In Exercises 7-12, describe all solutions of \(Ax = 0\) in parametric vector form, where \(A\) is row equivalent to the given matrix.

8. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 9}&5\\0&1&2&{ - 6}\end{array}} \right]\)

Short Answer

Expert verified

The general solution in the parametric vector form is \(x = {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}7\\6\\0\\1\end{array}} \right]\).

Step by step solution

01

Write the matrix as an augmented matrix

The augmented matrix \(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\) for the given matrix is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 9}&5&0\\0&1&2&{ - 6}&0\end{array}} \right]\)

02

Apply row operation

Perform an elementary row operation to produce the first augmented matrix.

Perform the sum of \(2\) times row 2 and row 1 at row 1.

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

03

Convert the matrix into the equation

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

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&{ - 7}&0\\0&1&2&{ - 6}&0\end{array}} \right]\)into the equation notation.

\[\begin{array}{c}{x_1} - 5{x_3} - 7{x_4} = 0\\{x_2} + 2{x_3} - 6{x_4} = 0\end{array}\]

04

Determine the basic variable and free variable of the system

The variables corresponding to the pivot columns in the matrix are calledbasic variables.The other variable is called a free variable.

\({x_1}\)and \({x_2}\) are basic variables, and \({x_3}\) and \({x_4}\) are free variables.

Thus, \({x_1} = 5{x_3} + 7{x_4},{x_2} = - 2{x_3} + 6{x_4}\).

05

Determine the general solution in the parametric vector form

Sometimes the parametric form of an equation is written as\(x = s{\mathop{\rm u}\nolimits} + t{\mathop{\rm v}\nolimits} \,\,\,\left( {s,t\,{\mathop{\rm in}\nolimits} \,\mathbb{R}} \right)\).

The general solution of \(Ax = 0\) in the parametric vector form can be represented as:

\(\begin{array}{c}x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5{x_3} + 7{x_4}}\\{ - 2{x_3} + 6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{7{x_4}}\\{6{x_4}}\\0\\{{x_4}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}7\\6\\0\\1\end{array}} \right]\end{array}\)

Thus, the general solution in the parametric vector form is \(x = {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}7\\6\\0\\1\end{array}} \right]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. (Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)).

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.

Construct a \(2 \times 3\) matrix \(A\), not in echelon form, such that the solution of \(Ax = 0\) is a line in \({\mathbb{R}^3}\).

Find the general solutions of the systems whose augmented matrices are given as

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

Suppose Ais an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free