Let \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}7\\5\\9\\7\end{array}} \right]\) 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.

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

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

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

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

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

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

The system of equations is shown below:

\(\begin{aligned}{c}{x_1} - \frac{7}{2}{x_4} &= 4\\{x_2} - \frac{9}{2}{x_4} &= 7\\{x_3} &= 1\end{aligned}\)

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.

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 \({x_4} = t\).

Substitute the value \({x_4} = t\) in the equation \({x_1} - \frac{7}{2}{x_4} = 4\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} - \frac{7}{2}\left( t \right) &= 4\\{x_1} &= 4 + \frac{{7t}}{2}\end{aligned}\)

Substitute the value \({x_4} = t\) in the equation \({x_2} - \frac{9}{2}{x_4} = 7\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} - \frac{9}{2}\left( t \right) &= 7\\{x_2} &= 7 + \frac{{9t}}{2}\end{aligned}\)

Obtain the vector in a parametric form by using \({x_1} = 4 + \frac{{7t}}{2}\), \({x_2} = 7 + \frac{{9t}}{2}\), \({x_3} = 1\), and \({x_4} = t\).

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{7t}}{2} + 4}\\{\frac{{9t}}{2} + 7}\\1\\t\end{array}} \right]\)

Or, it can be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{7{x_4}}}{2} + 4}\\{\frac{{9{x_4}}}{2} + 7}\\1\\{{x_4}}\end{array}} \right]\)

When \({x_4} = 0\), the solution is obtained as shown below:

\(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\7\\1\\0\end{array}} \right]\)

Thus, the solution is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\7\\1\\0\end{array}} \right]\) or \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}4\\7\\1\\0\end{array}} \right]\).

Therefore, bis in the range of the transformation. One of the choices for x is \(\left( {4,7,1,0} \right)\).