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Chapter 1: Q10E (page 1)

In Exercises 7鈥�10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

10. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

The solution set of a linear system contains one solution; i.e., \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{aligned}} \right)\).

Step by step solution

01

Rewrite the given augmented matrix of a linear system

The augmented matrix of a linear system is given as

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

02

Perform elementary row operations

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

To eliminate the \( - 4{x_4}\) term in the second equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Add 4 times the fourth row to the second row; i.e., \({R_2} \to {R_2} + 4{R_4}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\{0 + 4\left( 0 \right)}&{1 + 4\left( 0 \right)}&{0 + 4\left( 0 \right)}&{ - 4 + 4\left( 1 \right)}&{7 + 4\left( { - 3} \right)}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

03

Eliminate the fourth element of the first row

To eliminate the \(3{x_4}\) term in the first equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Subtract 3 times the fourth row from the first row; i.e., \({R_1} \to {R_1} - 3{R_4}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 - 3\left( 0 \right)}&{ - 2 - 3\left( 0 \right)}&{0 - 3\left( 0 \right)}&{3 - 3\left( 1 \right)}&{ - 2 - 3\left( { - 3} \right)}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&0&7\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

04

Eliminate the second element of the first row

To eliminate the \( - 2{x_2}\) term in the first equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&0&7\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Add 2 times the second row to the first row; i.e., \({R_1} \to {R_1} + 2{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 + 2\left( 0 \right)}&{ - 2 + 2\left( 1 \right)}&{0 + 2\left( 0 \right)}&{0 + 2\left( 0 \right)}&{7 + 2\left( { - 5} \right)}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&0&0&{ - 3}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

Convert the augmented matrix into equations to get \({x_1} = - 3,\,\,{x_2} = - 5,\,\,{x_3} = 6,\,\,\) and \({x_4} = - 3\).

Hence, the required solution is \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{aligned}} \right)\).

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Most popular questions from this chapter

a. Find the general flow pattern in the network shown in the figure.

b. Assuming that the flow must be in the directions indicated, find the minimum flows in the branches denoted by \({x_2}\), \({x_3}\), \({x_4}\) and \({x_5}\).

Find an equation involving \(g,\,h,\)and \(k\) that makes this augmented matrix correspond to a consistent system:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 4}&7&g\\0&3&{ - 5}&h\\{ - 2}&5&{ - 9}&k\end{array}} \right]\)

Suppose \(Ax = b\) has a solution. Explain why the solution is unique precisely when \(Ax = 0\) has only the trivial solution.

In Exercise 21, mark each statement True or False. Justify each answer.\(\) \(\)

21. a. A linear transformation is a special type of function.

b. If \({\bf{A}}\) is a \({\bf{3}} \times {\bf{5}}\)matrix and \({\bf{T}}\) is a transformation defined by\({\bf{T}}\left( {\bf{x}} \right) = {\bf{Ax}}\), then the domain of \({\bf{T}}\) is\({\mathbb{R}^{\bf{3}}}\).

c. If \({\bf{A}}\) is an \({\bf{m}} \times {\bf{n}}\) matrix, then the range of the transformation \({\bf{x}} \mapsto {\bf{Ax}}\) is\({\mathbb{R}^{\bf{m}}}\).

\(\) d. Every linear transformation is a matrix transformation.

e. A transformation \({\bf{T}}\) is linear if and only if \({\bf{T}}\left( {{{\bf{c}}_{\bf{1}}}{{\bf{v}}_{\bf{1}}} + {{\bf{c}}_{\bf{2}}}{{\bf{v}}_{\bf{2}}}} \right) = {{\bf{c}}_{\bf{1}}}{\bf{T}}\left( {{{\bf{v}}_{\bf{1}}}} \right) + {{\bf{c}}_{\bf{2}}}{\bf{T}}\left( {{{\bf{v}}_{\bf{2}}}} \right)\) for all \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) in the domain of \({\bf{T}}\) and for all scalars \({{\bf{c}}_{\bf{1}}}\) and\({{\bf{c}}_{\bf{2}}}\).\[\]

For Exercises 9 and 10, find all x in \({\mathbb{R}^{\bf{4}}}\) that are mapped into the zero vector by the transformation \({\bf{x}}| \to A{\bf{x}}\) for the given matrix A.

10. \(A = \left[ {\begin{array}{*{20}{c}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\)
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