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Q5Q

Expert-verifiedFound in: Page 1

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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)

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

(b)

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

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.

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

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.

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.

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.

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.

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

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.

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.

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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