Question

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

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

Short answer

The general solution in the parametric vector form is \(x = {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\0\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}4\\0\\0\\1\end{array}} \right]\).

Step-by-step solution

Write the matrix as an augmented matrix

The augmented matrix \(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\) for the given matrix is represented as:

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

Apply row operation

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

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

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

Convert the matrix into the equation

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

\[\begin{array}{c}{x_1} + 3{x_2} - 4{x_4} = 0\\0 = 0\end{array}\]

Determine the basic variable and free variable of the system

The variables corresponding to the pivot columns in the matrix are called basic variables.The other variable is called a free variable.

\({x_1}\) is a basic variable, and \({x_2},{x_3}\), and \({x_4}\) are free variables.

Thus, \({x_1} = - 3{x_2} + 4{x_4}\).

Determine the general solution in the parametric vector form

Sometimes the parametric form of an equation is written as\(x = s{\mathop{\rm u}\nolimits} + t{\mathop{\rm v}\nolimits} \,\,\,\left( {s,t\,{\mathop{\rm in}\nolimits} \,\mathbb{R}} \right)\).

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

\(\begin{array}{c}x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3{x_2} + 4{x_4}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3{x_2}}\\{{x_2}}\\0\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\0\\{{x_3}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{4{x_4}}\\0\\0\\{{x_4}}\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\0\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}4\\0\\0\\1\end{array}} \right]\end{array}\)

Thus, the general solution in the parametric vector form is \(x = {x_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\0\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}4\\0\\0\\1\end{array}} \right]\).