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Chapter 4: Q1E (page 191)

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

Short Answer

Expert verified

Vector \(x = \left( {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right)\)

Step by step solution

01

Use the definition

The coordinates of x relative to basis\({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\)are the weights\({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\)such that\(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then,\({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

02

Find x

By the above definition, you get

\(\begin{array}{c}x = 5\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}{ - 4}\\6\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{15}\\{ - 25}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 12}\\{18}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{15 - 12}\\{ - 25 + 18}\end{array}} \right)\\x = \left( {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right).\end{array}\)

03

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right)\).

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Most popular questions from this chapter

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

What would you have to know about the solution set of a homogenous system of 18 linear equations 20 variables in order to understand that every associated nonhomogenous equation has a solution? Discuss.

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)
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