Question

Suppose T: ℝ3→ℝ2 is a linear transformation. Let U and V be the vectors given below, and suppose that T(U) and T(V) are as given. Find T(3U+2V).

U =

[1]

[2]

[1]

V =

[1]

[3]

[2]

T(U) =

[−2]

[5]

T(V) =

[−2]

[7]

T(3U+2V) = ?

Answer 1

Answer:

$T(3u+2v)=(-10,29)$

Step-by-step explanation:

T is a linear transformation, hence it is homogeneous (T(cr)=cT(r) for all real c and r∈ℝ³) and additive (T(r+s)=T(r)+T(s), for all r,s∈ℝ³). Apply these properties with r=3u and s=2v to obtain:

$T(3u+2v)=T(3u)+T(2v)=3T(u)+2T(v)=3(-2,5)+2(-2,7)=(-6,15)+(-4,14)=(-10,29)$

We don't have an explicit definition of T, so it's more difficult to compute T(3u+2v) directly without using these properties.

Answer 2

The value of T(3U+2V) is (-10, 29)

How to determine T(3U + 2V)?

The given parameters are:

Linear transformation: T: ℝ3→ℝ2

T(U) = [−2] [5]

T(V) = [−2] [7]

To calculate T(3U+2V), we make use of the following expression:

T(3U+2V) = T(3U) + T(2V)

This gives

T(3U+2V) = 3T(U) + 2T(V)

Substitute known values

T(3U+2V) = 3 * (-2,5) + 2 * (-2,7)

Evaluate the quotient

T(3U+2V) = (-6,15) + (-4,14)

Combine the coordinates

T(3U+2V) = (-6 - 4, 15 + 14)

Evaluate the sum and differences

T(3U+2V) = (-10, 29)

Hence, the value of T(3U+2V) is (-10, 29)

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