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.

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

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:

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

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

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

Use the \({x_1}\) term from the first equation to eliminate the \( - 3{x_1}\) term from the second equation. Add \(3\) times row one to row two. Then, multiply row two by \( - \frac{1}{8}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&{ - 8}&{ - 16}&{ - 8}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&1&2&1\end{array}} \right]\)

Use the \({x_2}\) term from the second equation to eliminate the \( - 5{x_2}\) term from the first equation. Add 5 times row two to row one.

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

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

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&0&3&3\\0&1&2&1\end{array}} \right]\),in the equation notation.

\(\begin{array}{c}{x_1} + 0\left( {{x_2}} \right) + 3{x_3} = 3\\0\left( {{x_1}} \right) + {x_2} + 2{x_3} = 1\end{array}\)

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

Let, \({x_3} = t\).

Substitute the value \({x_3} = t\) in the equation \({x_1} + 3{x_3} = 3\) to obtain the general solution.

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

Substitute the value \({x_3} = t\) in the equation \({x_2} + 2{x_3} = 1\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} + 2\left( t \right) &= 1\\{x_2} &= 1 - 2t\end{aligned}\)

Obtain the vector in the parametric form by using \({x_1} = 3 - 3t\), \({x_2} = 1 - 2t\) and \({x_3} = t\).

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{3 - 3t}\\{1 - 2t}\\t\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\1\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3t}\\{ - 2t}\\t\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\1\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\end{array}} \right]\end{aligned}\)

Or it can be written as \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\1\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\end{array}} \right]\).

The solution in the parametric vector form is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\end{array}} \right]\).

This solution is not unique.