[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.

a)

Use code in the MATLAB to obtain the row-reduced echelon form of the matrix.

\[\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left[ {7\,\,\, - 9\,\,\,\, - 4\,\,\,\,5\,\,\,\,3\,\,\,\, - 3\,\,\, - 7;\, - 4\,\,\,\,6\,\,\,\,7\,\,\, - 2\,\,\,\, - 6\,\,\,\, - 5\,\,\,5;\,\,5\,\, - 7\,\,\, - 6\,\,\,\,5\,\,\, - 6\,\,\,2\,\,\,8;\, - 3\,\,\,\,5\,\,\,8\,\,\, - 1\,\,\,\, - 7\,\,\,\, - 4\,\,\,\,8;\,\,6\,\,\, - 8\,\,\,\, - 5\,\,\,\,4\,\,\,\,4\,\,\,\,9\,\,\,3} \right]\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{array}\]

\(A \sim \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{13}}{2}}&0&5&0&{ - 3}\\0&1&{\frac{{11}}{2}}&0&{\frac{1}{2}}&0&2\\0&0&0&1&{\frac{{ - 11}}{2}}&0&7\\0&0&0&0&0&1&1\\0&0&0&0&0&0&0\end{array}} \right]\)

The pivot columns of matrix \(A\) are a basis for \({\mathop{\rm Col}\nolimits} A\). Thus, matrix \(C\) has the following columns:

\(C = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&5&{ - 3}\\{ - 4}&6&{ - 2}&{ - 5}\\5&{ - 7}&5&2\\{ - 3}&5&{ - 1}&{ - 4}\\6&{ - 8}&4&9\end{array}} \right]\)

The pivot rows of the row-reduced echelon form of matrix \(A\) form a basis for row A. Therefore, matrix \(R\) has the following:

\(R = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{13}}{2}}&0&5&0&{ - 3}\\0&1&{\frac{{11}}{2}}&0&{\frac{1}{2}}&0&2\\0&0&0&1&{\frac{{ - 11}}{2}}&0&7\\0&0&0&0&0&1&1\end{array}} \right]\)

To determine the basis for \({\mathop{\rm Nul}\nolimits} A\), row reduce matrix A to the reduced echelon form. The solution to the equation \(A{\mathop{\rm x}\nolimits} = 0\) can be expressed in terms of the free variables as: \(\begin{array}{l}{{\mathop{\rm x}\nolimits} _1} = - \left( {\frac{{13}}{2}} \right){{\mathop{\rm x}\nolimits} _3} - 5{{\mathop{\rm x}\nolimits} _5} + 3{{\mathop{\rm x}\nolimits} _7}\\{{\mathop{\rm x}\nolimits} _2} = - \left( {\frac{{11}}{2}} \right){{\mathop{\rm x}\nolimits} _3} - \left( {\frac{1}{2}} \right){{\mathop{\rm x}\nolimits} _5} - 2{{\mathop{\rm x}\nolimits} _7}\\{{\mathop{\rm x}\nolimits} _4} = - \left( {\frac{{11}}{2}} \right){{\mathop{\rm x}\nolimits} _5} - 7{{\mathop{\rm x}\nolimits} _7}\\{{\mathop{\rm x}\nolimits} _6} = - {{\mathop{\rm x}\nolimits} _7}\end{array}\)

\({{\mathop{\rm x}\nolimits} _3}\), \({{\mathop{\rm x}\nolimits} _5},\) and \({{\mathop{\rm x}\nolimits} _7}\) are the free variables.

