Keypad and Pin 13

I have a keypad matrix set up on the Uno using pin 13 as one of the columns. Since the pin is the breakout for the on-board LED, the LED is always on on the board (or that’s what I’m thinking). Is this a problem for the Arduino?

#include <Keypad.h>

const byte rows = 4; //four rows
const byte cols = 4; //four columns

char keys[rows][cols] = {
  {'1','2','3','A'},
  {'4','5','6','B'},
  {'7','8','9','C'},
  {'*','0','#','D'}
};
byte rowPins[rows] = {9,8,7,6}; //connect to the row pinouts of the keypad
byte colPins[cols] = {13,12,11,10}; //connect to the column pinouts of the keypad
Keypad keypad = Keypad( makeKeymap(keys), rowPins, colPins, rows, cols );

//the pitch to play if the wrong code entered.
const int fail = 80;
const int piezoPin = 5;

const int ledPinRed = 4;
const int ledPinGreen = 3;

//set up an array to hold the code
char code[6] = {'1','2','3','4','5','6' };
//set up an array to hold the user's entry
char input[6];

boolean codeFlag = true;

void setup() {
  Serial.begin(9600);
  pinMode(ledPinRed, OUTPUT);
  pinMode(ledPinGreen, OUTPUT);
}
void loop() {
  //input the code.
  for (int i = 0; i < 6; i ++) {
    char key = keypad.waitForKey();
    input[i] = key;
    Serial.print(input[i]);
  }
  //debug: print the code when finished.
  for (int i = 0; i < 6; i++) {
    if (code[i] != input[i])
      codeFlag = false;
  }
  if (codeFlag == false) {
    Serial.println("\nACCESS DENIED"); 
    tone(piezoPin, fail, 1000);
    digitalWrite(ledPinRed, HIGH);
    delay(1000);
    digitalWrite(ledPinRed, LOW);
    // reset flag
    codeFlag = true;
  }
  else {
    Serial.println("\nACCESS GRANTED");
    tone(piezoPin, 144, 1000);
    digitalWrite(ledPinGreen, HIGH);
    delay(1000);
    digitalWrite(ledPinGreen, LOW);
  }
}

Nope, not a problem.