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.

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

Short Answer

Expert verified

The vector is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\).

Step by step solution

01

Definition of coordinate systems

For the basis of asubspaceH, 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\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\)with\({\left[ x \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right]\). So,\({c_1} = - 1\)and\({c_2} = 3\).

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

Then, it can be represented as shown below:

\(\begin{array}{l}{\bf{x}} = - 1\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}9\\3\end{array}} \right]\\{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\end{array}\)

Thus, the vector is\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{11}\\2\end{array}} \right]\).

The illustrated figure is shown below:

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Most popular questions from this chapter

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

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