Question

Rewrite the (numerical) matrix equation below in symbolic form as a vector equation, using symbols \({{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3},...\) for the vectors and \({c_1},{c_2},...\) for scalars. Define what each symbol represents, using the data given in the matrix equation.

\(\left( {\begin{array}{*{20}{c}}{ - 3}&5&{ - 4}&9&7\\5&8&1&{ - 2}&{ - 4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 3}\\2\\4\\{ - 1}\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\)

Short answer

The vectors are \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 3}\\5\end{array}} \right)\), \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}5\\8\end{array}} \right)\), \({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\), \({{\bf{v}}_4} = \left( {\begin{array}{*{20}{c}}9\\{ - 2}\end{array}} \right)\), \({{\bf{v}}_5} = \left( {\begin{array}{*{20}{c}}7\\{ - 4}\end{array}} \right)\), \({{\bf{v}}_6} = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\), and the scalars are \({c_1} = - 3\), \({c_2} = 2\), \({c_3} = 4\), \({c_4} = - 1\), and \({c_5} = 2\).

Step-by-step solution

Writing the definition of \(V{\bf{c}}\)

Consider the column of matrix \(V\) is represented as \(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{ \cdot \cdot \cdot }&{{{\bf{v}}_n}}\end{array}} \right)\), and vector c is represented as \(\left( {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

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

The matrix equation as a vector equation can be written as shown below:

\(\begin{array}{c}V{\bf{c}} = \left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{ \cdot \cdot \cdot }&{{{\bf{v}}_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_n}}\end{array}} \right)\\V{\bf{c}} = {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + \cdots + {c_n}{{\bf{v}}_n}\end{array}\)

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

Writing matrix V and vector c

Consider the equation \(\left( {\begin{array}{*{20}{c}}{ - 3}&5&{ - 4}&9&7\\5&8&1&{ - 2}&{ - 4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 3}\\2\\4\\{ - 1}\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\).

Here, \(V = \left( {\begin{array}{*{20}{c}}{ - 3}&5&{ - 4}&9&7\\5&8&1&{ - 2}&{ - 4}\end{array}} \right)\), \({\bf{c}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\2\\4\\{ - 1}\\2\end{array}} \right)\), and \(V{\bf{c}} = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\).

Writing the columns of matrix V 

Compare matrix \(V = \left( {\begin{array}{*{20}{c}}{ - 3}&5&{ - 4}&9&7\\5&8&1&{ - 2}&{ - 4}\end{array}} \right)\) with \(V = \left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{ \cdot \cdot \cdot }&{{{\bf{v}}_n}}\end{array}} \right)\).

So, \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 3}\\5\end{array}} \right)\), \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}5\\8\end{array}} \right)\), \({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\), \({{\bf{v}}_4} = \left( {\begin{array}{*{20}{c}}9\\{ - 2}\end{array}} \right)\), \({{\bf{v}}_5} = \left( {\begin{array}{*{20}{c}}7\\{ - 4}\end{array}} \right)\).

Also, \(V{\bf{c}} = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\), or \({{\bf{v}}_6} = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\).

Writing the entries for vector c

Compare vector \({\bf{c}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\2\\4\\{ - 1}\\2\end{array}} \right)\) with \({\bf{c}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

So, \({c_1} = - 3\), \({c_2} = 2\), \({c_3} = 4\), \({c_4} = - 1\), and \({c_5} = 2\).

Thus, the vectors are \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{ - 3}\\5\end{array}} \right)\), \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}5\\8\end{array}} \right)\), \({{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right)\), \({{\bf{v}}_4} = \left( {\begin{array}{*{20}{c}}9\\{ - 2}\end{array}} \right)\), \({{\bf{v}}_5} = \left( {\begin{array}{*{20}{c}}7\\{ - 4}\end{array}} \right)\), \({{\bf{v}}_6} = \left( {\begin{array}{*{20}{c}}8\\{ - 1}\end{array}} \right)\), and the scalars are \({c_1} = - 3\), \({c_2} = 2\), \({c_3} = 4\), \({c_4} = - 1\), and \({c_5} = 2\).