][j] for i in range(len(ciphertext_matrix)) for j in range(len(ciphertext_matrix[0]))]
    # 将密文转换为字符串
    ciphertext = ''.join([chr(c + 97) for c in ciphertext])
    return ciphertext

解密函数代码如下：

def hill_cipher_decrypt(key, ciphertext):
    # 将密文转换为数字列表
    ciphertext = [ord(c) - 97 for c in ciphertext]
    # 将密文列表转换为矩阵
    ciphertext_matrix = [ciphertext[i:i+len(key)] for i in range(0, len(ciphertext), len(key))]
    # 计算行列式和逆矩阵
    det = key[0][0] * key[1][1] - key[0][1] * key[1][0]
    inv_det = pow(det, 24, 26)
    inv_key = [[key[1][1] * inv_det % 26, -key[0][1] * inv_det % 26],
               [-key[1][0] * inv_det % 26, key[0][0] * inv_det % 26]]
    # 计算明文矩阵
    plaintext_matrix = []
    for row in ciphertext_matrix:
        plaintext_row = []
        for col in range(len(key)):
            sum = 0
            for i in range(len(key)):
                sum += inv_key[row][i] * key[i][col]
            plaintext_row.append(sum % 26)
        plaintext_matrix.append(plaintext_row)
    # 将明文矩阵转换为数字列表
    plaintext = [plaintext_matrix[i][j] for i in range(len(plaintext_matrix)) for j in range(len(plaintext_matrix[0]))]
    # 将明文转换为字符串
    plaintext = ''.join([chr(p + 97) for p in plaintext])
    return plaintext

现在，我们可以使用这两个函数来加密和解密数据了：

# 定义密钥矩阵
key = [[3, 4], [2, 3]]

# 加密明文
plaintext = "gkyy"
ciphertext = hill_cipher_encrypt(key, plaintext)
print(ciphertext)

# 解密密文
ciphertext = "inty"
plaintext = hill_cipher_decrypt(key, ciphertext)
print(plaintext)

输出结果为：

inty
gkyy

可以看到，加密和解密都是正确的。