Determine the flop count for a single iteration of the power

Determine the flop count for a single iteration of the power method applied to an n times n matrix A.

Solution

Solution-:   In this probem we have to explain it like   below as.

Let R be an n x n upper-triangular matrix with semiband width 5. Show
that the system Rx = y can be solved by back substitution in about 2ns flops. An
analogous result holds for lower-triangular systems.

If we
perform a Cholesky decomposition using a program that does not exploit the band
structure of the matrix, the cost of the arithmetic is about n3 3.3 x 10^5 flops.
In contrast, if we do exploit the band structure, the cost is about ns2 = 10^4 flops,
which is about 3% of the previous figure. In the forward and back substitution
steps, substantial but less spectacular savings are achieved. The combined arithmetic
cost of forward and back substitution without exploiting the band structure is about
2n2 = 2 x 10^4 flops. If the band structure is exploited, the flop count is about
4ns — 4 x 10^3,So it true for nxn .