Question

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

8. \({z_1}\left[ {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right] + {z_2}\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right] + {z_3}\left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right] + {z_4}\left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right]\)

Short answer

The vector equation \({z_1}\left[ {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right] + {z_2}\left[ {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right] + {z_3}\left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right] + {z_4}\left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right]\)in a matrix equation is written as \(\left[ {\begin{array}{*{20}{c}}4&{ - 4}&{ - 5}&3\\{ - 2}&5&4&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{z_1}}\\{{z_2}}\\{{z_3}}\\{{z_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right]\).

Step-by-step solution

Write the definition of \(A{\bf{x}}\)

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 the 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.

Obtain the columns of the matrix

Compare the given vector equation form \({z_1}\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right) + {z_2}\left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right) + {z_3}\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right) + {z_4}\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right)\)with the general equation\({z_1}{a_1} + {z_2}{a_2} + \cdots + {z_n}{a_n} = A{\bf{z}}\)form.

So, \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)\),\({{\bf{a}}_4} = \left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)\),and \(A{\bf{z}} = b = \left( {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right)\).

It shows that the equation is a linear combination of fourvectors \({z_1}\), \({z_2}\), \({z_3}\), and \({z_4}\).

Write matrix A and vector z

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

Write matrix A using fourcolumns \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)\),and\({{\bf{a}}_4} = \left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)\)and vector z using four entries \({z_1}\), \({z_2}\), \({z_3}\), and \({z_4}\).

So, \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)}\end{array}} \right)\), and \({\bf{z}} = \left( {\begin{array}{*{20}{c}}{{z_1}}\\{{z_2}}\\{{z_3}}\\{{z_4}}\end{array}} \right)\).

Write the vector equation into a matrix equation

By using matrix \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right)}\end{array}} \right)\), and vector \({\bf{z}} = \left( {\begin{array}{*{20}{c}}{{z_1}}\\{{z_2}}\\{{z_3}}\\{{z_4}}\end{array}} \right)\), the matrix equation can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}4&{ - 4}&{ - 5}&3\\{ - 2}&5&4&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{z_1}}\\{{z_2}}\\{{z_3}}\\{{z_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right)\)

Thus, the vector equation \({z_1}\left( {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right) + {z_2}\left( {\begin{array}{*{20}{c}}{ - 4}\\5\end{array}} \right) + {z_3}\left( {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right) + {z_4}\left( {\begin{array}{*{20}{c}}3\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right)\)can be written as a matrix equation as \(\left( {\begin{array}{*{20}{c}}4&{ - 4}&{ - 5}&3\\{ - 2}&5&4&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{z_1}}\\{{z_2}}\\{{z_3}}\\{{z_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\{13}\end{array}} \right)\).