Math Problem Statement

Let T: R^4 -> R^3 be a linear transformation such that T([1, 1, 0, -2]) = [2, 3, -1] and T([0, -1, 1, 1]) = [5, 0, 1]. Find T([1, 3, -2, -4]).

Solution

To solve this problem, we need to find the image of the vector [13−2−4]\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} under the linear transformation T:R4→R3T: \mathbb{R}^4 \to \mathbb{R}^3. We are given that

T[110−2]=[23−1]andT[0−111]=[501].T\begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \quad \text{and} \quad T\begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix}.

Step 1: Setting Up the Problem

Since TT is a linear transformation, any vector in R4\mathbb{R}^4 can be expressed as a linear combination of basis vectors. Let's express [13−2−4]\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} as a linear combination of [110−2]\begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} and [0−111]\begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix}, if possible.

Suppose: [13−2−4]=a[110−2]+b[0−111]\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} = a \begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} + b \begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix}

This gives us a system of equations for aa and bb:

a = 1 \\ a - b = 3 \\ b = -2 \\ -2a + b = -4 \end{cases}$$ ### Step 2: Solving for $$a$$ and $$b$$ From the equations: 1. From $$a = 1$$, we substitute into the second equation: $$1 - b = 3 \implies b = -2$$ 2. The third equation confirms $$b = -2$$. 3. The fourth equation is also consistent: $$-2(1) + (-2) = -4$$ So, we have $$a = 1$$ and $$b = -2$$. ### Step 3: Applying the Linear Transformation Now, we use linearity to find $$T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right)$$: $$T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = a T\left( \begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} \right) + b T\left( \begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix} \right)$$ Substituting $$a = 1$$ and $$b = -2$$, we get: $$T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = 1 \cdot \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} + (-2) \cdot \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix}$$ Calculating each component: $$= \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} + \begin{bmatrix} -10 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 - 10 \\ 3 + 0 \\ -1 - 2 \end{bmatrix} = \begin{bmatrix} -8 \\ 3 \\ -3 \end{bmatrix}$$ ### Final Answer $$T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = \begin{bmatrix} -8 \\ 3 \\ -3 \end{bmatrix}$$ Would you like more details on any step or have any questions? ### Related Questions 1. How do we verify if a transformation is linear? 2. What are other methods for finding the image of a vector under a linear transformation? 3. How can we represent linear transformations using matrices? 4. How do we determine if a set of vectors spans a vector space? 5. What is the significance of basis vectors in linear transformations? **Tip:** When solving systems of equations, always check if each equation is consistent with your solution to avoid errors in linear combinations.

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Math Problem Analysis

Mathematical Concepts

Linear Transformation
Linear Combinations
Vector Spaces
Systems of Linear Equations

Formulas

Linear transformation formula: T(a * v1 + b * v2) = a * T(v1) + b * T(v2)

Theorems

Linearity of Transformations
System of Equations Solving

Suitable Grade Level

Undergraduate Level (First-Year Linear Algebra)

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