Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Consider the equations
$| \begin{matrix} y + 2 k z = 0 \\ x + 2 y + 6 z = 2 \\ k x + 2 z = 1 \end{matrix} |$
where $k$ is an arbitrary constant.
a. For which values of the constant $k$ does this system have a unique solution?
b. When is there no solution?
c. When are there infinitely many solutions?

(a)The system of equations will have a unique solution for the value, $k \neq \frac{1}{2}, 1$ .

(b)The system of equations will have no solution for the value, $k = 1$ .

(c)The system of equations will have infinitely many solutions for the value, $k = \frac{1}{2}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Transform the equations into a matrix form

In the system of equations, the variables can be eliminated by performing arithmetic operations on the equations.

Represent the system of equations in the form of matrix.

$| \begin{matrix} 0 & 1 & 2 k & | & 0 \\ 1 & 2 & 6 & | & 2 \\ k & 0 & 2 & | & 1 \end{matrix} |$

Step 2: Perform the row operations

Perform the row Echelon operation to reduce the matrix.

$\begin{matrix} (\begin{matrix} 0 & 1 & 2 k & 0 \\ 1 & 2 & 6 & 2 \\ k & 0 & 2 & 1 \end{matrix}) = (\begin{matrix} k & 0 & 2 & 1 \\ 0 & 2 & \frac{6 k - 2}{k} & \frac{2 k - 1}{k} \\ 0 & 0 & \frac{2 k^{2} - 3 k + 1}{k} & - \frac{2 k - 1}{2 k} \end{matrix}) \\ = (\begin{matrix} 1 & 0 & 0 & \frac{1}{k - 1} \\ 0 & 1 & 0 & \frac{k}{k - 1} \\ 0 & 0 & 1 & - \frac{1}{2 (k - 1)} \end{matrix}) \end{matrix}$

Step 3: Solve the equations

Consider the third row of the original matrix.

$| \begin{matrix} 0 & 1 & 2 k & | & 0 \\ 1 & 2 & 6 & | & 2 \\ k & 0 & 2 & | & 1 \end{matrix} |$

Put $k = \frac{1}{2}$

$| \begin{matrix} 0 & 1 & 1 & | & 0 \\ 1 & 2 & 6 & | & 2 \\ \frac{1}{2} & 0 & 2 & | & 1 \end{matrix} |$

Perform the row Echelon operation to reduce the matrix.

$\begin{matrix} (\begin{matrix} 0 & 1 & 1 & 0 \\ 1 & 2 & 6 & 2 \\ \frac{1}{2} & 0 & 2 & 1 \end{matrix}) = (\begin{matrix} 1 & 2 & 6 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) \\ = (\begin{matrix} 1 & 0 & 4 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) \end{matrix}$

Step 4: Consider the conditions for finding the type of solutions

As the number of variables equals the number of equations, thus, the system has unique solution, except for the values, $k = \frac{1}{2}, 1$ .

When $k = 1$ , the third column equals to undefined value. Thus, for, $k = 1$ the system has no solution.

For, $k = \frac{1}{2}$ the system will have infinitely many solutions as the third row equals to zero.

Step 5: Final answer

(a)The system of equations will have a unique solution for the value, $k \neq \frac{1}{2}, 1$ .

(b)The system of equations will have no solution for the value, $k = 1$ .

(c)The system of equations will have infinitely many solutions for the value, $k = \frac{1}{2}$ .

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Consider the equations
$| \begin{matrix} y + 2 k z = 0 \\ x + 2 y + 6 z = 2 \\ k x + 2 z = 1 \end{matrix} |$
where $k$ is an arbitrary constant.
a. For which values of the constant $k$ does this system have a unique solution?
b. When is there no solution?
c. When are there infinitely many solutions?

Step by Step Solution

Step 1: Transform the equations into a matrix form

Step 2: Perform the row operations

Step 3: Solve the equations

Step 4: Consider the conditions for finding the type of solutions

Step 5: Final answer

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Consider the equations|y+2kz=0x+2y+6z=2kx+2z=1|wherek is an arbitrary constant.a. For which values of the constant kdoes this system have a unique solution?b. When is there no solution?c. When are there infinitely many solutions?

Step by Step Solution

Step 1: Transform the equations into a matrix form

Step 2: Perform the row operations

Step 3: Solve the equations

Step 4: Consider the conditions for finding the type of solutions

Step 5: Final answer

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