The green square is 18 and the red is 32
Answer:
$23.40
Step-by-step explanation:
Positive: 690
Negative: -30
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 
.
