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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

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

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

(a)

  1. For \(h = 12\) and \(k \ne 2\), the solution set of the system is empty.
  2. For \(h \ne 12\), the solution set of the system contains a unique solution.
  3. For \(h = 12\) and \(k = 2\), the solution set of the system contains infinitely many solutions.

(b)

  1. For \(3h + k = 0\), the solution set of the system is empty.
  2. For \(3h + k \ne 0\), the solution set of the system contains a unique solution.
  3. The system of equations cannot have infinitely many solutions for any value of h and k.
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

(a) Step 1: Apply the row operation

Convert the system of equations \({x_1} + 3{x_2} = k\) and \(4{x_1} + h{x_2} = 8\) into the augmented matrix as shown below:

\(\left( {\begin{aligned}{*{20}{c}}1&3&k\\4&h&8\end{aligned}} \right)\)

A basic principle of this section is that row operations do not affect the solution set of a linear system.

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

\(\left( {\begin{aligned}{*{20}{c}}1&3&k\\0&{12 - h}&{4k - 8}\end{aligned}} \right)\)

(i)  Step 2: Check if the solution set of the system is empty

For the system of equations to be inconsistent, the solution must not satisfy the system of equations.

Obtain the value of \(h\) for which the value of \(12 - h\) is 0.

\(\begin{aligned}{c}12 - h = 0\\h = 12\end{aligned}\)

Obtain the value of k for which the value of \(4k - 8\) is 0.

\(\begin{aligned}{c}4k - 8 = 0\\4k = 8\\k = 2\end{aligned}\)

For \(h = 12\) and \(k \ne 2\), the matrix becomes as shown below:

\(\left( {\begin{aligned}{*{20}{c}}1&3&k\\0&0&{4k - 8 \ne 0}\end{aligned}} \right)\)

So, the system is in the form of \(\left( 0 \right){x_2} = b\), where \(b \ne 0\), which cannot be possible.

It means, for \(h = 12\) and \(k \ne 2\), the solution set of the system is empty.

(ii)  Step 3: Check if the solution set of the system contains a unique solution

For the system of equations to be consistent, the solution must satisfy the system of equations.

For \(h \ne 12\), the matrix becomes as shown below:

\(\left( {\begin{aligned}{*{20}{c}}1&3&k\\0&{12 - h \ne 0}&{4k - 8}\end{aligned}} \right)\)

There are two pivot columns when \(h \ne 12\). Also, row two must give a solution.

Thus, for \(h \ne 12\), the solution set of the system contains a unique solution.

(iii) Step 4: Check if the solution set of the system contains infinitely many solutions

For \(h = 12\) and \(k = 2\), the matrix becomes as shown below:

\(\left( {\begin{aligned}{*{20}{c}}1&3&k\\0&0&0\end{aligned}} \right)\)

So, the system is in the form of \(\left( 0 \right){x_2} = 0\), where \(b = 0\). Also, it has one free variable.

Thus, for \(h = 12\) and \(k = 2\), the solution set of the system contains infinitely many solutions.

(b) Step 5: Apply the row operation

Convert the system of equations \( - 2{x_1} + h{x_2} = 1\) and \(6{x_1} + k{x_2} = - 2\) into the augmented matrix, as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{ - 2}&h&1\\6&k&{ - 2}\end{aligned}} \right)\)

A basic principle of this section is that row operations do not affect the solution set of a linear system.

Use the \( - 2{x_1}\) term in the first equation to eliminate the \(6{x_1}\) term from the second equation. Add 3 times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}{ - 2}&h&1\\0&{3h + k}&1\end{aligned}} \right)\)

(i)  Step 6: Check if the solution set of the system is empty

The system of equations is inconsistent if the solution does not satisfy it.

For \(3h + k = 0\), the matrix becomes as shown below:

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

The system is in the form of \(\left( 0 \right){x_2} = 1\), which is not possible.

Thus, for \(3h + k = 0\), the solution set of the system is empty.

(ii)  Step 7: Check if the solution set of the system contains a unique solution

The system of equations is consistent if the solution satisfies it.

For \(3h + k \ne 0\), the matrix becomes as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{ - 2}&h&1\\0&{3h + k \ne 0}&1\end{aligned}} \right)\)

There are two pivot columns when \(3h + k \ne 0\). Also, row two must give a solution.

Thus, for \(3h + k \ne 0\), the solution set of the system contains a unique solution.

(iii) Step 8: Check if the solution set of the system contains infinitely many solutions

From the above explanation, at \(3h + k \ne 0\), and 1 is on the right side of the equation.

Thus, for no values of h and k the solution set of the system contains infinitely many solutions.

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