Question

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.

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

Short answer

Vector \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}7\\3\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\1\end{array}} \right]\), and the solution is not unique.

Step-by-step solution

Write the concept for computing images under the transformation of vectors

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 rule of transformation \(T\left( {\bf{x}} \right)\). The matrix transformation is denoted as \({\bf{x}}| \to A{\bf{x}}\).

Obtain the augmented matrix

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:

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

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

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

Convert the augmented matrix into the row-reduced echelon form

Use the \(3{x_1}\) term from the second equation to eliminate the \( - 3{x_1}\) term from the fourth equation. Add \(3\) times row two to row four.

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

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.

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

Convert the augmented matrix into the row-reduced echelon form

Multiply row two by \(\frac{1}{2}\).

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

Use the \({x_2}\) term from the second equation to eliminate the \({x_2}\) term from the third equation. Add \( - 1\) times row two to row three.

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

Use the \({x_2}\) term from the second equation to eliminate the \({x_2}\) term from the fourth equation. Add \( - 1\) times row two to row four.

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

Use the \({x_2}\) term from the second equation to eliminate the \( - 2{x_2}\) term from the first equation. Add 2 times row two to row one.

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

Convert the matrix into an equation

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

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&0&3&7\\0&1&1&3\\0&0&0&0\\0&0&0&0\end{array}} \right]\),in the equation notation.

\(\begin{array}{c}{x_1} + 0\left( {{x_2}} \right) + 3{x_3} = 7\\0\left( {{x_1}} \right) + {x_2} + {x_3} = 3\end{array}\)

Separate the variables into free and basic types

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 a free variable.

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

Obtain the values of basic variables in the parametric form

Substitute the value \({x_3} = t\) in the equation \({x_1} + 3{x_3} = 7\) to obtain the general solution.

\(\begin{aligned}{c}{x_1} + 3\left( t \right) &= 7\\{x_1} &= 7 - 3t\end{aligned}\)

Substitute the value \({x_3} = t\) in the equation \({x_2} + {x_3} = 3\) to obtain the general solution.

\(\begin{aligned}{c}{x_2} + \left( t \right) &= 3\\{x_2} &= 3 - t\end{aligned}\)

Write the solution in the parametric form

Obtain the vector in the parametric form by using \({x_1} = 7 - 3t\), \({x_2} = 3 - t\), and \({x_3} = t\).

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{7 - 3t}\\{3 - t}\\t\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}7\\3\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3t}\\{ - t}\\t\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}7\\3\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\1\end{array}} \right]\end{aligned}\)

Or it can be written as \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\3\\0\end{array}} \right] + t\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\1\end{array}} \right]\).

So, the solution in the parametric vector form is \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\3\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\1\end{array}} \right]\).

This solution is not unique.