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In Exercises 1 and 2, find the vector x determined by the given

coordinate vector \({\left[ {\bf{x}} \right]_{\rm B}}\)and the given basis B. Illustrate your answer with a figure, as in the solution of Practice Problem 2.

1. \(B = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]} \right\}\), \({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\)

Short Answer

Expert verified

The vector is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\).

Step by step solution

01

Definition of coordinate systems

For the basis of asubspace H, let the set be\(B = \left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\). For theweights \({c_1},...,{c_p}\), the coordinate of x is represented as\({\bf{x}} = {c_1}{{\bf{b}}_1} + \cdots + {c_p}{{\bf{b}}_p}\).

The\(B\)-coordinatevector of x is represented as shown below:

\({\left[ {\bf{x}} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\ \vdots \\{{c_p}}\end{array}} \right]\)

02

Construct the augmented matrix

Compare\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\)with\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\). So,\({c_1} = 3\)and.

For theweights \({c_1}\)and\({c_2}\), and thevector x in H, it must satisfy the equation\({\bf{x}} = {c_1}{{\bf{b}}_1} + {c_2}{{\bf{b}}_2}\).

Consider the vectors\[{{\bf{b}}_1} = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\],\[{{\bf{b}}_2} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\], and\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\].

Then, it can be represented as shown below:

\(\begin{array}{l}{\bf{x}} = 3\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\end{array}\)

Thus, the vector is\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\).

The illustrated figure is shown below:

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Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

Suppose the last column of ABis entirely zero but Bitself has no column of zeros. What can you sayaboutthe columns of A?

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

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