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

7. \({x_1}\left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right]\)

Short Answer

Expert verified

The vector equation \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)in a matrix equation is written as \(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

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.

\({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\)

The left-hand side \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\) of the above equation is a 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.

02

Obtain the columns of the matrix

Compare the given vector equation form \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)with the general equation \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\) form.

So, \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\), and \(A{\bf{x}} = b = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

It shows that the equation is a linear combination of three vectors \({x_1}\), \({x_2}\), and \({x_3}\).

03

Write matrix Aand vector x

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 three columns \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), and\({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\) and vector x using three entries \({x_1}\), \({x_2}\), and \({x_3}\).

So, \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), and \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\).

04

Write the vector equation into a matrix equation

By using matrix \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), 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:

\(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)

Thus, the vector equation \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)can be written as a matrix equation as \(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

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Most popular questions from this chapter

In Exercise 22, mark each statement True or False. Justify each answer.

22. a. Every matrix transformation is a linear transformation.

b. The codomain of the transformation \({\bf{x}} \mapsto {\bf{Ax}}\) is the set of all linear combinations of the columns of \({\bf{A}}\).

c. If \({\bf{T}}:{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) is a linear transformation and if \({\bf{c}}\) is in \({\mathbb{R}^{\bf{m}}}\), then a uniqueness is 鈥淚s c in the range of T?鈥

d. A linear transformation preserves the operations of vector addition and scalar multiplication.

e. The superposition principle is a physical description of a linear transformation.

Let \(u = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) and \(v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\). Show that \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in Span \(\left\{ {u,v} \right\}\) for all \(h\) and\(k\).

In Exercises 13 and 14, determine if \(b\) is a linear combination of the vectors formed from the columns of the matrix \(A\).

13. \(A = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&2\\0&3&5\\{ - 2}&8&{ - 4}\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\\{ - 3}\end{array}} \right]\)

In Exercises 7-12, describe all solutions of \(Ax = 0\) in parametric vector form, where \(A\) is row equivalent to the given matrix.

12. \(\left( {\begin{array}{*{20}{c}}1&5&2&{ - 6}&9&0\\0&0&1&{ - 7}&4&{ - 8}\\0&0&0&0&0&1\\0&0&0&0&0&0\end{array}} \right)\)

In Exercises 19 and 20, find the parametric equation of the line

through a parallel to b.

19. \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\)

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