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Chapter 1: Q13E (page 1)

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

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

Short Answer

Expert verified

The general solution of the system is

\({x_1} = - 2 + 3{x_2} + {x_4}\)

\({x_2} = 1 - 4{x_5}\)

\({x_3}\)is free.

\({x_4} = 4 - 9{x_5}\)

\({x_5}\) is free.

Step by step solution

01

Convert the matrix into the equation

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

The given matrix is in the row echelon form.

Write the given matrix into the equation notation.

\(\begin{array}{l}{x_1} - 3{x_2} - {x_4} = - 2\\{x_2} - 4{x_5} = 1\\{x_4} + 9{x_5} = 4\end{array}\)

02

Determine the basic variables and free variables

The variables corresponding to the pivot columns in the matrix are called basic variables.

The other variables are called free variables.

The basic variables of the given matrix are \({x_1},{x_2},{x_4}\). The free variables are \({x_3},{x_5}\).

03

Write the general solution of the system

Thus, the general solution of the system is

\(\begin{array}{l}{x_1} = - 2 + 3{x_2} + {x_4}\\{x_2} = 1 + 4{x_5}\end{array}\)

\({x_3}\)is a free variable.

\({x_4} = 4 - 9{x_5}\)

\({x_5}\) is a free variable.

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Most popular questions from this chapter

Suppose \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) is a linearly independent set in \({\mathbb{R}^n}\). Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2}} \right\}\) is also linearly independent.

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 Exercise 23 and 24, make each statement True or False. Justify each answer.

23.

a. Another notation for the vector \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&3\end{array}} \right]\).

b. The points in the plane corresponding to \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 5}\\2\end{array}} \right]\) lie on a line through the origin.

c. An example of a linear combination of vectors \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) is the vector \(\frac{1}{2}{{\mathop{\rm v}\nolimits} _1}\).

d. The solution set of the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}&b\end{array}} \right]\) is the same as the solution set of the equation\({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + {x_3}{a_3} = b\).

e. The set Span \(\left\{ {u,v} \right\}\) is always visualized as a plane through the origin.

In Exercises 9, write a vector equation that is equivalent to

the given system of equations.

9. \({x_2} + 5{x_3} = 0\)

\(\begin{array}{c}4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

Determine the values(s) of \(h\) such that matrix is the augmented matrix of a consistent linear system.

17. \(\left[ {\begin{array}{*{20}{c}}2&3&h\\4&6&7\end{array}} \right]\)
See all solutions

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