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Chapter 1: Q9E (page 1)

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

the given system of equations.

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

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

Short Answer

Expert verified

The system of equations in the vector equation form is\({x_1}\left[ {\begin{array}{*{20}{c}}0\\4\\{ - 1}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\6\\3\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\).

Step by step solution

01

Write the system of equations in a single vector

Consider the system of equations as shown below:

\(\begin{array}{c}{x_2} + 5{x_3} = 0\\4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

Write the left-hand part and the right-hand part of the system of equations in the vector form as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_2} + 5{x_3}}\\{4{x_1} + 6{x_2} - {x_3}}\\{ - {x_1} + 3{x_2} - 8{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

02

Separate the left-hand side of the system of equations

The left-hand side of the vector equation contains \(\left[ {\begin{array}{*{20}{c}}{{x_2} + 5{x_3}}\\{4{x_1} + 6{x_2} - {x_3}}\\{ - {x_1} + 3{x_2} - 8{x_3}}\end{array}} \right]\). Write it as sum of vectors in terms of unknowns \({x_1}\), \({x_2}\), and \({x_3}\).

\(\left[ {\begin{array}{*{20}{c}}{0\left( {{x_1}} \right)}\\{4{x_1}}\\{ - {x_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{x_2}}\\{6{x_2}}\\{3{x_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - {x_3}}\\{ - 8{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

03

Take the unknowns common from each vector

It is observed that \({x_1}\) is the common in the vector \(\left[ {\begin{array}{*{20}{c}}{0\left( {{x_1}} \right)}\\{4{x_1}}\\{ - {x_1}}\end{array}} \right]\). Take \({x_1}\) as the common from the first vector as shown below:

\({x_1}\left[ {\begin{array}{*{20}{c}}0\\4\\{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{x_2}}\\{6{x_2}}\\{3{x_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - {x_3}}\\{ - 8{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

04

Take the unknowns’ common from each vector

Similarly, take \({x_2}\) as the common from the vector \(\left[ {\begin{array}{*{20}{c}}{{x_2}}\\{6{x_2}}\\{3{x_2}}\end{array}} \right]\), and \({x_3}\) as the common from the vector \(\left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - {x_3}}\\{ - 8{x_3}}\end{array}} \right]\). Take \({x_2}\) and \({x_3}\) as commons from each vector as shown below:

\({x_1}\left[ {\begin{array}{*{20}{c}}0\\4\\{ - 1}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\6\\3\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

Thus, the system of equations in the vector equation form is:

\({x_1}\left[ {\begin{array}{*{20}{c}}0\\4\\{ - 1}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}1\\6\\3\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

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