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In Exercises 5鈥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

Expert verified

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

01

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.

02

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 \(b = \left[ {\begin{array}{*{20}{c}}{ - 8}\\{16}\end{array}} \right]\).

03

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 \({x_4} = - 2\).

04

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

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Most popular questions from this chapter

In Exercises 11 and 12, determine if \({\rm{b}}\) is a linear combination of \({{\mathop{\rm a}\nolimits} _1},{a_2}\) and \({a_3}\).

11.\({a_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\0\end{array}} \right],{a_2} = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right],{a_3} = \left[ {\begin{array}{*{20}{c}}5\\{ - 6}\\8\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\6\end{array}} \right]\)

Find the general solutions of the systems whose augmented matrices are given as

12. \(\left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\).

Consider the problem of determining whether the following system of equations is consistent for all \({b_1},{b_2},{b_3}\):

\(\begin{aligned}{c}{\bf{2}}{x_1} - {\bf{4}}{x_2} - {\bf{2}}{x_3} = {b_1}\\ - {\bf{5}}{x_1} + {x_2} + {x_3} = {b_2}\\{\bf{7}}{x_1} - {\bf{5}}{x_2} - {\bf{3}}{x_3} = {b_3}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\). Then solve that problem.
  1. Define an appropriate matrix, and restate the problem using the phrase 鈥渃olumns of A.鈥
  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

In Exercises 15 and 16, fill in the missing entries of the matrix, assuming that the equation holds for all values of the variables

\(\left[ {\begin{array}{*{20}{c}}?&?&?\\?&?&?\\?&?&?\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3{x_1} - 2{x_3}}\\{4{x_1}}\\{{x_1} - {x_2} + {x_3}}\end{array}} \right]\)

In Exercise 19 and 20, choose \(h\) and \(k\) such that the system has

a. no solution

b. unique solution

c. many solutions.

Give separate answers for each part.

19. \(\begin{array}{l}{x_1} + h{x_2} = 2\\4{x_1} + 8{x_2} = k\end{array}\)

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