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.
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:
!(key == 'q')
Explanation:
Based on the description, the coded expression that would equate to this would be
!(key == 'q')
This piece of code basically states that "if key pressed is not equal to q", this is because the ! symbol represents "not" in programming. Meaning that whatever value the comparison outputs, it is swapped for the opposite. In this case, the user would press anything other than 'q' to continue, this would make the expression output False, but the ! operator makes it output True instead.
Answer:
it is a complete routine of life without it we cannot do time management
Explanation:
mark me brainliest ❤
def zipZapZop():
number = int(input("Enter the number: "))
dictionary = {3: "zip", 5: "zap", 7: "zop"}
amount = 0<em> #amount of non-divisible numbers by 3, 5 and 7</em>
<em> for key, value in dictionary.items():</em>
if(number%key == 0): <em>#key is the number</em>
print(value) <em>#value can be or zip, or zap, or zop</em>
else: amount += 1 #the number of "amount" increases every time, when the number is not divisible by 3, or 5, or 7
if(amount == 3): print(number) <em>#if the number is not by any of them, then we should print the number</em>
zipZapZop()
