Question

Consider the problem of determining whether the following system of equations is consistent for all \({b_1},{b_2},{b_3}\):

\(\begin{aligned}{c}{\bf{2}}{x_1} - {\bf{4}}{x_2} - {\bf{2}}{x_3} = {b_1}\\ - {\bf{5}}{x_1} + {x_2} + {x_3} = {b_2}\\{\bf{7}}{x_1} - {\bf{5}}{x_2} - {\bf{3}}{x_3} = {b_3}\end{aligned}\)

1. Define appropriate vectors, and restate the problem in terms of Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\). Then solve that problem.

2. Define an appropriate matrix, and restate the problem using the phrase “columns of A.”

3. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

Short answer

a. Vectorb is in the span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

b. Vector b is in the linear combination of the columns of matrix A.

c. Vector b is in the range of T.

Step-by-step solution

(a) Step 1: Write the system in the vector form

The system of equations\(2{x_1} - 4{x_2} - 2{x_3} = {b_1}\),\( - 5{x_1} + {x_2} + {x_3} = {b_2}\),and\(7{x_1} - 5{x_2} - 3{x_3} = {b_3}\) can be represented in the matrix equation form \(A{\bf{x}} = {\bf{b}}\) as follows:

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\{ - 5}&1&1\\7&{ - 5}&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{aligned}} \right)\)

Assume that the column vectors are \({{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}2\\{ - 5}\\7\end{aligned}} \right)\), \({{\bf{v}}_2} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\1\\{ - 5}\end{aligned}} \right)\), \({{\bf{v}}_3} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\\{ - 3}\end{aligned}} \right)\), and \({\bf{b}} = \left( {\begin{aligned}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{aligned}} \right)\).

Convert the matrix into the row-reduced echelon form

Row reduce matrix\(A = \left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\{ - 5}&1&1\\7&{ - 5}&{ - 3}\end{aligned}} \right)\).

Use the \(2{x_1}\) term in the first equation to eliminate the \( - 5{x_1}\) term from the second equation. Add \(\frac{5}{2}\) times row one to row two.Then, use the \(2{x_1}\) term in the first equation to eliminate the \(7{x_1}\) term from the third equation. Add \( - \frac{7}{2}\) times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\0&{ - 9}&{ - 4}\\7&{ - 5}&{ - 3}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\0&{ - 9}&{ - 4}\\0&9&4\end{aligned}} \right)\)

Add rows two and three.

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\0&{ - 9}&{ - 4}\\0&9&4\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\0&{ - 9}&{ - 4}\\0&0&0\end{aligned}} \right)\)

In the above augmented matrix, there is one free variable in the third equation, or there is no pivot position in each row. So, the system of equations is inconsistent.

Thus, the matrix of the columns does not span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

(b) Step 3: Write in terms of the columns of A

The system of equations\(2{x_1} - 4{x_2} - 2{x_3} = {b_1}\),\( - 5{x_1} + {x_2} + {x_3} = {b_2}\)and\(7{x_1} - 5{x_2} - 3{x_3} = {b_3}\), or matrix \(A = \left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\{ - 5}&1&1\\7&{ - 5}&{ - 3}\end{aligned}} \right)\) is shown below in the vector form:

\({x_1}\left( {\begin{aligned}{*{20}{c}}2\\{ - 5}\\7\end{aligned}} \right) + {x_2}\left( {\begin{aligned}{*{20}{c}}{ - 4}\\1\\{ - 5}\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\\{ - 3}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{aligned}} \right)\)

The row-reduced echelon form is shown below:

\(\left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\0&{ - 9}&{ - 4}\\0&0&0\end{aligned}} \right)\)

Here, the system of equations is inconsistent.

Thus, b is not in the linear combination of the columns of matrix A.

(c) Step 4: Check whether b is in the range of T

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

It can also be written as shown below:

\(T\left( {\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}2&{ - 4}&{ - 2}\\{ - 5}&1&1\\7&{ - 5}&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

Let the solution be\({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).

Then, the transformation becomes as shown below.

\(\begin{aligned}{c}T\left( {\left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\end{aligned}\)

So,\(T\left( {\bf{x}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).

Thus, the mapping transforms \({\mathbb{R}^3}\) onto \({\mathbb{R}^3}\).