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In Exercises 7–10, 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.

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

Short Answer

Expert verified

The linear system has no solution.

Step by step solution

01

Rewrite the augmented matrix

The augmented matrix of a linear system is given as

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

02

Perform elementary row operations

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

Ordinarily, the next step would be to interchange the third row and the fourth row to have 1 in the third row and the third column.

However, in this case, all three elements of the third row of the augmented matrix are zero.

03

Convert the third row into the equation form

The third row can be written in the equation form as

\(\begin{aligned}{c}0{x_1} + 0{x_2} + 0{x_3} = 1\\ \Rightarrow 0 = 1.\end{aligned}\)

This is a contradiction and is not a possible condition.

04

Conclusion

All elements of the third row of the matrix are zero, and it is well known that zero never equals one.

Thus, the solution set is empty, or the given system of linear equations has no solution.

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Most popular questions from this chapter

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

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

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.

30.\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\)

Construct three different augmented matrices for linear systems whose solution set is \({x_1} = - 2,{x_2} = 1,{x_3} = 0\).

Let \({\bf{u}}\) and \({\bf{v}}\) be vectors in\({\mathbb{R}^{\bf{n}}}\). It can be shown that the set \({\bf{P}}\) of all points in the parallelogram determined by \({\bf{u}}\) and \({\bf{v}}\) has the form \({\bf{av}} + {\bf{bv}}\), for \({\bf{0}} \le {\bf{a}} \le {\bf{1}}\), \({\bf{0}} \le {\bf{b}} \le {\bf{1}}\). Let \({\bf{T}}:{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{n}}}\) be a linear transformation. Explain why the image of a point in \({\bf{P}}\) under the transformation \({\bf{T}}\) lies in the parallelogram determined by \({\bf{T}}\left( {\bf{u}} \right)\) and \({\bf{T}}\left( {\bf{v}} \right)\).

In Exercises 5, write a system of equations that is equivalent to the given vector equation.

5. \({x_1}\left[ {\begin{array}{*{20}{c}}6\\{ - 1}\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\\{ - 5}\end{array}} \right]\)

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