Error : Matrix dimensions must agree

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Ahmad Sheikh
Ahmad Sheikh 2015-9-29
回答： Walter Roberson 2015-9-29
I dont know why it's giving can someone help ?
function derivative_x = AIAA(t,x)
derivative_x=zeros(5,1);
A1 =[ -5.9690 -41.3275 1.3080 -1.0463 -0.4869
-0.0078 -2.1282 0.9507 -20.8661 -0.0053
2.0858 49.5871 -10.9028 1.2224 -0.1164
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A2 =[ -5.9690 -41.3275 1.3080 -1.0463 -0.4869
-0.0078 -2.1282 0.9507 -20.8661 -0.0053
2.0858 49.5871 -10.9028 1.2224 -0.1164
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A3 =[ -0.2132 -41.1168 0.9679 -1.0683 0.0097
0.0051 -2.1951 1.0124 -20.8428 -0.0003
0.4030 45.4138 -6.9930 2.6577 -0.0198
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A4 =[ -0.2142 -41.1032 1.0609 -1.0410 0.0365
-0.0040 -2.1002 0.9922 -20.9089 0.0013
-0.1534 51.2034 -8.2651 -1.3891 0.0643
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A5 =[ 46.4754 -41.1787 -45.6425 -1.9833 -1.3092
-0.2261 -2.1446 1.3133 -20.8828 -0.0024
-33.7711 48.5222 29.6778 0.6092 0.4142
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A6 =[ -6.6983 -41.3450 1.3548 -1.0467 -0.5485
-0.0106 -2.1288 0.9347 -20.8662 -0.0060
2.2284 49.5581 -13.6275 1.2201 -0.1330
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A7 =[ -0.2534 -41.1302 0.9724 -1.0674 0.0073
0.0071 -2.1944 1.0233 -20.8428 -0.0001
0.5412 45.4620 -8.0555 2.6541 -0.0108
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A8 =[ -6.7972 -37.7358 -4.5678 -8.3716 13.8487
-0.3353 -1.9333 0.6742 -21.2896 0.7127
-17.5613 59.9487 -26.9974 -21.4951 37.5793
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A9 =[ 54.4509 -25.8776 76.5925 -9.7710 16.3861
0.0523 -2.1433 1.0735 -20.8769 -0.0662
-20.1909 -63.2368 -35.7359 4.3123 -11.0777
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A10 =[ 53.5818 -27.3231 75.3401 -9.8364 16.6126
-0.3772 -2.1922 0.4696 -20.8736 0.2516
-46.0467 -65.6067 -72.0813 4.5409 8.2318
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A11 =[ 2.2818 -32.6618 3.1819 -9.5013 1.4445
-0.0878 -2.5814 0.9246 -20.4504 -0.0835
-6.3412 -87.0851 -13.3051 30.2435 -5.7179
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A12 =[ 2.5007 -29.9855 3.4502 -11.1737 1.6224
0.0743 -1.6826 1.0627 -21.2935 0.0767
3.4620 -33.3368 -4.9838 -20.5278 3.9878
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A13 =[ 67.0151 -24.5698 94.2833 -9.6220 19.9493
0.0621 -2.1426 1.0873 -20.8758 -0.0798
-24.9889 -63.7592 -44.2180 4.3141 -13.4387
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A14 =[ 65.9082 -26.2945 92.6619 -9.7749 20.3256
-0.4541 -2.1999 0.3613 -20.8809 0.3027
-56.0385 -66.5172 -87.8631 4.0703 9.7567
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A15 =[ 2.9873 -32.6748 4.1806 -9.4956 1.6515
-0.1195 -2.5805 0.8903 -20.4506 -0.0945
-8.5833 -87.0188 -17.5545 30.2301 -6.4803
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
A16 =[ 3.3293 -29.7764 4.6105 -11.1986 1.8517
0.1045 -1.6745 1.0962 -21.2946 0.0867
4.9538 -32.9345 -5.1669 -20.5833 4.4969
0 0 1.0000 0 0
0 -2.6000 0 2.6000 0];
derivative_x = A1.*x + A2.*x + A3.*x + A4.*x + A5.*x + A6.*x + A7.*x + A8.*x +A9.*x + A10.*x + A11.*x + A12.*x + A13.*x + A14.*x + A15.*x + A16.*x;
while running the above funtion in another matlab file i get this error
clc;
clf;
Initial_Time=0;
Final_Time=10;
[x1,x2] = meshgrid(-2:0.1:2);
[t,x] = ode45(@AIAA, [Initial_Time Final_Time], [0.5 1 0.5 1]);
figure(1);
%subplot(3,1,1);
plot(t,x(:,1));
%axis([0 10 -2 2]);
xlabel('time')
ylabel('state x1k')
figure(2);
%subplot(3,1,2);
plot(t,x(:,2));
xlabel('time')
ylabel('state x2k')
%axis([0 10 -3 3]);

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回答（1 个）

Walter Roberson
Walter Roberson 2015-9-29
In order to be able to use .* between a 5 by 5 matrix and "x", such as A1 .* x, then x would have to be scalar or x would have to be a 5 x 5 matrix itself. As this is ode45, the size of the output has to be the same size as your "x" input, so if x was in fact a scalar then you would get an error because the 5 x 5 result of your calculation would not be the same size as x. Therefore your x would have to be a 5 x 5 matrix.
However, in ode45, the x input is taken as a column vector the same length as your x0. Your x0 is [0.5 1 0.5 1] which is length 4, so the parameter that your function will receive will be a 4 x 1 vector. That cannot possibly be .* with a 5 x 5 array.
If you were to pass in an x0 which was length 5 then you might possibly intend your code to be
derivative_x = A1*x + A2*x + A3*x + A4*x + A5*x + A6*x + A7*x + A8*x +A9*x + A10*x + A11*x + A12*x + A13*x + A14*x + A15*x + A16*x;
which is to say that you would use Matrix multiplication instead of element-by-element multiplication. A 5 x 5 array matrix-multiply a 5 x 1 vector would give a 5 x 1 result, which is exactly what you need.
Note that if you do mean matrix multiplication like that, then by the rules of matrix multiplication, A1*x + A2*x = (A2+A2)*x and so you might as well calculate the sum of your A1 to A16 and matrix-multiply the result by x.

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