i tried this.but the result that i obtained is different from actual X..i think the system has no unique solution..whether the chance of existence of unique solution depends on the dimension of the matrix or its elements? can anyone help me?
solve AX=B
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hi..i am working on OFDMA.i have represented the received vector in the form of AX=B where A has 1024*2048 elements ,X has 2048*1 elements so that the received vector has 1024*1 elements. But many of the elements in the X are zeros.So in order to solve the equation ,i only considered the non-zero elements in X and corresponding elements in A .so now i have 1024*1023 elements in A,1023*1 elements in X.How can i solve it? i used pinv command.But the result obtained is not matching with my actual X..Can anyone help me?
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Torsten
le 8 Fév 2016
You can easily check:
If both A*X and A*X_actual give B, the system does not have a unique solution.
Best wishes
Torsten.
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