Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications, which if a multiplication takes sec would take these times: n 10 15 20 25
Time 0.004 sec 22 min 77 years 0.5.109years

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

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If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

answer

step-by-step explanation: