Question

Let \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 7}\\{13}\\{ - 5}\end{array}} \right]\) and let A be the matrix in Exercise 38. Is b in the range of the transformation \({\bf{x}}| \to A{\bf{x}}\)? If so, find an x whose image under the transformation is b.

Short answer

Yes, bis in the range of the transformation. One of the choices for x is \(\left( { - 2, - 4,5,1} \right)\), whose image under the transformation is b.

Step-by-step solution

Convert the augmented matrix into the row reduction echelon form

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

Obtain the augmented matrix\(\left[ {\begin{array}{*{20}{c}}A&b\end{array}} \right]\).

\(\left[ {\begin{array}{*{20}{c}}{ - 9}&{ - 4}&{ - 9}&4&{ - 7}\\5&{ - 8}&{ - 7}&6&{ - 7}\\7&{11}&{16}&{ - 9}&{13}\\9&{ - 7}&{ - 4}&5&{ - 5}\end{array}} \right]\)

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ { - 9{\rm{ }} - 4{\rm{ }} - 9{\rm{ }}4{\rm{ }} - 7;{\rm{ }}5{\rm{ }} - 8{\rm{ }} - 7{\rm{ }}6{\rm{ }} - 7;{\rm{ }}7{\rm{ }}11{\rm{ }}16{\rm{ }} - 9{\rm{ 13}};{\rm{ }}9{\rm{ }} - 7{\rm{ }} - 4{\rm{ }}5{\rm{ }} - {\rm{5}}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{ - 9}&{ - 4}&{ - 9}&4&{ - 7}\\5&{ - 8}&{ - 7}&6&{ - 7}\\7&{11}&{16}&{ - 9}&{13}\\9&{ - 7}&{ - 4}&5&{ - 5}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{3/4}&{ - 5/4}\\0&1&0&{5/4}&{ - 11/4}\\0&0&1&{ - 7/4}&{13/4}\\0&0&0&0&0\end{array}} \right]\)

Write the augmented matrix into the system of equations

The system of equations is shown below:

\(\begin{array}{c}{x_1} + \frac{3}{4}{x_4} = - \frac{5}{4}\\{x_2} + \frac{5}{4}{x_4} = - \frac{{11}}{4}\\{x_3} - \frac{7}{4}{x_4} = \frac{{13}}{4}\end{array}\)

From the above system of equations, it is observed that the augmented matrix is a consistent system.

Thus, bis in the range of the transformation.

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 basic variables in parametric forms

Substitute the value \({x_4} = t\) in the equation \({x_1} + \frac{3}{4}{x_4} = - \frac{5}{4}\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} + \frac{3}{4}\left( t \right) &= - \frac{5}{4}\\{x_1} &= - \frac{5}{4} - \frac{{3t}}{4}\end{aligned}\)

Substitute the value \({x_4} = t\) in the equation \({x_2} + \frac{5}{4}{x_4} = - \frac{{11}}{4}\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} + \frac{5}{4}\left( t \right) &= - \frac{{11}}{4}\\{x_2} &= - \frac{{11}}{4} - \frac{{5t}}{4}\end{aligned}\)

Substitute the value \({x_4} = t\) in the equation \({x_3} - \frac{7}{4}{x_4} = \frac{{13}}{4}\) to obtain the general solution.

\(\begin{aligned}{c}{x_3} - \frac{7}{4}\left( t \right) &= \frac{{13}}{4}\\{x_3} &= \frac{{13}}{4} + \frac{{7t}}{4}\end{aligned}\)

Write the solution in a parametric form

Obtain the vector in a parametric form by using \({x_1} = - \frac{5}{4} - \frac{{3t}}{4}\), \({x_2} = - \frac{{11}}{4} - \frac{{5t}}{4}\), \({x_3} = \frac{{13}}{4} + \frac{{7t}}{4}\), and \({x_4} = t\).

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - \frac{5}{4} - \frac{{3t}}{4}}\\{ - \frac{{11}}{4} - \frac{{5t}}{4}}\\{\frac{{13}}{4} + \frac{{7t}}{4}}\\t\end{array}} \right]\)

Or, it can be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - \frac{5}{4} - \frac{{3{x_4}}}{4}}\\{ - \frac{{11}}{4} - \frac{{5{x_4}}}{4}}\\{\frac{{13}}{4} + \frac{{7{x_4}}}{4}}\\{{x_4}}\end{array}} \right]\)

When \({x_4} = 1\), the solution is obtained as shown below:

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{ - \frac{5}{4} - \frac{3}{4}}\\{ - \frac{{11}}{4} - \frac{5}{4}}\\{\frac{{13}}{4} + \frac{7}{4}}\\1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 4}\\5\\1\end{array}} \right]\end{aligned}\)

Thus, the solution is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 4}\\5\\1\end{array}} \right]\) or \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 4}\\5\\1\end{array}} \right]\).

Therefore, bis in the range of the transformation. One of the choices for x is \(\left( { - 2, - 4,5,1} \right)\).