In Exercises 5 and 6, follow the method of Examples 1 and 2 to write the solution set of the given homogeneous system in parametric vector form.

6. \(\begin{array}{c}{x_1} + 3{x_2} - 5{x_3} = 0\\{x_1} + 4{x_2} - 8{x_3} = 0\\ - 3{x_1} - 7{x_2} + 9{x_3} = 0\end{array}\)

Anaugmented matrix for a system of equations is a matrix of numbers in which eachrow represents theconstants from one equation, and eachcolumn represents all thecoefficients for a single variable.

The augmented matrix \(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\) for the given system of equations \({x_1} + 3{x_2} - 5{x_3} = 0,{x_1} + 4{x_2} - 8{x_3} = 0\) and \( - 3{x_1} - 7{x_2} + 9{x_3} = 0\) is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 5}&0\\1&4&{ - 8}&0\\{ - 3}&{ - 7}&9&0\end{array}} \right]\)

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

Perform the sum of \( - 1\) times row 1 and row 2 at row 2 and the sum of \(3\) times row 1 and row 3 at row 3.

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

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

Perform the sum of \( - 2\) times row 2 and row 3 at row 3.

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

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

Perform the sum of \( - 3\) times row 2 and row 1 at row 1.

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

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 \(\left[ {\begin{array}{*{20}{c}}1&0&4&0\\0&1&{ - 3}&0\\0&0&0&0\end{array}} \right]\)into the equation notation.

\[\begin{array}{c}{x_1} + 4{x_3} = 0\\{x_2} - 3{x_3} = 0\end{array}\]

Thus, \({x_1} = - 4{x_3},{x_2} = 3{x_3}\) and \({x_3}\) is a free variable.

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

\(\begin{array}{c}x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4{x_3}}\\{3{x_3}}\\{{x_3}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\\1\end{array}} \right]\end{array}\)

Thus, the general solution in the parametric vector form is \(x = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\\1\end{array}} \right]\).