Question

In Exercises 37 and 38, the given matrix determines a linear

transformation \(T\). Find all x such that \(T\left( {\bf{x}} \right) = 0\).

37. \(\left[ {\begin{array}{*{20}{c}}4&{ - 2}&5&{ - 5}\\{ - 9}&7&{ - 8}&0\\{ - 6}&4&5&3\\5&{ - 3}&8&{ - 4}\end{array}} \right]\)

Short answer

The values of x such that \(T\left( {\bf{x}} \right) = 0\) are the multiples of \(\left( {7,9,0,2} \right)\).

Step-by-step solution

Convert the matrix into the row reduction echelon form

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}4&{ - 2}&5&{ - 5}\\{ - 9}&7&{ - 8}&0\\{ - 6}&4&5&3\\5&{ - 3}&8&{ - 4}\end{array}} \right]\).

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {4{\rm{ }} - 2{\rm{ 5 }} - {\rm{5}};{\rm{ }} - 9{\rm{ }}7{\rm{ }} - 8{\rm{ 0}};{\rm{ }} - {\rm{6 4 5 3}};{\rm{ 5 }} - 3{\rm{ 8 }} - {\rm{4 }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}4&{ - 2}&5&{ - 5}\\{ - 9}&7&{ - 8}&0\\{ - 6}&4&5&3\\5&{ - 3}&8&{ - 4}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 7/2}\\0&1&0&{ - 9/2}\\0&0&1&0\\0&0&0&0\end{array}} \right]\)

Write the matrix into the system of equations

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:

\(\begin{array}{c}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 7/2}\\0&1&0&{ - 9/2}\\0&0&1&0\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\)

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

\(\begin{aligned}{c}{x_1}\left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 7/2}\\{ - 9/2}\\0\\0\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( { - 7/2} \right){x_4}}\\{\left( 0 \right){x_1} + {x_2} + \left( 0 \right){x_3} + \left( { - 9/2} \right){x_4}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + {x_3} + \left( 0 \right){x_4}}\\{\left( 0 \right){x_1} + \left( 0 \right){x_2} + \left( 0 \right){x_3} + \left( 0 \right){x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{aligned}\)

So, the system of equations is obtained as shown below:

\(\begin{aligned}{c}{x_1} - \frac{7}{2}{x_4} &= 0\\{x_2} - \frac{9}{2}{x_4} &= 0\\{x_3} &= 0\end{aligned}\)

Separate the variables into free and basic variables

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 the basic variables, and \({x_4}\) is a free variable.

Let \({x_4} = t\).

Obtain the value of the basic variables in parametric forms

Substitute the value \({x_4} = t\) in the equation \({x_1} - \frac{7}{2}{x_4} = 0\) to obtain the general solution.

\(\begin{array}{l}{x_1} - \frac{7}{2}\left( t \right) = 0\\{x_1} = \frac{{7t}}{2}\end{array}\)

Substitute the value \({x_4} = t\) in the equation \({x_2} - \frac{9}{2}{x_4} = 0\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} - \frac{9}{2}\left( t \right) &= 0\\{x_2} &= \frac{{9t}}{2}\end{aligned}\)

Write the solution in a parametric form

Obtain the vector in a parametric form by using \({x_1} = \frac{{7t}}{2}\), \({x_2} = \frac{{9t}}{2}\), \({x_3} = 0\), and \({x_4} = t\).

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{7t/2}\\{9t/2}\\0\\t\end{array}} \right]\\ &= t\left[ {\begin{array}{*{20}{c}}{7/2}\\{9/2}\\0\\1\end{array}} \right]\end{aligned}\)

Or, it can be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \frac{{{x_4}}}{2}\left[ {\begin{array}{*{20}{c}}7\\9\\0\\2\end{array}} \right]\)

So, the solution in the parametric vector form is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \frac{{{x_4}}}{2}\left[ {\begin{array}{*{20}{c}}7\\9\\0\\2\end{array}} \right]\) or \({\bf{x}} = \frac{{{x_4}}}{2}\left[ {\begin{array}{*{20}{c}}7\\9\\0\\2\end{array}} \right]\).

Thus, the values of x such that \(T\left( {\bf{x}} \right) = 0\) are the multiples of \(\left( {7,9,0,2} \right)\).