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Chapter 1: Q14E (page 1)

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

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

Short Answer

Expert verified

The general solution of the system is

\({x_1} = - 5 + 2{x_2} + 5{x_3} + 6{x_4}\)

\({x_2} = 2 + 6{x_3} + 3{x_4}\)

\({x_3}\)is free.

\({x_4}\)is free.

\({x_5} = 0\).

Step by step solution

01

Convert the given matrix into the equation

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

Write the given matrix into the equation notation.

\(\begin{aligned}{c}{x_1} - 2{x_2} - 5{x_3} - 6{x_4} = - 5\\{x_2} - 6{x_3} - 3{x_4} = 2\\{x_5} = 0\end{aligned}\)

02

Determine the basic variable and free variable

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 \({x_1},{x_2},{x_5}\). The free variables are \({x_3},{x_4}\).

03

Write the general solution of the system

Thus, the general solution of the system is

\(\begin{aligned}{c}{x_1} = - 5 + 2{x_2} + 5{x_3} + 6{x_4}\\{x_2} = 2 + 6{x_3} + 3{x_4}\end{aligned}\)

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

\({x_4}\)is a free variable.

\({x_5} = 0\).

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Most popular questions from this chapter

Follow the method of Example 3 to describe the solutions of the following system in parametric vector form. Also, give a geometric description of the solution set and compare it to that in Exercise 5.

\(\begin{aligned}{c}{x_1} + 3{x_2} + {x_3} = 1\\ - 4{x_1} - 9{x_2} + 2{x_3} = - 1\\ - 3{x_2} - 6{x_3} = - 3\end{aligned}\)

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