Whats the answer for these linear transformations questions?

Question

Q1: use function to find (a) the image of v and (b) the preimage of w.

1-T(v₁,v₂)=(v₁+v₂,v₁-v₂),v=(3,-14),w=(3,19)

2-T(v₁,v₂,v₃)=(v₂-v₁,v₁+v₂,2v₁),v=(2,3,0),w=(-11,-1,10)

Q2:determine whether the fuction is a linear transformation(explain).

1-T:R²→R²,T(x,y)=(x,1)

2-T:R³→R³,T(x,y,z)=(x+y,x-y,z)

Q3:let T:R³→R³ be a linear transformation such that T(1,0,0)=(2,4,-1),T(0,1,0)=(1,3,-2),and T(0,0,1)=(0,-2,2) Find :

1-T(0,3,-1)

Q4:the linear transformation T:Rⁿ→R^m is defined by T(v)=Av.Find the dimensions of Rⁿ and R^m.

1-A= ⌈ 0 1 -2 1⌉

-1 4 5 0

⌊ 0 1 3 1⌋.

2-A= ⌈ 1 2 ⌉

-2 4

⌊-2 2 ⌋.

Answer 1

Q1.1

T(v₁, v₂) = (v₁ + v₂, v₁ - v₂)

T(3, -14) = (3 - 14, 3+14) = (-11, 17).

w = (3, 19) = T(v₁, v₂). Therefore, we get a system of two linear equations

v₁ + v₂ = 3

v₁ - v₂ = 19.

The solution is (v₁, v₂) =(11, -8).

Q1.2

T(v₁, v₂, v₃) = (v₂ - v₁, v₁ + v₂, 2v₁)

T(2, 3, 0) = (3 - 2, 3 + 2, 2^.2) = (1, 5, 4).

w = (-11, -1, 10) = (v₂ - v₁, v₁ + v₂, 2v₁). Therefore, we get a system of three linear equations

-v₁ + v₂ = -11

v₁ + v₂ = -1

2v₁ = 10.

The solution is v₁ = 5, v₂ = -6 and v₃ can be any number because it not a variable of the equations.

Q2.1

Let us assume that T(x, y) = (x, 1) is a linear transformation. Then

T(ax, ay) = aT(x, y) = a(x, 1) = (ax, a).

However, by the definition

T(ax, ay) = (ax, 1).

This contradiction proves that T(x, y) = (x, 1) is not a linear transformation.

Q2.2

It is easy to check that for T(x, y, z)=(x+y, x-y, z)

T(aR₁ + bR₂) = aT(R₁) + bT(R₂).

Here R₁ = (x₁, y₁, z₁) and R₂ = (x₂, y₂, z₂). Therefore, the transformation is linear.

Q3.

T(0, 3, -1) = 0^. T(1, 0, 0) + 3 T(0, 1, 0) - T(0, 0, 1) = 3(1, 3, -2) - (0, -2, 2) = (3, 11, -8).

Q4.1

A is a 3x4 matrix. Hence, it should apply to 4-dimensional vector and produce 3-dimensional vector. Thus n = 4 and m = 3.

Q4.2

A is 3x2 matrix. Hence n = 2 and m = 3.