Question

Given matrix B= A + `3A∧3` + `5A∧5`+...+`99A∧99`, How to find matrix B, given answer is 2500A, how to do it?

Answer 1

This result is not true in general.

Explanation:

We seek:

`bb(B) = bb(A) + 3bb(A)^3+5bb(A)^5 +...+ 99bb(A)^99 \ \ \` (50 terms)

The solution to this probelm is completely dependant upon the matrix `bb(A).`

If we choose `bb(A)=bb(I)`; then:

`bb(B)= bb(I) + 3bb(I)^3+5bb(I)^5 +...+ 99bb(I)^99`

If we look at the coefficients `{1,3,5,7,...,99}` then we see that they form an AP with `a=1.d=2,n=50`, so their sum is given by:

`S_n=n/2(2a+(n-1)d) = 25(2+49.2)=25(2+98)=25.100=2500`

So we get that:

`bb(B) = 2500 \ bb(I) = 2500 \ bb(A)`

Which is consistent with the quoted result.

However, Consider the matrix given by

`bb(A) = ( (1,1), (0,1 ) )` and we find that `bb(A)^n = ( (1,n), (0,1 ) )`

Making the sum:

`bb(B) = bb(A) + 3bb(A)^3+5bb(A)^5 +...+ 99bb(A)^99`

`\ \ \ \ = ( (1,1), (0,1 ) ) + 3( (1,3), (0,1 ) )+5( (1,5), (0,1 ) )+...+99( (1,99), (0,1 ) )`

`\ \ \ \ = ( (1,1), (0,1 ) ) + ( (3,3^2), (0,3 ) )+( (5,5^2), (0,5 ) )+...+( (99,99^2), (0,99 ) )`

And with this matrix we have:

`2500 \ bb(A) = 2500 \ ( (1,1), (0,1 ) ) = ( (2500,2500), (0,2500 ) )`

And without any further calculation it is apparent that:

`bb(B )!= 2500 \ bb(A)`