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Chapter 1: Q1.1-15E (page 1)

Determine if the systems in Exercises 15 and 16 are consistent.

Do not completely solve the systems.

15.\[\begin{array}{c}{x_1}\,\,\,\,\,\,\,\,\,\,\, + 3{x_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\\{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x_4} = 3\\ - 2{x_2} + \,3{x_3}\,\,\, + 2{x_4} = 1\\3{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 7{x_4} = - 5\end{array}\]

Short Answer

Expert verified

The given system is consistent.

Step by step solution

01

Write the augmented matrix of the system

To express a system in theaugmented matrixform, extract the coefficients of the variables and the constants and place these entries in the column of the matrix.

The given system of equations is as follows:

\[\begin{array}{c}{x_1}\,\,\,\,\,\,\,\,\,\,\, + 3{x_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\\{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x_4} = 3\\ - 2{x_2} + \,3{x_3}\,\,\, + 2{x_4} = 1\\3{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 7{x_4} = - 5\end{array}\]

So, the augmented matrix for the given system can be written as follows:

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\3&0&0&7&{ - 5}\end{array}} \right]\]

02

Reduce the augmented matrix to a triangular matrix

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

To eliminate the \[3{x_1}\] term from the fourth equation, perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\3&0&0&7&{ - 5}\end{array}} \right]\] as shown below.

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

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\{3 - 3\left( 1 \right)}&{0 - 3\left( 0 \right)}&{0 - 3\left( 3 \right)}&{7 - 3\left( 0 \right)}&{ - 5 - 3\left( 2 \right)}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

03

Apply the row operation

Use the \[{x_2}\] term in the second equation to eliminate the \[ - 2{x_2}\] term from the third equation. Perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\] as shown below.

Add 2 times the second row to the third row; i.e., \({R_3} \to {R_3} + 2{R_2}\).

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\{0 + 2\left( 0 \right)}&{ - 2 + 2\left( 1 \right)}&{3 + 2\left( 0 \right)}&{2 + 2\left( { - 3} \right)}&{1 + 2\left( 3 \right)}\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

04

Apply the row operation

Use the \[3{x_3}\] term in the third equation to eliminate the \[ - 9{x_3}\] term from the fourth equation. Perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\] as shown below.

Add 3 times the third row to the fourth row; i.e., \({R_4} \to {R_4} + 3{R_3}\).

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\{0 + 3\left( 0 \right)}&{0 + 3\left( 0 \right)}&{ - 9 + 3\left( 3 \right)}&{7 + 3\left( { - 4} \right)}&{ - 11 + 3\left( 7 \right)}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&0&{ - 5}&{10}\end{array}} \right]\]

05

Convert the augmented matrix back to the system of equations

From the obtained augmented matrix, the system of equations can be written as follows:

\[\begin{array}{c}{x_1}\,\,\,\,\,\,\,\,\,\,\, + 3{x_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\\{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x_4} = 3\\\,3{x_3}\,\,\, - 4{x_4} = 7\\ - 5{x_4} = 10\end{array}\]

A unique value of \[{x_4}\] can be obtained from the fourth equation.

If \[{x_4}\] is substituted by its unique value in the second and third equations, the unique values of \[{x_2}\] and \[{x_3}\] can be calculated. Thus, substituting \[{x_3}\] by its value in the first equation, you will get a unique value of\[{x_1}\].

Since all the values can be uniquely determined, a unique solution exists for the given system.

Hence, the given system is consistent.

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Most popular questions from this chapter

Find all the polynomials of degreeโ‰ค2[a polynomial of the formf(t)=a+bt+ct2] whose graph goes through the points (1,3)and(2,6),such that f'(1)=1[wheref'(t)denotes the derivative].

Consider the problem of determining whether the following system of equations is consistent:

\(\begin{aligned}{c}{\bf{4}}{x_1} - {\bf{2}}{x_2} + {\bf{7}}{x_3} = - {\bf{5}}\\{\bf{8}}{x_1} - {\bf{3}}{x_2} + {\bf{10}}{x_3} = - {\bf{3}}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of linear combinations. Then solve that problem.
  1. Define an appropriate matrix, and restate the problem using the phrase โ€œcolumns of A.โ€
  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

Consider a dynamical systemwith two components. The accompanying sketch shows the initial state vectorxโ†’0and two eigen vectorsฯ…1โ†’โ€Šโ€Šandโ€Šโ€Šฯ…2โ†’of A (with eigen values ฮป1โ†’andฮป2โ†’respectively). For the given values ofฮป1โ†’andฮป2โ†’, draw a rough trajectory. Consider the future and the past of the system.

ฮป1โ†’=0.9,ฮป2โ†’=0.9

Suppose an experiment leads to the following system of equations:

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{249}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.843\end{aligned}\) (3)

  1. Solve system (3), and then solve system (4), below, in which the data on the right have been rounded to two decimal places. In each case, find the exactsolution.

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{25}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.8{\bf{4}}\end{aligned}\) (4)

  1. The entries in (4) differ from those in (3) by less than .05%. Find the percentage error when using the solution of (4) as an approximation for the solution of (3).
  1. Use your matrix program to produce the condition number of the coefficient matrix in (3).

Let \({{\bf{a}}_1}\) \({{\bf{a}}_2}\), and b be the vectors in \({\mathbb{R}^{\bf{2}}}\) shown in the figure, and let \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}\end{aligned}} \right)\). Does the equation \(A{\bf{x}} = {\bf{b}}\) have a solution? If so, is the solution unique? Explain.
See all solutions

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