[M] Let \(A\) be the matrix in Exercise 35. Construct a matrix \(C\) whose columns are the pivot columns of \(A\), and construct a matrix \(R\) whose rows are the nonzero rows of the reduced echelon form of \(A\). Compute \(CR\), and discuss what you see.

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

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

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

Compute \(CR\) as shown below:

\(\begin{aligned} CR &= \left[ {\begin{array}{*{20}{c}}7&{ - 9}&5&{ - 3}\\{ - 4}&6&{ - 2}&{ - 5}\\5&{ - 7}&5&2\\{ - 3}&5&{ - 1}&{ - 4}\\6&{ - 8}&4&9\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{\frac{{13}}{2}}&0&5&0&{ - 3}\\0&1&{\frac{{11}}{2}}&0&{\frac{1}{2}}&0&2\\0&0&0&1&{\frac{{ - 11}}{2}}&0&7\\0&0&0&0&0&1&1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\\ &= A\end{aligned}\)

Thus, \(A = CR\).