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Linear Algebra and its Applications
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Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Consider the problem of determining whether the following system of equations is consistent:

\(\begin{aligned}{c}{\bf{4}}{x_1} - {\bf{2}}{x_2} + {\bf{7}}{x_3} = - {\bf{5}}\\{\bf{8}}{x_1} - {\bf{3}}{x_2} + {\bf{10}}{x_3} = - {\bf{3}}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of linear combinations. Then solve that problem.

  1. Define an appropriate matrix, and restate the problem using the phrase “columns of A.”

  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

a. Vector b is in the span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

b. Vector b is in linear combinations of the columns of matrix A.

c. Vector b is in the range of T.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

(a) Step 1: Write the system in the vector form

The system of equations \(4{x_1} - 2{x_2} + 7{x_3} = - 5\) and \(8{x_1} - 3{x_2} + 10{x_3} = - 3\) can be represented in the matrix equation form \(A{\bf{x}} = {\bf{b}}\) as follows:

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\)

Assume that the column vectors are \({{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}4\\8\end{aligned}} \right)\), \({{\bf{v}}_2} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\), \({{\bf{v}}_3} = \left( {\begin{aligned}{*{20}{c}}7\\{10}\end{aligned}} \right)\), and \({\bf{b}} = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\).

Step 2: Convert the system of equations into the row-reduced echelon form

Convert the system of equations \(4{x_1} - 2{x_2} + 7{x_3} = - 5\) and \(8{x_1} - 3{x_2} + 10{x_3} = - 3\) into the augmented matrix as shown below:

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\8&{ - 3}&{10}&{ - 3}\end{aligned}} \right)\)

Use the \(4{x_1}\) term in the first equation to eliminate the \(8{x_1}\) term from the second equation. Add \( - 2\) times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\0&1&{ - 4}&7\end{aligned}} \right)\)

In the above augmented matrix, there are no free variables. It means the system of equations is consistent.

Thus, b is in the span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

(b) Step 3: Write in terms of columns of A

The system of equations \(4{x_1} - 2{x_2} + 7{x_3} = - 5\) and \(8{x_1} - 3{x_2} + 10{x_3} = - 3\) or the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\8&{ - 3}&{10}&{ - 3}\end{aligned}} \right)\) in the vector form is shown below:

\({x_1}\left( {\begin{aligned}{*{20}{c}}4\\8\end{aligned}} \right) + {x_2}\left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{*{20}{c}}7\\{10}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\)

From the above row-reduced echelon form,

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\0&1&{ - 4}&7\end{aligned}} \right)\).

The system of equations is consistent.

Thus, b is in linear combinations of the columns of matrix A.

(c) Step 4: If b is in the range of T

Consider the transformation \(T\left( {\bf{x}} \right) = A{\bf{x}}\).

It can also be written as shown below:

\(T\left( {\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

Let the solution be \({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)\).

Then, the transformation becomes:

\(\begin{aligned}{c}T\left( {\left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4\left( {9/4} \right) - 14 + 0}\\{8\left( {9/4} \right) - 21 + 0}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\\ = {\bf{b}}.\end{aligned}\)

So, \(T\left( {\bf{x}} \right) = {\bf{b}}\).

Thus, vector b is in the range of T.

Most popular questions for Math Textbooks

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

15. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\)

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Let \(u = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) and \(v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\). Show that \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in Span \(\left\{ {u,v} \right\}\) for all \(h\) and\(k\).

Question: If A is a non-zero matrix of the form, [a-bba] then the rank of A must be 2.

Determine the value(s) of \(a\) such that \(\left\{ {\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right)} \right\}\) is linearly independent.
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