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Linear Algebra and its Applications
Found in: Page 1
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

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.

The vectors are linearly independent for all values of \(a\), except \(a = 2\) and \(a = - 1\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the vector in the augmented matrix form

Write the vector in the augmented matrix form.

\(\begin{aligned}{l}{x_1}{{\mathop{\rm v}\nolimits} _1} + {x_2}{{\mathop{\rm v}\nolimits} _2} = {{\mathop{\rm v}\nolimits} _3}\\{x_1}\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right) + {x_2}\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\end{aligned}} \right)\,....\left( * \right)\\\left( {\begin{aligned}{*{20}{c}}1&a&0\\a&{a + 2}&0\end{aligned}} \right)\end{aligned}\)

Step 2: Apply the row operation

At row two, multiply row one by \(a\) and subtract it from row two.

\(\begin{aligned}{l}\left( {\begin{aligned}{*{20}{c}}1&a&0\\0&{a + 2 - {a^2}}&0\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}1&a&0\\0&{\left( {2 - a} \right)\left( {1 + a} \right)}&0\end{aligned}} \right)\end{aligned}\)

Step 3: Determine the value of \(a\) 

The columns of matrix \(A\) are linearly independent if and only if the equation \(Ax = 0\) has only a trivial solution.

There is a non-trivial solution for equation (*) if and only if \(\left( {2 - a} \right)\left( {1 + a} \right) = 0\).

Thus, the vectors are linearly independent for all values of \(a\), except \(a = 2\) and \(a = - 1\).

Most popular questions for Math Textbooks

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\8\end{array}} \right],\) and \({\rm{y = }}\left[ {\begin{array}{*{20}{c}}h\\{ - 5}\\{ - 3}\end{array}} \right]\). For what values(s) of \(h\) is \(y\) in the plane generated by \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)

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.

16. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\3\end{array}} \right]\)

If A is an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

In Exercises 23 and 24, key statements from this section are either quoted directly, restated slightly (but still true), or altered in some way that makes them false in some cases. Mark each statement True or False, and justify your answer.(If true, give the approximate location where a similar statement appears, or refer to a definition or theorem. If false, give the location of a statement that has been quoted or used incorrectly, or cite an example that shows the statement is not true in all cases.) Similar true/false questions will appear in many sections of the text.

24.

a. Elementary row operations on an augmented matrix never change the solution set of the associated linear system.

b. Two matrices are row equivalent if they have the same number of rows.

c. An inconsistent system has more than one solution.

d. Two linear systems are equivalent if they have the same solution set.

Determine h and k such that the solution set of the system (i) is empty, (ii) contains a unique solution, and (iii) contains infinitely many solutions.

a. \({x_1} + 3{x_2} = k\)

\(4{x_1} + h{x_2} = 8\)

b. \( - 2{x_1} + h{x_2} = 1\)

\(6{x_1} + k{x_2} = - 2\)
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