/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 In Exercises \(1-4,\) find a lea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-4,\) find a least-squares solution of \(A \mathbf{x}=\mathbf{b}\) by (a) constructing the normal equations for \(\hat{\mathbf{x}}\) and (b) solving for \(\hat{\mathbf{x}}\) . $$ A=\left[\begin{array}{rr}{2} & {1} \\ {-2} & {0} \\ {2} & {3}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{-5} \\ {8} \\\ {1}\end{array}\right] $$

Short Answer

Expert verified
The least-squares solution is \( \hat{\mathbf{x}} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \).

Step by step solution

01

Set Up the Normal Equation

The normal equation for the least-squares solution is given by \( A^T A \hat{\mathbf{x}} = A^T \mathbf{b} \). Start by calculating \( A^T \), the transpose of matrix \( A \). The transpose is given by \( A^T = \begin{bmatrix} 2 & -2 & 2 \ 1 & 0 & 3 \end{bmatrix} \).
02

Compute A^T A

Multiply \( A^T \) by \( A \) to find \( A^T A \). This gives:\[A^T A = \begin{bmatrix} 2 & -2 & 2 \ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \ -2 & 0 \ 2 & 3 \end{bmatrix} = \begin{bmatrix} 12 & 8 \ 8 & 10 \end{bmatrix}\]
03

Compute A^T b

Multiply \( A^T \) by \( \mathbf{b} \) to find \( A^T \mathbf{b} \):\[A^T \mathbf{b} = \begin{bmatrix} 2 & -2 & 2 \ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} -5 \ 8 \ 1 \end{bmatrix} = \begin{bmatrix} -20 \ -14 \end{bmatrix}\]
04

Solve for \( \hat{\mathbf{x}} \)

Combine results from previous two steps into the normal equation \( A^T A \hat{\mathbf{x}} = A^T \mathbf{b} \). This becomes:\[\begin{bmatrix} 12 & 8 \ 8 & 10 \end{bmatrix} \hat{\mathbf{x}} = \begin{bmatrix} -20 \ -14 \end{bmatrix}\]Now, solve this system using the matrix equation method or via substitution/elimination to find \( \hat{\mathbf{x}} = \begin{bmatrix} -2 \ 1 \end{bmatrix} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Equations
In the realm of mathematics, normal equations are essential when tackling least-squares solutions for a system of linear equations. These equations come into play when we want to find the best approximate solution to the system when no exact solution exists. This happens often due to inconsistencies in the given system.
Constructing a normal equation involves taking the transpose of the matrix, denoted as \( A^T \), and forming the equation \( A^T A \hat{\mathbf{x}} = A^T \mathbf{b} \). This equation helps to find the vector \( \hat{\mathbf{x}} \), which represents the best-fit solution in the least squares sense.
The process starts by finding \( A^T \), the transpose of matrix \( A \). Then, you multiply it with \( A \) to get \( A^T A \), which is always a square and symmetric matrix. This setup allows us to proceed towards solving \( \hat{\mathbf{x}} \), the least-squares solution.
Matrix Multiplication
Matrix multiplication is a fundamental operation utilized in many mathematical computations, including finding least-squares solutions. When performing matrix multiplication, it's crucial to understand the rules to ensure the matrices are compatible.
For instance, when multiplying two matrices \( A \) and \( B \), the number of columns in \( A \) must equal the number of rows in \( B \). Each element of the resulting matrix is computed by taking the dot product of the corresponding row of the first matrix and the column of the second matrix.
  • Calculate \( A^T A \) by multiplying the transpose of the matrix \( A \) with \( A \) itself.
  • Also, compute \( A^T \mathbf{b} \) by multiplying the transpose of \( A \) with vector \( \mathbf{b} \).
These computations are indispensable for setting up the normal equations.
Systems of Equations
Systems of equations can be thought of as a collection of equations that are solved together. Each equation applies to the same set of variables, and the solutions are the points where all the equations intersect.
In scenarios where perfect solutions aren't feasible due to overdetermined systems, least-squares solutions enable finding the best approximate solutions. This is achieved by minimizing the sum of the squares of the differences between the given values and those predicted by the linear model.
The normal equations, derived using matrix multiplication as previously described, lead to a system that allows us to extract the least-squares solution. The goal is to solve \( A^T A \hat{\mathbf{x}} = A^T \mathbf{b} \), typically by methods such as substitution or elimination, which are conventional techniques in solving linear systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 17 and \(18,\) all vectors and subspaces are in \(\mathbb{R}^{n} .\) Mark each statement True or False. Justify each answer. a. If \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is an orthogonal basis for \(W,\) then mul- tiplying \(\mathbf{v}_{3}\) by a scalar \(c\) gives a new orthogonal basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, c \mathbf{v}_{3}\right\\} .\) b. The Gram-Schmidt process produces from a linearly in- dependent set \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{p}\right\\}\) an orthogonal set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) with the property that for each \(k,\) the vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\) span the same subspace as that spanned by \(\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}\) c. If \(A=Q R,\) where \(Q\) has orthonormal columns, then \(R=Q^{T} A\)

