你可以用一个函数对数组进行排序，该函数返回偶数值的true。

const
    move = (array, fn) => array.sort((a, b) => fn(b) - fn(a)),
    array = [5, 1, 2, 4, 7, 9];

move(array, v => v % 2 === 0);

console.log(...array);

如果您不介意性能较低的解决方案，filter在这里可以非常清楚：

let matriz = [
    [5, 1, 2, 4, 7, 9],
    [2, 4, 3, 1, 7, 9],
    [1, 2, 3, 4, 5, 6],
]

for (let i = 0; i < matriz.length; i++) {
  matriz[i] = matriz[i].filter(x => x%2 == 0)
                        .concat(
                         matriz[i].filter(x => x%2 != 0))
}

console.log(matriz)

(the问题是，您需要遍历每个内部数组两次，而只需一次即可）

如果您只是使用第一个偶数元素切换索引

let matriz = [
[5, 1, 2, 4, 7, 9],
[2, 4, 3, 1, 7, 9],
[1, 2, 3, 4, 5, 6],]

for (let i = 0; i < matriz.length; i++) {
  let innerArrLength = matriz[i].length
  for (let j = 0; j < innerArrLength; j++) {
    if (matriz[i][j] % 2 === 0) {
        if(j === 0){
          break
        }else{
          const firstIndex = matriz[i][0];
          matriz[i][0] = matriz[i][j];
          matriz[i][j] = firstIndex
          break
        }
    } 
 }
}
console.log(matriz)

将第一个偶数置于最前面

let matriz = [ 
 [5, 1, 2, 4, 7, 9],
 [2, 4, 3, 1, 7, 9],
 [1, 2, 3, 4, 5, 6],]

 for (let i = 0; i < matriz.length; i++) {
  let innerArrLength = matriz[i].length
  for (let j = 0; j < innerArrLength; j++) {
    if (matriz[i][j] % 2 === 0) {
        if(j === 0){
          break
        }else{
          const element = matriz[i][j];
          delete matriz[i][j]
          matriz[i].unshift(element)
          break
        }
    } 
 }
}
console.log(matriz)