In Exercises 3–6, with T defined by \(T\left( {\bf{x}} \right) = A{\bf{x}}\), find a vector x whose image under T is b, and determine whether x is unique.

4. \(A = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2\\0&1&{ - 4}\\3&{ - 5}&{ - 9}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}6\\{ - 7}\\{ - 9}\end{array}} \right]\)

The multiplication of matrix\(A\)of the order\(m \times n\)and vector x gives a new vector defined as\(A{\bf{x}}\)or b.

This concept is defined by the transformation rule \(T\left( {\bf{x}} \right)\). The matrix transformation is denoted as \({\bf{x}}| \to A{\bf{x}}\).

Consider the transformation\(T\left( {\bf{x}} \right) = A{\bf{x}} = b\).

So,\(T\left( {\bf{x}} \right) = A{\bf{x}} = b\)can be represented as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2\\0&1&{ - 4}\\3&{ - 5}&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6\\{ - 7}\\{ - 9}\end{array}} \right]\)

Write the augmented matrix\(\left[ {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right]\)as shown below:

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

Use the \({x_1}\) term from the first equation to eliminate the \(3{x_1}\) term from the third equation. Add \( - 3\) times row one to row three.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&6\\0&1&{ - 4}&{ - 7}\\3&{ - 5}&{ - 9}&{ - 9}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&6\\0&1&{ - 4}&{ - 7}\\0&4&{ - 15}&{ - 27}\end{array}} \right]\)

Use the \({x_2}\) term from the second equation to eliminate the \(4{x_2}\) term from the third equation. Add \( - 4\) times row two to row three.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&6\\0&1&{ - 4}&{ - 7}\\0&4&{ - 15}&{ - 27}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 3}&2&6\\0&1&{ - 4}&{ - 7}\\0&0&1&1\end{array}} \right]\)

Use the \({x_3}\) term from the third equation to eliminate the \( - 4{x_3}\) term from the second equation. Add \(4\) times row three to row two.

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

Use the \({x_2}\) term from the second equation to eliminate the \( - 3{x_2}\) term from the first equation. Add 3 times row two to row one.

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

To obtain the solution, convert the augmented matrix into the system of equations.

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&0&0&{ - 5}\\0&1&0&{ - 3}\\0&0&1&1\end{array}} \right]\),in the equation notation.

\(\begin{array}{c}{x_1} + 0\left( {{x_2}} \right) + 0\left( {{x_3}} \right) = - 5\\0\left( {{x_1}} \right) + {x_2} + 0\left( {{x_3}} \right) = - 3\\0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) + {x_3} = 1\end{array}\)

According to the pivot positions in the obtained matrix, \({x_1}\), \({x_2}\), and \({x_3}\) are the basic variables, and there are no free variables.

Thus, the solutions are as shown below:

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 3}\\1\end{array}} \right]\)

Thus, the solution is unique.