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The system of equations\(4{x_1} - 2{x_2} + 7{x_3} = - 5\)and\(8{x_1} - 3{x_2} + 10{x_3} = - 3\)can be represented in the matrix equation form \(A{\bf{x}} = {\bf{b}}\) as follows:

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\)

Assume that the column vectors are \({{\bf{v}}_1} = \left( {\begin{aligned}{*{20}{c}}4\\8\end{aligned}} \right)\), \({{\bf{v}}_2} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\), \({{\bf{v}}_3} = \left( {\begin{aligned}{*{20}{c}}7\\{10}\end{aligned}} \right)\), and \({\bf{b}} = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\).

Convert the system of equations\(4{x_1} - 2{x_2} + 7{x_3} = - 5\)and\(8{x_1} - 3{x_2} + 10{x_3} = - 3\)into the augmented matrix as shown below:

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\8&{ - 3}&{10}&{ - 3}\end{aligned}} \right)\)

Use the \(4{x_1}\) term in the first equation to eliminate the \(8{x_1}\) term from the second equation. Add \( - 2\) times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\0&1&{ - 4}&7\end{aligned}} \right)\)

In the above augmented matrix, there are no free variables. It means the system of equations is consistent.

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

The system of equations\(4{x_1} - 2{x_2} + 7{x_3} = - 5\)and\(8{x_1} - 3{x_2} + 10{x_3} = - 3\)or the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\8&{ - 3}&{10}&{ - 3}\end{aligned}} \right)\) in the vector form is shown below:

\({x_1}\left( {\begin{aligned}{*{20}{c}}4\\8\end{aligned}} \right) + {x_2}\left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right) + {x_3}\left( {\begin{aligned}{*{20}{c}}7\\{10}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\)

From the above row-reduced echelon form,

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7&{ - 5}\\0&1&{ - 4}&7\end{aligned}} \right)\).

The system of equations is consistent.

Thus, b is in linear combinations of the columns of matrix A.

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

It can also be written as shown below:

\(T\left( {\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

Let the solution be\({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)\).

Then, the transformation becomes:

\(\begin{aligned}{c}T\left( {\left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)} \right) = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}&7\\8&{ - 3}&{10}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{9/4}\\7\\0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4\left( {9/4} \right) - 14 + 0}\\{8\left( {9/4} \right) - 21 + 0}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\{ - 3}\end{aligned}} \right)\\ = {\bf{b}}.\end{aligned}\)

So,\(T\left( {\bf{x}} \right) = {\bf{b}}\).

Thus, vector b is in the range of T.