Answer:
(A) Set A is linearly independent and spans 
. 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:
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. 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 .
Number 6 the y intercept is 50 because that’s the point where it crosses the y axis and 7 I believe it would be D= 200
The volume of a sphere refers to the number of cubic units that will exactly fill a sphere. The volume of a sphere can be found or calculate by using the formula V=4/3πr^3, where r represents the radius of the figure.
In this exercise is given that a sphere has a radius of 4 centimeters and it is asked to find its volume and use 3.14 as the value of π or pi. The first step would be substitute the values into the previous mention formula.
V=4/3πr^3
V=4/3(3.14)(4 cm)^3
V=4/3)(3.14)(64 cm^3)
V=267.9 cm^3
The volume of the sphere is 267.9 cubic centimeters.
Maybe it the mixed number form. 2 1/2?
