Answer:
C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.
Step-by-step explanation:
The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.
- There is an n×n matrix C such that CA=
. - There is an n×n matrix D such that AD=.
- The equation Ax=0 has only the trivial solution x=0.
- A is row-equivalent to the n×n identity matrix .
- For each column vector b in
, the equation Ax=b has a unique solution. - The columns of A span .
Therefore the statement:
If there is an n X n matrix D such that AD=I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix:
- The equation Ax=0 has only the trivial solution x=0.
- A is row-equivalent to the n×n identity matrix .
The correct option is C.
9.28 plus 2 is 11.28. 1/4 of 18 is 4.50. Add 4.50 to 9.28 and your total is $15.78.
The equation would be:
9.28+18(.25)+2=c
Answer:
For w = 18 units perimeter is minimum
P = 2(18 + w)
Step-by-step explanation:
Given;
Area of the rectangle = 324 units²
P is the perimeter
w is the width
Let L be the length of the rectangle
therefore,
P = 2(L + w) ............(1)
also,
Lw = 324
or
L = 
..........(2)
substituting 2 in 1
P = 
now,
for minimizing the perimeter
= 0
or
= 0
or
= 0
or
= -1
or
w² = 324
or
w = 18 units
For w = 18 units perimeter is minimum
therefore,
from 2
L = 
or
L = 18 units
objective function for P is:
P = 2(18 + w)
Answer:
I dont know how to but 0.987 as a fraction is 987/1000
Step-by-step explanation:
Answer:
1) 1 element
2) 13 elements
3) 22 elements
4) 40 elements
Step-by-step explanation:
1) Only one element will have no tails: the event that all the coins are heads.
2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.
3) There are 
elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are 
possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22
4) We can have:
- A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
- One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)
For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)
For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).
This gives us a total of 18+22 = 40 possibilities.
