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Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}{ - 9}&{ - 4}&{ - 9}&4\\5&{ - 8}&{ - 7}&6\\7&{11}&{16}&{ - 9}\\9&{ - 7}&{ - 4}&5\end{array}} \right]\).

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ { - 9{\rm{ }} - 4{\rm{ }} - 9{\rm{ }}4;{\rm{ }}5{\rm{ }} - 8{\rm{ }} - 7{\rm{ }}6;{\rm{ }}7{\rm{ }}11{\rm{ }}16{\rm{ }} - 9;{\rm{ }}9{\rm{ }} - 7{\rm{ }} - 4{\rm{ }}5{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{ - 9}&{ - 4}&{ - 9}&4\\5&{ - 8}&{ - 7}&6\\7&{11}&{16}&{ - 9}\\9&{ - 7}&{ - 4}&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{3/4}\\0&1&0&{5/4}\\0&0&1&{ - 7/4}\\0&0&0&0\end{array}} \right]\)

It is observed that there are four columns in the given matrix, which means there should be four entries in vector x.

Thus, the equation \(A{\bf{x}} = 0\) can be written as shown below:

\(\begin{aligned} A{\bf{x}} &= 0\\\left[ {\begin{array}{*{20}{c}}1&0&0&{3/4}\\0&1&0&{5/4}\\0&0&1&{ - 7/4}\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{aligned}\)

Write the above matrix equation in the system of equations as shown below:

\(\begin{aligned}{c}{x_1}\left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{3/4}\\{5/4}\\{ - 7/4}\\0\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( {3/4} \right){x_4}}\\{\left( 0 \right){x_1} + {x_2} + \left( 0 \right){x_3} + \left( {5/4} \right){x_4}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + {x_3} + \left( { - 7/4} \right){x_4}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{aligned}\)

So, the system of equations is shown below:

\(\begin{array}{c}{x_1} + \frac{3}{4}{x_4} = 0\\{x_2} + \frac{5}{4}{x_4} = 0\\{x_3} - \frac{7}{4}{x_4} = 0\end{array}\)

From the above equations, \({x_1}\), \({x_2}\), and \({x_3}\) correspond to the pivot positions. So, \({x_1}\), \({x_2}\), and \({x_3}\) are basic variables, and \({x_4}\) is a free variable.

Let \({x_4} = t\).

Substitute the value \({x_4} = t\) in the equation \({x_1} + \frac{3}{4}{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} + \frac{3}{4}\left( t \right) &= 0\\{x_1} &= - \frac{{3t}}{4}\end{aligned}\)

Substitute the value \({x_4} = t\) in the equation \({x_2} + \frac{5}{4}{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} + \frac{5}{4}\left( t \right) &= 0\\{x_2} &= - \frac{{5t}}{4}\end{aligned}\)

Substitute the value \({x_4} = t\) in the equation \({x_3} - \frac{7}{4}{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_3} - \frac{7}{4}\left( t \right) &= 0\\{x_3} &= \frac{{7t}}{4}\end{aligned}\)

Obtain the vector in a parametric form by using \({x_1} = - \frac{{3t}}{4}\), \({x_2} = - \frac{{5t}}{4}\), \({x_3} = \frac{{7t}}{4}\), and \({x_4} = t\).

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{ - 3t/4}\\{ - 5t/4}\\{ - 7t/4}\\t\end{array}} \right]\\ &= t\left[ {\begin{array}{*{20}{c}}{ - 3/4}\\{ - 5/4}\\{ - 7/4}\\1\end{array}} \right]\end{aligned}\)

Or, it can be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \frac{{{x_4}}}{4}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\\{ - 7}\\4\end{array}} \right]\)

So, the solution in the parametric vector form is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \frac{{{x_4}}}{4}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\\{ - 7}\\4\end{array}} \right]\) or \({\bf{x}} = \frac{{{x_4}}}{4}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\\{ - 7}\\4\end{array}} \right]\).

Thus, the values of x such that \(T\left( {\bf{x}} \right) = 0\) are the multiples of \(\left( { - 3, - 5, - 7,4} \right)\).