Question: In Exercises 9 and 10, write the system first as a vector equation

and then as a matrix equation.

9.\(3{x_1} + {x_2} - 5{x_3} = 9\)

\({x_2} + 4{x_3} = 0\)

Consider the system of equations as shown below:

\(\begin{array}{c}3{x_1} + {x_2} - 5{x_3} = 9\\{x_2} + 4{x_3} = 0\end{array}\)

Write the left-hand part and right-hand part of the system of equations in the vector form as shown below:

\(\left( {\begin{array}{*{20}{c}}{3{x_1} + {x_2} - 5{x_3}}\\{{x_2} + 4{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\)

The left-hand side of the vector equation contains \(\left( {\begin{array}{*{20}{c}}{3{x_1} + {x_2} - 5{x_3}}\\{{x_2} + 4{x_3}}\end{array}} \right)\). Separate the vector into three different vectors in terms of unknowns \({x_1}\), \({x_2}\), and \({x_3}\).

\(\left( {\begin{array}{*{20}{c}}{3{x_1}}\\{0\left( {{x_1}} \right)}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 5{x_3}}\\{4{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\)

It is observed that \({x_1}\) is common in the vector \(\left( {\begin{array}{*{20}{c}}{3{x_1}}\\{0\left( {{x_1}} \right)}\end{array}} \right)\). Take \({x_1}\) as the common from the first vector as shown below:

\({x_1}\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 5{x_3}}\\{4{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\)

Similarly, take \({x_2}\) as the common from the vector \(\left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_2}}\end{array}} \right)\), and \({x_3}\) as the common from the vector \(\left( {\begin{array}{*{20}{c}}{ - 5{x_3}}\\{4{x_3}}\end{array}} \right)\). Take \({x_2}\) and \({x_3}\) as the common from each vector as shown below:

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

Thus, the system of equations in the vector equation form is:

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

Compare the given vector equation form \({x_1}\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\)with the general equation\({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\)form.

So, \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)\), \(A{\bf{x}} = b = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\).

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

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

By using matrix \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)}\end{array}} \right)\) 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:

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

Thus, the vector equation \({x_1}\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}1\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\)can be written as a matrix equation as \(\left( {\begin{array}{*{20}{c}}3&1&{ - 5}\\0&1&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9\\0\end{array}} \right)\).