Question

In Exercises 5–8, use the definition ofAx to write the matrix

equation as a vector equation, or vice versa.

5. \(\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\)

Short answer

The matrix equation as a vector equation is\(5 \cdot \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1 \cdot \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3 \cdot \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2 \cdot \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\).

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.

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

\(\begin{array}{c}A{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\\b = {x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\end{array}\)

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 and vector x

Consider the equation \(\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\).

Here, \(A = \left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right]\), and \(b = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\).

Write matrix A columns and vector x entries

Here, \({{\bf{a}}_1} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right]\), \({{\bf{a}}_2} = \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right]\), \({{\bf{a}}_3} = \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right]\), \({{\bf{a}}_4} = \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right]\), \({x_1} = 5\), \({x_2} = - 1\), \({x_3} = 3\), and \({x_4} = - 2\).

Use the definition to write the matrix equation as a vector equation

Write the matrix equation \(\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\) as a vector equationby using the definition as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\\5 \cdot \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1 \cdot \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3 \cdot \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2 \cdot \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\end{array}\)

Thus, the matrix equation \(\left[ {\begin{array}{*{20}{c}}5&1&{ - 8}&4\\{ - 2}&{ - 7}&3&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\{ - 1}\\3\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\)can be written as a vector equation as \(5 \cdot \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right] - 1 \cdot \left[ {\begin{array}{*{20}{c}}1\\{ - 7}\end{array}} \right] + 3 \cdot \left[ {\begin{array}{*{20}{c}}{ - 8}\\3\end{array}} \right] - 2 \cdot \left[ {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\).