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Linear Algebra and its Applications
Found in: Page 1
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Solve each system in Exercises 1–4 by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure.

  1. \(\begin{aligned}{c}{x_1} + 5{x_2} = 7\\ - 2{x_1} - 7{x_2} = - 5\end{aligned}\)

The values are \({x_1} = - 8\) and\({x_2} = 3.\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Convert the given system into an augmented matrix

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

Thus, the augmented matrix for the given system of equations \({x_1} + 5{x_2} = 7\) and \( - 2{x_1} - 7{x_2} = - 5\) is represented as follows:

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

Step 2: Apply an elementary row operation

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

To obtain the solution of the system of equations, you must eliminate one of the variables, \({x_1}\) or\({x_2}\).

Use the \({x_1}\) term in the first equation to eliminate the \( - 2{x_1}\) term from the second equation. Perform an elementary row operation on the matrix \(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{ - 2}&{ - 7}&{ - 5}\end{aligned}} \right)\) as shown below.

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

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

After performing the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&3&9\end{aligned}} \right)\)

Step 3: Apply an elementary row operation

To obtain 1 as the coefficient of\({x_2}\), perform an elementary row operation on the matrix \(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&3&9\end{aligned}} \right)\) as shown below.

Multiply the second row by\(\frac{1}{3};\) i.e., \({R_2} \to \frac{1}{3}{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}1&5&7\\{\frac{0}{3}}&{\frac{3}{3}}&{\frac{9}{3}}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes

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

Step 4: Apply an elementary row operation

Use the \({x_2}\) term in the second equation to eliminate the \(5{x_2}\) term from the first equation. Perform an elementary row operation on the matrix \(\left( {\begin{aligned}{*{20}{c}}1&5&7\\0&1&3\end{aligned}} \right)\) as shown below.

Add \( - 5\) times the second row to the first row; i.e., \({R_1} \to {R_1} - 5{R_2}\).

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

After performing the row operation, the matrix becomes

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

Step 5: Convert the matrix into the equation

Again, you have to convert the augmented matrix into the system of equations to obtain the solution of the system of equations.

Write the obtained matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&{ - 8}\\0&1&3\end{aligned}} \right)\) into the equation notation:

\(\begin{aligned}{c}{x_1} + 0\left( {{x_2}} \right) = - 8\\0\left( {{x_1}} \right) + {x_2} = 3\end{aligned}\)

Step 6: Obtain the solution of the system of equations

Now, obtain the solution of the system of equations by equating \({x_1}\) to \( - 8\) and \({x_2}\) to \(3\):

\(\begin{aligned}{c}{x_1} = - 8\\{x_2} = 3\end{aligned}\)

Thus, the required values for the given system of equations are \({x_1} = - 8\) and \({x_2} = 3\).

Most popular questions for Math Textbooks

In Exercises 15 and 16, list five vectors in Span \(\left\{ {{v_1},{v_2}} \right\}\). For each vector, show the weights on \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) used to generate the vector and list the three entries of the vector. Do not make a sketch.

15. \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}7\\1\\{ - 6}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\0\end{array}} \right]\)

In Exercises 31, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

31. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\4&{ - 1}&3&{ - 6}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\0&7&{ - 1}&{ - 6}\end{array}} \right]\)

In Exercises 23 and 24, key statements from this section are either quoted directly, restated slightly (but still true), or altered in some way that makes them false in some cases. Mark each statement True or False, and justify your answer. (If true, give the approximate location where a similar statement appears, or refer to a definition or theorem. If false, give the location of a statement that has been quoted or used incorrectly, or cite an example that shows the statement is not true in all cases.) Similar true/false questions will appear in many sections of the text.

23.

a. Every elementary row operation is reversible.

b. A \(5 \times 6\)matrix has six rows.

c. The solution set of a linear system involving variables \({x_1},\,{x_2},\,{x_3},........,{x_n}\)is a list of numbers \(\left( {{s_1},\, {s_2},\,{s_3},........,{s_n}} \right)\) that makes each equation in the system a true statement when the values \ ({s_1},\, {s_2},\, {s_3},........,{s_n}\) are substituted for \({x_1},\,{x_2},\,{x_3},........,{x_n}\), respectively.

d. Two fundamental questions about a linear system involve existence and uniqueness.

Suppose T and S satisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that S is a linear transformation. (Hint: Given u, v in \({\mathbb{R}^n}\), let \({\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\). Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \(T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \). Why? Apply S to both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).)

If A is a 2×2 matrix with eigenvalues 3 and 4 and if localid="1668109698541" u→ is a unit eigenvector of A, then the length of vector Alocalid="1668109419151" u→ cannot exceed 4.
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