Editorial - Matlan - Soal TKA 2

Problem 1

ƒ(x)=m*x3+4*x2-n*x+2

mod(ƒ(x),x+2)=12

mod(ƒ(x),x-1)=3

mod(ƒ(x),2*x2-2*x+1)=?

ƒ(1)=3=m+4-n+2

m-n=-3

ƒ(-2)=12=-8*m+16+2*n+2

8*m-2*n=6

4*m-n=3

3*m=6

m=2

-n=-3-2=-5

n=5

ƒ(x)=2*x3+4*x2-5*x+2

Divide ƒ(x) by 2*x2-2*x+1

We find that the remainder is -1

Problem 5

P=[[3,-2],[n+1,3]]

Q=[[1,m-2],[3,4]]

R=[[4,5],[6,7]]

P+Q=R

T=(P^t)+Q2-R

det(T)=?

n+1+3=6

n=2

-2+m-2=5

m=9

T=[[,],[,]]

Problem 6

A=[[3,-1],[4,2]]

(B^t)=[[5,4],[2,2]]

det((A⋅B)(-1))=?

B=[[5,2],[4,2]]

det((A⋅B)(-1))=1/det(A⋅B)

A⋅B=[[3,-1],[4,2]]*[[5,2],[4,2]]=[[11,4],[28,12]]

det(A⋅B)=1/(11⋅12-28⋅4)=1/20

det((A⋅B)(-1))=20

Problem 9

y=a*(x-4)2+10

-6=16*a+10

16*a=-16

a=-1

y=(X-4)2+10

Scale by 2, centered on origin

1/2*y=-(x/2-4)2+10

y=-2*(x/2-4)2+20

y=-2⋅(x2)/4+8*x-32+20

y=-12+8*x+(-x2)/2

Problem 10

ƒ(x)=a*bx+c

a+c=2

a*b+c=6

a*b2+c=18

Just by observation, b=3

a+c=2

3*a+c=6

2*a=4

a=2

c=0

∴a+b+c=2+3+0=5

Problem 11

ƒ(x)=3(x-2)

Translated by [[-2],[3]]

[[(x^′)],[(y^′)]]=[[x],[y]]+[[-2],[3]]

Substitute.

y-3=3x

y=3x+3=g(x)

Rapidfire some subproblems.

g(1)=6 True

g(0)=3 False

g(-1) +g(2)=1/3+3+9+3≠15 False

Subproblem 11.4

g(ƒ(2))=12

g(ƒ(2))=g(1)=6

The statement is false.

Subproblem 11.5

ƒ(g(1))=81

g(1)=6

ƒ(6)=34=81

The statement is true.

Problem 18

Subproblem 18.1

DE=a+c

DE=DA+DH

True.

Subproblem 18.2

DE=2√(,2)

True, just by pythagoras.

Subproblem 18.3

DB+BF=2√(,3)

DB=(2,2,0)

BF=(0,0,2)

DB+BF=(2,2,2)=√(,4+4+4)=2√(,3)

True!

Subproblem 18.4

DC+BF+BA=c-a=DH-DA

[[0],[2],[0]]+[[0],[0],[2]]+[[-2],[0],[0]]=[[0],[0],[2]]-[[],[],[]]

IDK ABOUT VECTORS BRO

Problem 19

PQ=PC+CQ=CA/2+(2*CB)/3

1/2*[[-2],[-8],[6]]+2/3*[[-15],[3],[9]]=[[-11],[-2],[9]]

|PQ|=√(,121+4+81)=√(,206)