Ax1 = b,
Ax2 = x1
Ax3 = x2,
Ax4 = x3,
Ax = x4,
how to solve this sequence A and b are defined
A = diag(2*ones(1,n)) - diag(ones(1,n-1),1) - diag(ones(1,n-1),-1)
b = [0:1:n/2-1 n/2-1:-1:0]

1 Kommentar

James Tursa
James Tursa am 6 Mär. 2015
Please explain in more detail. Does Ax1 mean A*x1, so that the line Ax1 = b means A*x1 = b and you are solving for x1? Then you use that to solve for x1, x3, x4, and ultimately x?

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James Tursa
James Tursa am 6 Mär. 2015
Bearbeitet: James Tursa am 6 Mär. 2015

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Plugging everything in I get
A^5 * x = b
So the solution would be
x = A^5 \ b
However, your expressions for A and b don't seem right since the dimensions are incompatible for the linear expression. E.g.,
>> n = 3
n =
3
>> A = diag(2*ones(1,n)) - diag(ones(1,n-1),1) - diag(ones(1,n-1),-1)
A =
2 -1 0
-1 2 -1
0 -1 2
>> b = [0:1:n/2-1 n/2-1:-1:0]'
b =
0
0.500000000000000
>> A^5\b
Error using \
Matrix dimensions must agree.

2 Kommentare

HIGH
HIGH am 6 Mär. 2015
Bearbeitet: HIGH am 6 Mär. 2015
Thanks for the reply. Actually A and b are correct and I figured the X's out. but I need to write a function for "upper bidiagonal system" to find these X'using the attached m files. I tried couple of ways but yet having trouble writing the function.
Ohh I am sorry, n should be equal to 10.. n=10
HIGH
HIGH am 6 Mär. 2015
n=10 sorry

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