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Chapter 6: Q1E (page 331)

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

1. \(A = \left[ {\begin{aligned}{{}{}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{1}}}&{\bf{3}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{4}}\\{\bf{1}}\\{\bf{2}}\end{aligned}} \right]\)

Short Answer

Expert verified

(a) \(\left[ {\begin{aligned}{{}{}}{ - 4}\\{11}\end{aligned}} \right]\)

(b)\(\left[ {\begin{aligned}{{}{}}3\\2\end{aligned}} \right]\)

Step by step solution

01

(a) Step 1: Find the products \({A^T}A\) and \({A^T}{\bf{b}}\)

Find the product \({A^T}A\).

\(\begin{aligned}{}{A^T}A &= \left[ {\begin{aligned}{{}{}}{ - 1}&2&{ - 1}\\2&{ - 3}&3\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - 1}&2\\2&{ - 3}\\{ - 1}&3\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}6&{ - 11}\\{ - 11}&{22}\end{aligned}} \right]\end{aligned}\)

Find the product \({A^T}{\bf{b}}\).

\(\begin{aligned}{}{A^T}{\bf{b}} &= \left[ {\begin{aligned}{{}{}}{ - 1}&2&{ - 1}\\2&{ - 3}&3\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}4\\1\\2\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - 4}\\{11}\end{aligned}} \right]\end{aligned}\)

02

Find the solution by constructing the normal equations

The normal equations can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} = {A^T}{\bf{b}}\\\left[ {\begin{aligned}{{}{}}6&{ - 11}\\{ - 11}&{22}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right] = \left[ {\begin{aligned}{{}{}}{ - 4}\\{11}\end{aligned}} \right]\end{aligned}\)

03

(b) Step 3: Find the component \({\bf{\hat x}}\)

The component \({\bf{\hat x}}\) can be calculated as:

\(\begin{aligned}{}{\bf{\hat x}} &= {\left( {{A^T}A} \right)^{ - 1}}\left( {{A^T}{\bf{b}}} \right)\\ &= {\left[ {\begin{aligned}{{}{}}6&{ - 11}\\{ - 11}&{22}\end{aligned}} \right]^{ - 1}}\left[ {\begin{aligned}{{}{}}{ - 4}\\{11}\end{aligned}} \right]\\ &= \frac{1}{{11}}\left[ {\begin{aligned}{{}{}}{22}&{11}\\{11}&6\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - 4}\\{11}\end{aligned}} \right]\\ &= \frac{1}{{11}}\left[ {\begin{aligned}{{}{}}{33}\\{22}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}3\\2\end{aligned}} \right]\end{aligned}\)

The \({\bf{\hat x}}\) component is \(\left[ {\begin{aligned}{{}{}}3\\2\end{aligned}} \right]\).

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Most popular questions from this chapter

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

3. \(\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

A Householder matrix, or an elementary reflector, has the form \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\) where u is a unit vector. (See Exercise 13 in the Supplementary Exercise for Chapter 2.) Show that Q is an orthogonal matrix. (Elementary reflectors are often used in computer programs to produce a QR factorization of a matrix A. If A has linearly independent columns, then left-multiplication by a sequence of elementary reflectors can produce an upper triangular matrix.)

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

2. \(\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{}}5\\6\\{ - 7}\end{aligned}} \right)\)
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