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Chapter 2: Q2.8-9Q (page 93)

9: With A and p as in Exercise 7, determine if p is in Nul A.

Short Answer

Expert verified

p is not in Nul A.

Step by step solution

01

State the values of A and p as in Exercise 7

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}2\\{ - 8}\\6\end{array}} \right],{{\mathop{\rm v}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\8\\{ - 7}\end{array}} \right],{{\mathop{\rm v}\nolimits} _3} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\6\\{ - 7}\end{array}} \right],{\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}6\\{ - 10}\\{11}\end{array}} \right]\), and \(A = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _2}}&{{{\mathop{\rm v}\nolimits} _3}}\end{array}} \right]\).

02

Write matrix A using the vectors

Write matrix A in the form \[\left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _2}}&{{{\mathop{\rm v}\nolimits} _3}}\end{array}} \right]\], as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&{ - 4}\\{ - 8}&8&6\\6&{ - 7}&{ - 7}\end{array}} \right]\)

03

Determine whether p is in Nul A

The null spaceof matrix A is the set Nul Aof all solutions of the homogeneous equation\(Ax = 0\).

Calculate \(A{\mathop{\rm p}\nolimits} \), as shown below:

\(\begin{array}{c}A{\mathop{\rm p}\nolimits} = \left[ {\begin{array}{*{20}{c}}2&{ - 3}&{ - 4}\\{ - 8}&8&6\\6&{ - 7}&{ - 7}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}6\\{ - 10}\\{11}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{12 + 30 - 44}\\{ - 48 - 80 + 66}\\{36 + 70 - 77}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 62}\\{29}\end{array}} \right]\end{array}\)

Since \[A{\mathop{\rm p}\nolimits} \ne 0\], p is not in Nul A.

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Most popular questions from this chapter

In exercises 11 and 12, mark each statement True or False. Justify each answer.

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

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 matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).
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