Variable x must be of size [2 1]. It is currently of size [2 2]. Check where the variable is assigned a value.

All lines except the 8th comes up as correct but line 8 comes up as incorrect and states the reason "Variable x must be of size [2 1]. It is currently of size [2 2]. Check where the variable is assigned a value."

How can i change this code to get a [2 1] matrix instead of [2 2] ?

Answer 1

[KSSV]

A = [1 2 4 5 7 ; 9 2 1 4 13] ;
B = [-11 2 ; 3 -1 ; 2 0; 5 11 ; -21 1];
C = A*B ;
D = B*A ;
s = det(C) ; 
t = det(D) ; 

x = inv(C).*[8 -1].^t ; 

[Sithum Nanayakkara]

This code also gives the same 2x2 matrix as the one i got with my code. :(

i want the result to be a 2x1 matrix.

thanks.

[KSSV]

That case t should be a transpose......try

x = inv(C)*[8 -1]'; 

[Sithum Nanayakkara]

Thank You so much !

Making it the transpose worked, the question is pretty confusing beacuse t matrix and the transpose function looks the same.

Thanks again, really grateful.

[Reza Ghasemi]

SO what was the final step u did sithum? n how does ur equation look like?

[Sithum Nanayakkara]

Reza, This is the final answer:

A = [1,2,4,5,7;9,2,1,4,13];
B = [-11,2;3,-1;2,0;5,11;-21,1];
C = A*B;
D = B*A;
s = det(C);
t = det(D);
x = inv(C)*[8 -1]';

[Reza Ghasemi]

thank you so much sithum your a legend

[Reza Ghasemi]

[Reza Ghasemi]

have you done this section? the last part, i was wondering if u could help me with that.

[Sithum Nanayakkara]

Here :

D =  [-3,-18,1,-47,3;2,4,2,11,1;1,4,3,10,8;1,32,4,69,5;7,-40,5,-101,-1];
b = [11;3;7;-17;1];
x = linsolve(D,b) ;

[KSSV]

clc; clear all ;
D = [-3 -18 1 -47 3 ; 
     2 4 2 11 1 ;
     1 4 3 10 8 ;
     1 32 4 69 5 ;
     7 -40 5 -101 -1] ;
b = [11 ; 3 ;7 ;-17 ;1] ;
x = D\b