基于二维数组类的矩阵运算方法的实现

以下是基于二维数组类的矩阵运算方法的实现代码，包括矩阵加法、矩阵减法、矩阵乘法和矩阵转置：

public class Matrix {
    private int rows; // 矩阵行数
    private int cols; // 矩阵列数
    private double[][] data; // 矩阵数据

    public Matrix(int rows, int cols) {
        this.rows = rows;
        this.cols = cols;
        this.data = new double[rows][cols];
    }

    // 矩阵加法
    public Matrix add(Matrix other) {
        if (this.rows != other.rows || this.cols != other.cols) {
            throw new IllegalArgumentException("Matrix sizes do not match.");
        }

        Matrix result = new Matrix(rows, cols);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                result.data[i][j] = this.data[i][j] + other.data[i][j];
            }
        }
        return result;
    }

    // 矩阵减法
    public Matrix subtract(Matrix other) {
        if (this.rows != other.rows || this.cols != other.cols) {
            throw new IllegalArgumentException("Matrix sizes do not match.");
        }

        Matrix result = new Matrix(rows, cols);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                result.data[i][j] = this.data[i][j] - other.data[i][j];
            }
        }
        return result;
    }

    // 矩阵乘法
    public Matrix multiply(Matrix other) {
        if (this.cols != other.rows) {
            throw new IllegalArgumentException("Invalid matrix sizes for multiplication.");
        }

        Matrix result = new Matrix(this.rows, other.cols);
        for (int i = 0; i < this.rows; i++) {
            for (int j = 0; j < other.cols; j++) {
                double sum = 0;
                for (int k = 0; k < this.cols; k++) {
                    sum += this.data[i][k] * other.data[k][j];
                }
                result.data[i][j] = sum;
            }
        }
        return result;
    }

    // 矩阵转置
    public Matrix transpose() {
        Matrix result = new Matrix(cols, rows);
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                result.data[j][i] = this.data[i][j];
            }
        }
        return result;
    }
}

以上是一个简单的实现。需要注意的是，矩阵运算需要满足一定的条件，如加法和减法的矩阵大小必须相同，乘法的左矩阵列数必须等于右矩阵行数。在实际应用中，还需要考虑更多的情况。