Suppose T: ℝ2→ℝ3 is a linear transformation. Let U, V and W be the vectors given below, and suppose that T(U) and T(V) are as given. Find T(W).

U = [1, 1]^T
V = [3, 4]^T
W = [-10, -13]^T
T(U) = [0, 0, 1]^T
T(V) = [-5, -2, 6]^T

To find T(W), we need to understand how linear transformations work.

A linear transformation T takes a vector from one vector space to another. In this case, T is a linear transformation from ℝ2 to ℝ3.

To find T(W), we can use the linearity property of T.

The linearity property states that for any vectors U and V, and any scalar k, T(kU + V) = kT(U) + T(V).

In this case, we know T(U) and T(V), so we can use these values to find T(W).

Let's start by finding the scalar k. We know U = [1, 1]^T, and T(U) = [0, 0, 1]^T. Using the linearity property, we can say:

T(2U) = 2T(U)
⇒ T([2, 2]^T) = [0, 0, 2]^T

Since T([2, 2]^T) is equal to [0, 0, 2]^T, we can conclude that k = 2.

Now, let's find T(W) using the linearity property:

T(W) = T(-10U) + T(-13V)
= -10T(U) + (-13)T(V)
= -10[0, 0, 1]^T + (-13)[-5, -2, 6]^T
= [0, 0, -10] + [65, 26, -78]
= [65, 26, -88]

Therefore, T(W) = [65, 26, -88]^T.

To find T(W), we need to use the fact that T is a linear transformation.

For any vector X, we can write it as a linear combination of basis vectors. Let's say X = aU + bV, where a and b are scalars. Then, T(X) = T(aU + bV).

Since T is a linear transformation, we know that T(aU + bV) = aT(U) + bT(V).

Now, let's substitute the given values into the equation. We have T(W) = aT(U) + bT(V), where W = -10U - 13V.

Substituting the values of U, V, T(U), and T(V), we have T(W) = a[0, 0, 1]^T + b[-5, -2, 6]^T.

Now, we need to find the values of a and b. To do this, we can use the fact that W = -10U - 13V.

Substituting the values of U and V, we get [-10, -13]^T = a[1, 1]^T + b[3, 4]^T.

This gives us a system of equations:

-10 = a + 3b
-13 = a + 4b

We can solve this system of equations to find the values of a and b.

From the first equation, we have a = -10 - 3b.

Substituting this into the second equation, we get -13 = (-10 - 3b) + 4b.

Simplifying, we have -13 = -10 + b.

Now, we can solve for b by subtracting -10 from both sides: -13 + 10 = b.

This gives us b = -3.

Substituting this value of b into the first equation, we have a = -10 - 3(-3).

Simplifying, we get a = -10 + 9 = -1.

Therefore, a = -1 and b = -3.

Now, we can substitute these values of a and b into T(W) = aT(U) + bT(V).

T(W) = (-1)[0, 0, 1]^T + (-3)[-5, -2, 6]^T.

Simplifying, we get T(W) = [0, 0, -1] + [15, 6, -18].

Adding the corresponding elements, we have T(W) = [15, 6, -19]^T.

Therefore, T(W) = [15, 6, -19]^T.