Python solution using stack principle:

def is_matched(expression):
    dic = {'{':'}','[':']','(':')'}
    lst =[]
    for i in expression:
        if i in dic.keys():
            lst.append(dic[i])
        elif len(lst)>0 and i==lst[-1]:
            lst.pop()
        else: 
            return False
    return len(lst) == 0