91Ó°ÊÓ

Consider matrix\(A = \left[ {\begin{array}{*{20}{c}}5&2&0&{ - 8}&{ - 8}\\4&1&2&{ - 8}&{ - 9}\\5&1&3&5&{19}\\{ - 8}&{ - 5}&6&8&5\end{array}} \right]\).

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {5\,\,\,2\,\,\,0\,\,\, - 8\,\,\, - 8;\,\,4\,\,\,1\,\,\,2\,\,\, - 8\,\,\, - 9;\,\,5\,\,\,1\,\,\,3\,\,\,5\,\,\,19;\, - 8\,\,\, - 5\,\,\,6\,\,\,8\,\,\,5} \right]\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}5&2&0&{ - 8}&{ - 8}\\4&1&2&{ - 8}&{ - 9}\\5&1&3&5&{19}\\{ - 8}&{ - 5}&6&8&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{60}&{122}\\0&1&0&{ - 154}&{ - 309}\\0&0&1&{ - 47}&{ - 94}\\0&0&0&0&0\end{array}} \right]\)

Mark the pivot column in the row reduced echelon form of the matrix as shown below.

Thepivot columnsof matrix \(A\) form a basis for the column space of A.

Mark the pivot columns of matrix A as shown below.

The basis for Col A is as shown below:

\[\left[ {\begin{array}{*{20}{c}}5\\4\\5\\{ - 8}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}2\\1\\1\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\2\\3\\6\end{array}} \right]\]

Write the augmented matrix \(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\) as shown below

\(\left[ {\begin{array}{*{20}{c}}5&2&0&{ - 8}&{ - 8}&0\\4&1&2&{ - 8}&{ - 9}&0\\5&1&3&5&{19}&0\\{ - 8}&{ - 5}&6&8&5&0\end{array}} \right]\)

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {5\,\,\,2\,\,\,0\,\,\, - 8\,\,\, - 8\,\,\,0;\,\,4\,\,\,1\,\,\,2\,\,\, - 8\,\,\, - 9\,\,\,0;\,\,5\,\,\,1\,\,\,3\,\,\,5\,\,\,19\,\,0;\, - 8\,\,\, - 5\,\,\,6\,\,\,8\,\,\,5\,\,\,0} \right]\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}5&2&0&{ - 8}&{ - 8}&0\\4&1&2&{ - 8}&{ - 9}&0\\5&1&3&5&{19}&0\\{ - 8}&{ - 5}&6&8&5&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{60}&{122}&0\\0&1&0&{ - 154}&{ - 309}&0\\0&0&1&{ - 47}&{ - 94}&0\\0&0&0&0&0&0\end{array}} \right]\)

Convert the matrix into the system of the equation as shown below.

Therefore, the solution of \(Ax = 0\) is:

\(\begin{array}{l}{x_1} = - 60{x_4} - 122{x_5}\\{x_2} = 154{x_4} + 309{x_5}\\{x_3} = 47{x_4} + 94{x_5}\end{array}\)

Here, \({x_4}\,{\mathop{\rm and}\nolimits} \,\,{x_5}\) are free variables.

Write the solution of \(Ax = 0\) in the parametric vector form as shown below.

\[\begin{array}{c}{\mathop{\rm x}\nolimits} = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\\{{{\mathop{\rm x}\nolimits} _3}}\\{{{\mathop{\rm x}\nolimits} _4}}\\{{{\mathop{\rm x}\nolimits} _5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 60{{\mathop{\rm x}\nolimits} _4} - 122{x_5}}\\{154{{\mathop{\rm x}\nolimits} _4} + 309{x_5}}\\{47{x_4} + 94{{\mathop{\rm x}\nolimits} _5}}\\{{{\mathop{\rm x}\nolimits} _4}}\\{{{\mathop{\rm x}\nolimits} _5}}\end{array}} \right]\\ = {{\mathop{\rm x}\nolimits} _4}\left[ {\begin{array}{*{20}{c}}{ - 60}\\{154}\\{47}\\1\\0\end{array}} \right] + {{\mathop{\rm x}\nolimits} _5}\left[ {\begin{array}{*{20}{c}}{ - 122}\\{309}\\{97}\\0\\1\end{array}} \right]\\ = {{\mathop{\rm x}\nolimits} _4}{\mathop{\rm u}\nolimits} + {{\mathop{\rm x}\nolimits} _5}{\mathop{\rm v}\nolimits} \end{array}\]

Nul A coincides with the set of a linear combination of u and v. That is, \(\left\{ {{\mathop{\rm u}\nolimits} ,v} \right\}\) generates Nul A.

Thus, \(\left\{ {{\mathop{\rm u}\nolimits} ,v} \right\}\) is a basis for Nul A.