Repeat Exercise 43 with the matrices A and B from Exercise 42. Then give an explanation for what you discover, assuming that B was constructed as specified.

To identify the pivot and the pivot position, observe the matrix’s leftmost column (nonzero column), that is, the pivot column. At the top of this column, 12 is the pivot.

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 6}&{ - 3}&7&{10}\\{ - 7}&{ - 6}&4&7&{ - 9}&5\\9&9&{ - 9}&{ - 5}&5&{ - 1}\\{ - 4}&{ - 3}&1&6&{ - 8}&9\\8&7&{ - 5}&{ - 9}&{11}&{ - 8}\end{array}} \right]\).

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}12{\rm{ }}10{\rm{ }} - 6{\rm{ }} - 3{\rm{ }}7{\rm{ }}10;{\rm{ }} - 7 - 6{\rm{ }}4{\rm{ }}7{\rm{ }} - 9{\rm{ }}5;{\rm{ }}9{\rm{ }}9{\rm{ }} - 9{\rm{ }} - 5{\rm{ }}5{\rm{ }} - 1;{\rm{ }}\\ - 4{\rm{ }} - 3{\rm{ }}1{\rm{ }}6{\rm{ }} - 8{\rm{ }}9;{\rm{ }}8{\rm{ }}7{\rm{ }} - 5{\rm{ }} - 9{\rm{ }}11{\rm{ }} - 8\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 6}&{ - 3}&7&{10}\\{ - 7}&{ - 6}&4&7&{ - 9}&5\\9&9&{ - 9}&{ - 5}&5&{ - 1}\\{ - 4}&{ - 3}&1&6&{ - 8}&9\\8&7&{ - 5}&{ - 9}&{11}&{ - 8}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&2&0&2&0\\0&1&{ - 3}&0&{ - 2}&0\\0&0&0&1&{ - 1}&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{array}} \right]\)

Mark the nonzero leading entries in columns 1, 2, 4, and 6.

Now, mark the pivot columns of the given matrix as shown below:

Construct matrix B by using 1, 2, 4, and 6 pivot columns of the matrixas shown below:

\(B = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}\\{ - 7}&{ - 6}&7&5\\9&9&{ - 5}&{ - 1}\\{ - 4}&{ - 3}&6&9\\8&7&{ - 9}&{ - 8}\end{array}} \right]\)

Let one of the dependent vectors be\({\bf{v}} = \left[ {\begin{array}{*{20}{c}}{ - 14}\\9\\{ - 18}\\1\\{ - 11}\end{array}} \right]\).

Obtain the augmented matrix\(\left[ {\begin{array}{*{20}{c}}B&{\bf{v}}\end{array}} \right]\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&{ - 14}\\{ - 7}&{ - 6}&7&5&9\\9&9&{ - 5}&{ - 1}&{ - 18}\\{ - 4}&{ - 3}&6&9&1\\8&7&{ - 9}&{ - 8}&{ - 11}\end{array}} \right]\)

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}12{\rm{ }}10{\rm{ }} - 3{\rm{ 10 }} - {\rm{14}};{\rm{ }} - 7 - 6{\rm{ 7 5 9}};{\rm{ }}9{\rm{ }}9{\rm{ }} - 5{\rm{ - 1 }} - {\rm{18}};{\rm{ }}\\ - 4{\rm{ }} - 3{\rm{ 6 9 1}};{\rm{ }}8{\rm{ }}7{\rm{ }} - 9{\rm{ }} - {\rm{8 }} - 11\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&{ - 14}\\{ - 7}&{ - 6}&7&5&9\\9&9&{ - 5}&{ - 1}&{ - 18}\\{ - 4}&{ - 3}&6&9&1\\8&7&{ - 9}&{ - 8}&{ - 11}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&{ - 2}\\0&0&1&0&1\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

Thus, every linearly dependent column in A is in the set spanned by the columns of B because there are four pivot elements.