Sent a picture of the solution to the problem (s).
 
        
        
        
Answer:
(A) Set A is linearly independent and spans  . Set is a basis for
. Set is a basis for  .
.
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors  ![A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D1%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%201%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%201%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20) , we are to decide which of the given statements is true:
 , we are to decide which of the given statements is true:
In Matrix ![A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D%281%29%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%20%281%29%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%20%281%29%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20) , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column.
 , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column.  has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans
 has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans  .
.
Therefore Set A is linearly independent and spans  . Thus it is basis for
. Thus it is basis for  .
.
 
        
             
        
        
        
Answer:
<h2>A.

</h2>
Step-by-step explanation:
The given formula is

Where  and
 and 
To solve for  , first we need to move
, first we need to move 

Then, we move the term 1

Finally, we move the factor 

Replacing given values, we have

Thereofre, the right answer is A.
 
        
             
        
        
        
Answer:
1
Step-by-step explanation:

 
        
             
        
        
        
Answer:
if you mulitply (19*19*27*10*10) you get: 
974700