Therefore, matrix \(N\) is obtained as shown below:

\(N = \left[ {\begin{array}{*{20}{c}}{ - \frac{{13}}{2}}&{ - 5}&3\\{ - \frac{{11}}{2}}&{ - \frac{1}{2}}&{ - 2}\\1&0&0\\0&{\frac{{11}}{2}}&{ - 7}\\0&1&0\\0&0&{ - 1}\\0&0&1\end{array}} \right]\)

b)

Matrix \({A^T}\)is obtainedas shown below:

\({A^T} = \left[ {\begin{array}{*{20}{c}}7&{ - 4}&5&{ - 3}&6\\{ - 9}&6&{ - 7}&5&{ - 8}\\{ - 4}&7&{ - 6}&8&{ - 5}\\5&{ - 2}&5&{ - 1}&4\\3&{ - 6}&{ - 6}&{ - 7}&4\\{ - 3}&{ - 5}&2&{ - 4}&9\\{ - 7}&5&8&8&3\end{array}} \right]\)

Consider matrix \({{\mathop{\rm A}\nolimits} ^T} = \left[ {\begin{array}{*{20}{c}}7&{ - 4}&5&{ - 3}&6\\{ - 9}&6&{ - 7}&5&{ - 8}\\{ - 4}&7&{ - 6}&8&{ - 5}\\5&{ - 2}&5&{ - 1}&4\\3&{ - 6}&{ - 6}&{ - 7}&4\\{ - 3}&{ - 5}&2&{ - 4}&9\\{ - 7}&5&8&8&3\end{array}} \right]\).

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

\[\begin{array}{l} > > {{\mathop{\rm A}\nolimits} ^T} = \left[ {7\,\,\, - 4\,\,\,\,5\,\,\, - 3\,\,\,6;\,\, - 9\,\,\,6\,\,\, - 7\,\,\,5\,\,\, - 8;\, - 4\,\,\,7\,\,\, - 6\,\,\,8\,\,\, - 5;\,5\,\,\, - 2\,\,\,\,5\,\,\, - 1\,\,\,4;\,3\,\,\, - 6\,\,\, - 6\,\,\, - 7\,\,\,4;\, - 3\,\,\, - 5\,\,\,2\,\,\, - 4\,\,\,9;\,\, - 7\,\,\,5\,\,\,8\,\,\,8\,\,\,3} \right]\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {{{\mathop{\rm A}\nolimits} ^T}} \right)\end{array}\]

\({{\mathop{\rm A}\nolimits} ^T} \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{\frac{{ - 2}}{{11}}}\\0&1&0&0&{\frac{{ - 41}}{{11}}}\\0&0&1&0&0\\0&0&0&1&{\frac{{28}}{{11}}}\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right]\)

Thus, the solution to the equation \({A^T}{\mathop{\rm x}\nolimits} = 0\) can be expressed in terms of the free variables as shown below:

\(\begin{array}{l}{{\mathop{\rm x}\nolimits} _1} = \left( {\frac{2}{{11}}} \right){{\mathop{\rm x}\nolimits} _5}\\{{\mathop{\rm x}\nolimits} _2} = \left( {\frac{{41}}{{11}}} \right){{\mathop{\rm x}\nolimits} _5}\\{{\mathop{\rm x}\nolimits} _3} = 0\\{{\mathop{\rm x}\nolimits} _4} = - \left( {\frac{{28}}{{11}}} \right){{\mathop{\rm x}\nolimits} _5}\end{array}\)

\({{\mathop{\rm x}\nolimits} _5}\)is a free variable.

Therefore, matrix \(M\) is obtained as shown below:

\(M = \left[ {\begin{array}{*{20}{c}}{\frac{2}{{11}}}\\{\frac{{41}}{{11}}}\\0\\{ - \frac{{28}}{{11}}}\\1\end{array}} \right]\)

Matrix \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) is a \(7 \times 7\) matrix since the columns of \({R^T}\) and \(N\) are in \({\mathbb{R}^7}\) and \(\dim {\mathop{\rm Row}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = 7\). Matrix \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\) is \(5 \times 5\) since the columns of \(C\) and \(M\) are in \({\mathbb{R}^5}\) and \(\dim {\mathop{\rm Col}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} {A^T} = 5\). Both \(S\) and \(T\) are invertible since their columns are linearly independent.

Thus, it is verified that \(S\) and \(T\) are invertible.