x=[0 1 10] p=[100 80] how can I get A A=[3x(1) 2x(1) x(1) 1 0 0 0 0; 2x(1) x(1) 1 0 0 0 0 0; 3x(2) 2x(2) x(2) 1 -3x(2) -2x(2) -x(2) -1; 2x(2) x(2) 1 0 -2x(2) -x(2) -1 0; p(1)x(1) p(1) 0 0 -p(2)x(2) -p(2) 0 0; p(1) 0 0 0 -p(2) 0 0 0; 0 0 0 0 -p(2)x(3) -p(2) 0 0; 0 0 0 0 -p(3) 0 0 0; ]
In article <fkus3t$11h$1@fred.mathworks.com>, Jim lei <redlightlike@mathworks.com> wrote: >x=[0 1 10] >p=[100 80] >how can I get A > >A=[3x(1) 2x(1) x(1) 1 0 0 0 0; > 2x(1) x(1) 1 0 0 0 0 0; > 3x(2) 2x(2) x(2) 1 -3x(2) -2x(2) -x(2) -1; > 2x(2) x(2) 1 0 -2x(2) -x(2) -1 0; > p(1)x(1) p(1) 0 0 -p(2)x(2) -p(2) 0 0; > p(1) 0 0 0 -p(2) 0 0 0; > 0 0 0 0 -p(2)x(3) -p(2) 0 0; > 0 0 0 0 -p(3) 0 0 0; >] A=[3*x(1) 2*x(1) x(1) 1 0 0 0 0; ... 2*x(1) x(1) 1 0 0 0 0 0; ... 3*x(2) 2*x(2) x(2) 1 -3*x(2) -2*x(2) -x(2) -1; ... 2*x(2) x(2) 1 0 -2*x(2) -x(2) -1 0; ... p(1)*x(1) p(1) 0 0 -p(2)*x(2) -p(2) 0 0; ... p(1) 0 0 0 -p(2) 0 0 0; ... 0 0 0 0 -p(2)*x(3) -p(2) 0 0; ... 0 0 0 0 -p(3) 0 0 0 ... ] -- "Beware of bugs in the above code; I have only proved it correct, not tried it." -- Donald Knuth
"Jim lei" <redlightlike@mathworks.com> wrote in message <fkus3t$11h$1@fred.mathworks.com>... x=linapce(0,0.57 n) p=(100 80,n-1) how can I get A A=[ x(1) x(1) x(1) 1 0 0 0 0; > x(1) x(1) 1 0 0 0 0 0; > x(2) x(2) x(2) 1 -x(2) -x(2) -x(2) -1; > x(2) x(2) 1 0 -x(2) -x(2) -1 0; > p(1)*x(1) p(1) 0 0 -p(2)*x(2) -p(2) 0 0; > p(1) 0 0 0 -p(2) 0 0 0; > 0 0 0 0 -p(2)*x(3) -p(2) 0 0; > 0 0 0 0 -p(2) 0 0 0; > ]
"Jim lei" <redlightlike@mathworks.com> wrote in message <fkuvn9$d68$1@fred.mathworks.com>... > "Jim lei" <redlightlike@mathworks.com> wrote in message > <fkus3t$11h$1@fred.mathworks.com>... > > > ok, so which is it: x=[0 1 10] p=[100 80] how can I get A A=[3x(1) 2x(1) x(1) 1 0 0 0 0; 2x(1) x(1) 1 0 0 0 0 0; 3x(2) 2x(2) x(2) 1 -3x(2) -2x(2) -x(2) -1; 2x(2) x(2) 1 0 -2x(2) -x(2) -1 0; p(1)x(1) p(1) 0 0 -p(2)x(2) -p(2) 0 0; p(1) 0 0 0 -p(2) 0 0 0; 0 0 0 0 -p(2)x(3) -p(2) 0 0; 0 0 0 0 -p(3) 0 0 0; ] or this: x=linapce(0,0.57 n) p=(100 80,n-1) A=[ x(1) x(1) x(1) 1 0 0 0 0; x(1) x(1) 1 0 0 0 0 0; x(2) x(2) x(2) 1 -x(2) -x(2) -x(2) -1; x(2) x(2) 1 0 -x(2) -x(2) -1 0; p(1)*x(1) p(1) 0 0 -p(2)*x(2) -p(2) 0 0; p(1) 0 0 0 -p(2) 0 0 0; 0 0 0 0 -p(2)*x(3) -p(2) 0 0; 0 0 0 0 -p(2) 0 0 0; ] in the first case it almost would have worked except you only had 2 elements for p but required a p(3) in the definition of A. in the second case you have lost it completely. linapce i don't recognize as a function in matlab, the arguments passed to it are incorrectly formatted, and the value assigned to p is a badly formatted command... so i recommend taking a deep breath, slow down, and start over. Oh, and forget the !!! in the subject, your lack of proper planning doesn't make this any more important or interesting to anyone here.