Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that
. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then
.
C) Consider
. This set is orthogonal because
, but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in
. Then the columns of A form an orthonormal set. We have that
. To see this, note than the component
of the product
is the dot product of the i-th row of
and the jth row of
. But the i-th row of
is equal to the i-th column of
. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set
and suppose that there are coefficients a_i such that
. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then
then
.
Answer:
75
Step-by-step explanation:
f(1) = 7
f(n) = 3f(n-1) + 3
So what you are trying to do here is find the recursive value (that's what it is called) for f(n). Computers love this sort of thing, but we humans have to do it slowly and carefully.
So let's try f(2)
That means that f(2) = 3*f(n-1) + 3
but if f(2) is used it means that you have to know f(2-1) which is just f(1) and we know that.
so f(2) = 3*f(1)+3
f(2) = 3*7 + 3
f(2) = 21 + 3
f(2) = 24
Now do it again. We now know F(2), so we should be able to find f(3)
f(3) = 3*f(3 - 1) + 3
f(3) = 3*f(2) + 3 We know that f(2) = 24
f(3) = 3* 24 + 3
f(3) = 72 + 3
f(3) = 75
Answer:
y= (-7/16)
Step-by-step explanation:
Multiply (3/4)(1/4)= (3/16)
y= (3/16)-(5/8)
y= (-7/16)
I hope this is right, im not 100% sure