用java语言对矩阵进行初等行变换的计算并使结果成为最简行列式且中间矩阵都是整数矩阵

要使结果成为最简行列式，需要在进行初等行变换的同时，记录每次变换的过程，并在最后进行化简。以下是一个示例代码：

public static void elementaryRowOperations(int[][] matrix, int row1, int row2, int k, List<String> process) {
    // 将第row1行乘以k加到第row2行上
    for (int j = 0; j < matrix[0].length; j++) {
        matrix[row2][j] += k * matrix[row1][j];
    }
    // 记录变换过程
    process.add("R" + (row2 + 1) + " = R" + (row2 + 1) + " + " + k + "R" + (row1 + 1));
}

public static void swapRows(int[][] matrix, int row1, int row2, List<String> process) {
    // 交换第row1行和第row2行
    int[] temp = matrix[row1];
    matrix[row1] = matrix[row2];
    matrix[row2] = temp;
    // 记录变换过程
    process.add("R" + (row1 + 1) + " <-> R" + (row2 + 1));
}

public static int[][] rowEchelonForm(int[][] matrix, List<String> process) {
    int n = matrix.length;
    int m = matrix[0].length;
    int lead = 0;
    for (int r = 0; r < n; r++) {
        if (lead >= m) {
            break;
        }
        int i = r;
        while (matrix[i][lead] == 0) {
            i++;
            if (i == n) {
                i = r;
                lead++;
                if (lead == m) {
                    break;
                }
            }
        }
        if (matrix[i][lead] != 0) {
            swapRows(matrix, i, r, process);
            int lv = matrix[r][lead];
            for (int j = 0; j < m; j++) {
                matrix[r][j] /= lv;
            }
            for (int i1 = r + 1; i1 < n; i1++) {
                int k = matrix[i1][lead];
                elementaryRowOperations(matrix, r, i1, -k, process);
            }
        }
        lead++;
    }
    return matrix;
}

public static int[][] reducedRowEchelonForm(int[][] matrix, List<String> process) {
    matrix = rowEchelonForm(matrix, process);
    int n = matrix.length;
    int m = matrix[0].length;
    for (int r = n - 1; r >= 0; r--) {
        int lead = 0;
        for (int j = 0; j < m; j++) {
            if (matrix[r][j] != 0) {
                lead = j;
                break;
            }
        }
        if (matrix[r][lead] != 0) {
            for (int i = 0; i < r; i++) {
                int k = matrix[i][lead];
                elementaryRowOperations(matrix, r, i, -k, process);
            }
        }
    }
    return matrix;
}

public static void main(String[] args) {
    int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    List<String> process = new ArrayList<String>();
    matrix = reducedRowEchelonForm(matrix, process);
    // 输出变换过程
    for (String p : process) {
        System.out.println(p);
    }
    // 输出最简行列式
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix[0].length; j++) {
            if (matrix[i][j] != 0) {
                System.out.print(matrix[i][j] + " ");
                break;
            }
        }
        System.out.println();
    }
}

以上代码中的`elementaryRowOperations`和`swapRows`方法与之前的示例代码相同。`rowEchelonForm`方法实现了将矩阵转化为行阶梯形式的函数，`reducedRowEchelonForm`方法实现了将矩阵转化为最简行列式的函数。在`main`方法中，我们对一个3x3的矩阵进行了行变换，并输出变换过程和最简行列式。