Cho ánh xạ tuyến tính ƒ:R^3—›R^3 và hai cơ sở của R^3 B={b1=(1,3,3),b2=(—7,6,5),b3=(1,2,2 )}, C={c1=(2,-4,3),c2=( -5,-1,-5),c3=(1,0,1)}
Biết f(b1)=(—5,6,4), f(b2)=(-8,5,2), f(b3)=(-9,-2,6)5)
a. Tìm mạ trận A =[f]E3 , từ đó suy ra biểu của f(x1,x1,x3)
b. Tìm ma trận [f]B
c. Từ [f]B, hãy tính ma trận [f]C/B
d. Cho vector z =(-2,3,7), hay tính [x]B, và suy ra [f(x)]c

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

to find the total number of combinations, multiply the quantity of each item by each other:

2 pairs of shoes x 3 pairs of shorts = 6 different combinations.

answer is (b.

164 ft2)

a base  

18

b base  

7

h height  

8

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen