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Chapter 1: Q5Q (page 1)

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

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

Short Answer

Expert verified

(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.

Step by step solution

01

(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.

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

02

(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.

03

(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.

04

(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.

05

(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.

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

06

(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.

07

(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.

08

(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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Most popular questions from this chapter

Use the accompanying figure to write each vector listed in Exercises 7 and 8 as a linear combination of u and v. Is every vector in \({\mathbb{R}^2}\) a linear combination of u and v?

8.Vectors w, x, y, and z

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice that u - vis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

4. u and v as in Exercise 2

In Exercises 10, write a vector equation that is equivalent tothe given system of equations.

10. \(4{x_1} + {x_2} + 3{x_3} = 9\)

\(\begin{array}{c}{x_1} - 7{x_2} - 2{x_3} = 2\\8{x_1} + 6{x_2} - 5{x_3} = 15\end{array}\)

The following equation describes a Givens rotation in \({\mathbb{R}^3}\). Find \(a\) and \(b\).

\(\left( {\begin{aligned}{*{20}{c}}a&0&{ - b}\\0&1&0\\b&0&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{4}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{\bf{2}}\sqrt {\bf{5}} }\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)
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