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

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 \(A = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _2}}&{{{\mathop{\rm v}\nolimits} _3}}\end{array}} \right]\).

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:

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

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.