In Exercises 5–8, use the definition of Ax to write the matrixequation as a vector equation, or vice versa.

7. \({x_1}\left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right]\)

It is known that the column of matrix \(A\) is represented as \(\left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\), and vector x is represented as \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right)\).

According to the definition, the weights in a linear combination of matrix A columns are represented by the entries in vector x.

\({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\)

The left-hand side \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\) of the above equation is a linear combination of vectors \({x_1},{x_2},...,{x_n}\).

Thus, the vector equation in the matrix form is written as:

\(A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right) = b\)

The number of columns in matrix\(A\)should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

Compare the given vector equation form \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)with the general equation \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\) form.

So, \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\), and \(A{\bf{x}} = b = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

It shows that the equation is a linear combination of three vectors \({x_1}\), \({x_2}\), and \({x_3}\).

According to the definition,the number of columns in matrix\(A\)should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

Write matrix A using three columns \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), and\({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\) and vector x using three entries \({x_1}\), \({x_2}\), and \({x_3}\).

So, \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), and \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\).

By using matrix \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), and vector \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\), the matrix equation can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)

Thus, the vector equation \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)can be written as a matrix equation as \(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).