91Ó°ÊÓ

It is known that \(\frac{{2{n^3}}}{3} + \frac{{{n^2}}}{2} - \frac{{7n}}{6}\) flops are required for the reduction to the echelon form for \(n \times \left( {n + 1} \right)\) matrix. If \(n\) is greater than or equal to 30, then there are approximately \(\frac{{2{n^3}}}{3}\) flops. Approximately \({n^2}\) flops are needed to reduce the echelon form further.

It is given that \(n = 30\). So, substitute 30 for \(n\) in \(\frac{{2{n^3}}}{3}\) to obtain the number of flops for \(n = 30\).

\(\begin{array}{c}\frac{{2{n^3}}}{3} = \frac{{2{{\left( {30} \right)}^3}}}{3}\\ = 18000\end{array}\)

Thus, when \(n = 30\) the reduction to echelon form takes 18,000 flops.

Substitute 30 for \(n\) in \({n^2}\) to obtain the number of flops for \(n = 30\), for further reduction.

\(\begin{array}{c}{n^2} = {\left( {30} \right)^2}\\ = 900\end{array}\)

Thus, when \(n = 30\), the further reduction to echelon form takes 900 flops.

The fraction involved with the backward phase is as shown below:

\(\frac{{900}}{{18000}} = 0.048\)

It is given that \(n = 300\). So, substitute 300 for \(n\) in \(\frac{{2{n^3}}}{3}\) to obtain the number of flops for \(n = 300\).

\(\begin{array}{c}\frac{{2{n^3}}}{3} = \frac{{2{{\left( {300} \right)}^3}}}{3}\\ = 18000000\end{array}\)

Thus, when \(n = 300\), the reduction to the echelon form takes 18,000,000 flops.

Substitute 300 for \(n\) in \({n^2}\) to obtain the number of flops for \(n = 300\), for further reduction.

\(\begin{array}{c}{n^2} = {\left( {300} \right)^2}\\ = 90000\end{array}\)

Thus, when \(n = 300\), the further reduction to the echelon form takes 90,000 flops.

The fraction involved with the backward phase is as shown below:

\(\frac{{90000}}{{18000000}} = 0.005\)

Thus, the fraction for \(n = 30\) is 0.048, and for \(n = 300\), it is 0.005.