I don't see any error on your code, everything is well written and simple. Except trying putting PUBLIC on your Class displayer. If you encounter the same problem again try to rebuild your whole code and try to recompile the whole code. Even try restarting your personal computer.
The best option describes the purpose of configuring Native Supplicant Profile on the Cisco ISE is it enforces the use of MSCHAPv2 or EAP-TLS for 802.1X authentication.
Answer:
The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)
Then satisfying this theorem the system is consistent and has one single solution.
Explanation:
1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)

Then the system is consistent and has a unique solution.
<em>E.g.</em>

2) Writing it as Linear system


3) The Rank (A) is 3 found through Gauss elimination


4) The rank of (A|B) is also equal to 3, found through Gauss elimination:
So this linear system is consistent and has a unique solution.
Answer:
Explanation:
The following Python program uses a combination of dictionary, list, regex, and loops to accomplish what was requested. The function takes a file name as input, reads the file, and saves the individual words in a list. Then it loops through the list, adding each word into a dictionary with the number of times it appears. If the word is already in the dictionary it adds 1 to its count value. The program was tested with a file named great_expectations.txt and the output can be seen below.
import re
def wordCount(fileName):
file = open(fileName, 'r')
wordList = file.read().lower()
wordList = re.split('\s', wordList)
wordDict = {}
for word in wordList:
if word in wordDict:
wordDict[word] = wordDict.get(word) + 1
else:
wordDict[word] = 1
print(wordDict)
wordCount('great_expectations.txt')
The elements in a string type array will be initialized to "Null".