In Exercises 21 and \(22,\) all vectors and subspaces are in \(\mathbb{R}^{n} .\) Mark each statement True or False. Justify each answer. a. If \(W\) is a subspace of \(\mathbb{R}^{n}\) and if \(\mathbf{v}\) is in both \(W\) and \(W^{\perp}\) then \(\mathbf{v}\) must be the zero vector. b. In the Orthogonal Decomposition Theorem, each term in formula \((2)\) for \(\hat{\mathbf{y}}\) is itself an orthogonal projection of \(\mathbf{y}\) onto a subspace of \(W\) c. If \(\mathbf{y}=\mathbf{z}_{1}+\mathbf{z}_{2},\) where \(\mathbf{z}_{1}\) is in a subspace \(W\) and \(\mathbf{z}_{2}\) is in \(W^{\perp},\) then \(\mathbf{z}_{1}\) must be the orthogonal projection of \(\mathbf{y}\) onto \(W .\)

Let \(\overline{x}=\frac{1}{n}\left(x_{1}+\cdots+x_{n}\right)\) and \(\overline{y}=\frac{1}{n}\left(y_{1}+\cdots+y_{n}\right) .\) Show that the least-squares line for the data \(\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\) must pass through \((\overline{x}, \overline{y}) .\) That is, show that \(\overline{x}\) and \(\overline{y}\) satisfy the linear equation \(\overline{y}=\hat{\beta}_{0}+\hat{\beta}_{1} \overline{x}\) .[Hint: Derive this equation from the vector equation \(\mathbf{y}=X \hat{\boldsymbol{\beta}}+\boldsymbol{\epsilon} .\) Denote the first column of \(X\) by \(\mathbf{1}\) . Use the fact that the residual vector \(\epsilon\) is orthogonal to the column space of \(X\) and hence is orthogonal to \(\mathbf{1} . ]\)

Compute the quantities in Exercises \(1-8\) using the vectors $$ \mathbf{u}=\left[\begin{array}{r}{-1} \\ {2}\end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l}{4} \\ {6}\end{array}\right], \quad \mathbf{w}=\left[\begin{array}{r}{3} \\ {-1} \\ {-5}\end{array}\right], \quad \mathbf{x}=\left[\begin{array}{r}{6} \\ {-2} \\ {3}\end{array}\right] $$ \(\mathbf{u} \cdot \mathbf{u}, \mathbf{v} \cdot \mathbf{u},\) and \(\frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}}\)

In Exercises \(7-10,\) let \(W\) be the subspace spanned by the \(\mathbf{u}^{\prime}\) 's, and write \(\mathbf{y}\) as the sum of a vector in \(W\) and a vector orthogonal to \(W\) $$ \mathbf{y}=\left[\begin{array}{r}{4} \\ {3} \\ {3} \\\ {-1}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l}{1} \\ {1} \\\ {0} \\ {1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-1} \\\ {3} \\ {1} \\ {-2}\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{r}{-1} \\ {0} \\ {1} \\\ {1}\end{array}\right] $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.