Matrix dimensions must agree.

Hi, i still new to matlab.

I've got an error that says "Matrix dimensions must agree."

The error is a line 4, "y = (0.943*(x-9).^2)-(24.99*t+6113.91); "

what do i need to do to correct the error?

x   =   9:0.1:15;
N   =   length(x)-1;
y   =   (0.943*(x-9).^2)-(24.99*t+6113.91);   



%Spline Equations
%The equations are substituted into the code.

%x1 (-(1.37)*t.^2)+6000;                              0<=t<=9
%x2 (0.943*(x-9).^2)-(24.99*t+6113.91);               9<=t<=15
%x3 (-(0.6054)*(x-15).^2)-((13.34*t)+5975.1);         15<=t<=39
%x4 (-(6.1557)*(x-39).^2)-((960.74*t)+7254.97);       39<=t<=49
%x5 (5.622*(x-49).^2)-((183.8*t)+12673.95);           49<=t<=57
%x6 (2.4008*(x-57).^2)-((93.89*t)+7903.9);            57<=t<=69
%x7 (-(12.91)*(x-69).^2)-((36.27*t)+4273.63);         69<=t<=74
%x8 (3.091*(x-74).^2)-((92.92*t)+8324.598);           74<=t<=81
%x9 (0.2117*(x-81).^2)-((49.64*t)+4969.84);           81<=t<=96
%x10 (-(5.0625)*(x-96).^2)-((43.28*t)+4407.744);      96<=t<=100

% The unknowns are 3*N with ao=0 "Linear Spline"
% Filling Matrix from point matching
V   =   [0;zeros(2*N,1);zeros(N-1,1)];
Z   =   zeros(length(V),length(V));
j=1;
f=1;
for i=2:2:2*N    
    Z(i,f:f+2)          =   [x(j)^2 x(j) 1];
    V(i)                =   y(j);
    j                   =   j+1;
    Z(i+1,f:f+2)        =   [x(j)^2 x(j) 1];  
    V(i+1)              =   y(j);
    f                   =   f+3;
end
% Filling Matrix from smoothing condition
j=1;
l=2;
for i=2*N+2:3*N
    
    Z(i,j:j+1)            =   [2*x(l) 1];
    Z(i,j+3:j+4)          =   [-2*x(l) -1];
    j                     =   j+3;
    l                     =   l+1;
end
% Adjusting the value of a1 to be zero "Linear Spline"
Z(1,1)=1;
% Inverting and obtaining the coeffiecients, Plotting
Coeff       =       Z\V;
j=1;
hold on;
for i=1:N
    curve=@(l) Coeff(j)*l.^2+Coeff(j+1)*l+Coeff(j+2);
    ezplot(curve,[x(i),x(i+1)]);
    hndl=get(gca,'Children');
    set(hndl,'LineWidth',2);
    hold on
    j=j+3;
end
scatter(x,y,50,'r','filled')
grid on;
xlim([min(x)-2 max(x)+2]);
ylim([min(y)-2 max(y)+2]);
xlabel('x');
ylabel('y');
title('Quadratic Spline')

[Walter Roberson]

We do not know the definition of t, but whatever it is is not the same size as x which is 1 x 61.

Are you trying to create a matrix of results, one output for each (x(J),t(K)) pair ?

Answer 1

[John D'Errico]

Look at line 4.

y   =   (0.943*(x-9).^2)-(24.99*t+6113.91); 

We know what x is, that is, a vector.

x   =   9:0.1:15;

But what is t? Is it possible that t is something else, already defined by you?

Now go back and read the error message. Then read the line of code you wrote.

[Nevin Pang]

my error was writing the wrong alphabet t instead of x which the code uses.

thank you for your help!!!

[Nevin Pang]

after running the code, i'd like to produce a graph with multiple results from my code but im unsure how to use the hold on function (ie; where to put within the code)

(mulitple results as in, i'm using the limits from 0-9, 9-15 etc as listed in the above code as t and the equations x1 thru 10.

figure (1)
title('Quadratic Spline')
xlim([min(t)-2 max(t)+2]);
ylim([min(y)-2 max(y)+2]);
xlabel('x');
ylabel('y');
grid on 
grid minor
hold on

i placed a hold on at the end but i am not able to collate the results on one graph.

any help would be appreciated!

thank you!!

[Walter Roberson]

If you have the symbolic toolbox you could use piecewise()