Determine whether each set equipped with the given operations is a vector space. For those that are not vector spaces identify the vector space axioms that fail
1. The set of all 2 X 2 invertible matrices with the standard matrix addition and scalar multiplication.
2. The set of all real-line functions f defined everywhere on the real line and such that f( 1)=0 with the addition and multiplication operations

1) It is not a vector space

2) It is a vector space

Step-by-step explanation:

1) This is not a vector space. There is no neuter element for the addition, since the null matrix is not invertible.

2) This is a vector space. Lets denote the set given by this item A. A is, In fact, this is a subspace of the vector space given by the real-line funcitons defined in all the real line. Note that if k is a real number and f is an element of A, then

k*A is an element of A, because k*f is defined everywere (kf(x) = k* (f(x)) )

and k*f(1) = k*0 = 0.

If f,g are elements of A, then f+g is an element of A: f+g(x) = f(x) + g(x) is defined everywhere and f+g(1) = f(1)+g(1) = 0+0 = 0.

Also, the zero function is definced everywhere and in 1 it takes the value 0. Since A is a subspace of the vector space given by the real line functions, then it s indeed a vector space.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)