91Ó°ÊÓ

Consider the function \(y=f(x)\) defined for \(2 \times 1\) matrices \(x\) by \(y=A x,\) where $$ A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $$ Plot \(f(x)\) together with \(x\) in each case below. How would you describe the action of \(f ?\) (a) \(x=\left(\begin{array}{l}1 \\ 1\end{array}\right)\) (b) \(x=\left(\begin{array}{l}2 \\ 0\end{array}\right)\) (c) \(x=\left(\begin{array}{l}4 \\ 3\end{array}\right)\) (d) \(x=\left(\begin{array}{r}2 \\ -2\end{array}\right)\) Answer: $$f\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{c} x_{1}+x_{2} \\ x_{2} \end{array}\right)$$ A. $$f\left(\begin{array}{l} 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ (GRAPH CAN'T COPY) B. $$f\left(\begin{array}{l} 2 \\ 0 \end{array}\right)=\left(\begin{array}{l} 2 \\ 0 \end{array}\right)$$ (GRAPH CAN'T COPY) C. $$f\left(\begin{array}{l} 4 \\ 3 \end{array}\right)=\left(\begin{array}{l} 7 \\ 3 \end{array}\right)$$ (GRAPH CAN'T COPY) D. $$f\left(\begin{array}{c} 2 \\ -2 \end{array}\right)=\left(\begin{array}{c} 0 \\ -2 \end{array}\right)$$

Step by step solution

01

Understand the Function

The function \( y = f(x) \) is defined by the matrix equation \( y = Ax \), where \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \). This means for any input matrix \( x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \), the output \( y \) is given by performing the matrix multiplication: \( \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \).

02

Perform Matrix Multiplication

For each given input \( x \), multiply it by matrix \( A \) to find \( y = Ax \). Matrix multiplication for \( y = Ax \) results in: \( \begin{bmatrix} 1 \cdot x_1 + 1 \cdot x_2 \ 0 \cdot x_1 + 1 \cdot x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \ x_2 \end{bmatrix} \).

03

Step 3A: Compute for \( x = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)

Substitute \( x_1 = 1 \) and \( x_2 = 1 \) into the function: \( f\left(\begin{bmatrix} 1 \ 1 \end{bmatrix}\right) = \begin{bmatrix} 1 + 1 \ 1 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \end{bmatrix} \).

04

Step 3B: Compute for \( x = \begin{bmatrix} 2 \\ 0 \end{bmatrix} \)

Substitute \( x_1 = 2 \) and \( x_2 = 0 \) into the function: \( f\left(\begin{bmatrix} 2 \ 0 \end{bmatrix}\right) = \begin{bmatrix} 2 + 0 \ 0 \end{bmatrix} = \begin{bmatrix} 2 \ 0 \end{bmatrix} \).

05

Step 3C: Compute for \( x = \begin{bmatrix} 4 \\ 3 \end{bmatrix} \)

Substitute \( x_1 = 4 \) and \( x_2 = 3 \) into the function: \( f\left(\begin{bmatrix} 4 \ 3 \end{bmatrix}\right) = \begin{bmatrix} 4 + 3 \ 3 \end{bmatrix} = \begin{bmatrix} 7 \ 3 \end{bmatrix} \).

06

Step 3D: Compute for \( x = \begin{bmatrix} 2 \\ -2 \end{bmatrix} \)

Substitute \( x_1 = 2 \) and \( x_2 = -2 \) into the function: \( f\left(\begin{bmatrix} 2 \ -2 \end{bmatrix}\right) = \begin{bmatrix} 2 + (-2) \ -2 \end{bmatrix} = \begin{bmatrix} 0 \ -2 \end{bmatrix} \).

07

Analyze the Function's Action

From each computation, we see that the transformation (f) can be described as adding the two components of the input vector and keeping the second component unchanged. Therefore, the function shifts horizontally by \(x_2\) units while keeping the vertical component \(x_2\) constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication

Matrix multiplication is a fundamental concept in linear algebra, especially significant when talking about linear transformations. When we multiply a matrix by a vector, as in our exercise, we are essentially transforming the vector. The matrix in question, \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \), when multiplied with a vector \( x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \), results in a new vector, \( y = \begin{bmatrix} x_1 + x_2 \ x_2 \end{bmatrix} \).
This new vector is derived as follows: each element of the resulting vector is a sum of the products of corresponding elements from the row of the matrix and the incoming vector. For example:

• The first element, \( x_1 + x_2 \), results from multiplying the first row of \( A \) with vector \( x \).
• The second element, \( x_2 \), arises from the second row of \( A \). Here, you notice that only \( x_2 \) affects the result, based on the placement of '1' and '0' in the matrix row.

Understanding this multiplication process helps visualize how each element of the transformation is formed, and why these elements change the way they do.

Matrix Representation

In linear algebra, a matrix serves as a compact way to represent a variety of transformations. For the exercise provided, the matrix \( A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \) is more than just a box of numbers; it's a visual tool that shows how linear transformations are structured.

• Each row in the matrix applies a specific rule to the vector components (terms of the transformation).
• The way numbers are arranged determines how inputs are mapped to outputs. For example, the addition of the first two elements \( (x_1 + x_2) \) as the result of the transformation is a direct result of the first row of matrix \( A \).

By arranging numbers in such a form, matrices allow us to encapsulate multiple operations in a streamlined manner. They make it clear how each component of the input vector will influence the outcome by simply applying linear combinations of the original vector components. When we look at such matrices, we gain insight into how a specific linear transformation is expected to act on any given vector.

Vector Transformation

Vector transformation involves mapping one vector to another through operations, which often employ matrices. In our exercise, transforming the vector involves understanding the function \( f(x) = Ax \).
The transformation \( f(x) \) ideally describes an action where:

• The first element of the output is the sum of the vector's components, \( x_1 + x_2 \).
• The second component remains unchanged from the input vector.

This results in changes in direction and magnitude of the vector—a crucial concept in understanding vector transformations.For practical purposes:

• When plotting, the original vector is typically shown along with its transformed version for a visual comparison.
• This comparison helps describe the action as a shift or stretch, depending on the row operations of the transformation matrix.

Grasping vector transformations via matrix application offers a wider perspective on how we can rotate, shift, or even alter the scale of a vector. It's a cornerstone for fields that rely on advanced vector manipulations, like computer graphics and physics.