用MATLAB求已知向量[4+5i, 6, 7-1.5j, 5+9j],以其为对角线元素生成对角矩阵。
时间: 2024-02-05 12:13:14 浏览: 64
可以使用MATLAB的diag函数来生成对角矩阵。具体操作如下:
首先创建已知向量:
```
v = [4+5i, 6, 7-1.5j, 5+9j];
```
然后使用diag函数生成对角矩阵:
```
D = diag(v);
```
最终生成的对角矩阵D为:
```
D =
4.0000 + 5.0000i 0 0 0
0 6.0000 0 0
0 0 7.0000 - 1.5000i 0
0 0 0 5.0000 + 9.0000i
```
其中,对角线上的元素即为已知向量v的元素。
相关问题
max Π=6(5I+500L)+(1-L)(1000+5I+9000L)-0.5I²-4000L²求解使得M最大的I,L值,用python解决这一问题
要求解使得M最大化的I和L值,我们可以使用Python中的优化算法来解决这个问题。在这里,我们将使用SciPy库中的optimize模块来进行最优化。
首先,我们需要定义目标函数Π(I, L)。然后,我们可以使用`scipy.optimize.minimize`函数来找到使得目标函数最大化的I和L值。
下面是用Python代码解决这个问题的示例:
```python
from scipy.optimize import minimize
# 定义目标函数
def objective(x):
I, L = x[0], x[1]
return -6*(5*I+500*L)+(1-L)*(1000+5*I+9000*L)-0.5*I**2-4000*L**2
# 设定初始值
x0 = [0, 0]
# 定义约束条件
constraints = ({'type': 'ineq', 'fun': lambda x: x[0]}, # I >= 0
{'type': 'ineq', 'fun': lambda x: x[1]}) # L >= 0
# 进行优化
result = minimize(objective, x0, constraints=constraints)
# 输出结果
I_optimal, L_optimal = result.x[0], result.x[1]
max_profit = -result.fun
print(f"最大利润: {max_profit}")
print(f"I的最优值: {I_optimal}")
print(f"L的最优值: {L_optimal}")
```
这段代码会输出最大利润以及对应的最优的I和L值。请注意,由于目标函数是非凸的,所以可能存在多个局部最优解。因此,此解法只能找到局部最优解,而不一定是全局最优解。
使用C++语言编写一段代码,实现将字符串2i+3+5i-(1-i)*(2+i) 拆分成“2i”,"+","3+5i","-","(","1-i",")","*","(","2+i",")" 这11个字串
```cpp
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
std::vector<std::string> split(const std::string &str, const std::string &delimiter) {
std::vector<std::string> result;
size_t start = 0;
size_t end = str.find(delimiter);
while (end != std::string::npos) {
result.push_back(str.substr(start, end - start));
start = end + delimiter.length();
end = str.find(delimiter, start);
}
result.push_back(str.substr(start, end));
return result;
}
int main() {
std::string input = "2i+3+5i-(1-i)*(2+i)";
std::vector<std::string> tokens = split(input, "+-*/()");
for (const auto &token : tokens) {
if (token == "2i") {
std::cout << "2i" << std::endl;
} else if (token == "+") {
std::cout << "+" << std::endl;
} else if (token == "3") {
std::cout << "3" << std::endl;
} else if (token == "5i") {
std::cout << "5i" << std::endl;
} else if (token == "-") {
std::cout << "-" << std::endl; } else if (token == "(") {
std::cout << "(" << std::endl;
} else if (token == "1") {
std::cout << "1" << std::endl;
} else if (token == "-") {
std::cout << "-" << std::endl;
} else if (token == ")") {
std::cout << ")" << std::endl;
} else if (token == "*") {
std::cout << "*" << std::endl;
} else if (token == "2") {
std::cout << "2" << std::endl;
} else if (token == "+") {
std::cout << "+" << std::endl;
} else if (token == "i") {
std::cout << "i" << std::endl;
} else if (token == ")") {
std::cout << ")" << std::endl;
} else if (token == "*") {
std::cout << "*" << std::endl; } else if (token == "(") {
std::cout << "(" << std::endl;
} else if (token == "1") {
std::cout << "1" << std::endl;
} else if (token == "-") {
std::cout << "-" << std::endl;
} else if (token == ")") {
std::cout << ")" << std::endl;
} else if (token == "*") {
std::cout << "*" << std::endl;
} else if (token == "2") {
std::cout << "2" << std::endl;
} else if (token == "+") {
std::cout << "+" << std::endl;
} else if (token == "i") {
std::cout << "i" << std::endl;
} else if (token == ")") {
std::cout << ")" << std::endl;
} else if (token == "*") {
std::cout << "*" << std::endl;
} else if (token == "(") {
std::cout << "(" << std::endl;
} else if (token == "1") {
std::cout << "1" << std::endl;
} else if (token == "-") {
std::cout << "-" << std::endl;
} else if (token == ")") {
std::cout << ")" << std::endl;
} else if (token == "*") {
std::cout << "*" << std
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