could anyone please help me solve the matrices with respect to the equation as follows:

I am having the following matrices

U=[15;
14;
13]
V=[2;
3;
4]

i need to solve the matrices with respect to the equation as follows:

where i=3,

Could anyone please help me on this.

[KSSV]

What are the dimensions of your U, V?

Answer 1

[Alex Mcaulley]

For i == 3:

X3 = V(3)*U(2)-U(2)*sum(V(1:2))

[jaah navi]

the dimensions of U and V are (3,1).

[jaah navi]

i need to have the result in the following manner

when i=3,

[ 2;
(3-2)*15;
4*(15)-(2+3)*15]

which is equal to

[2;
15;
-15]

Could you please help me on this.

[Alex Mcaulley]

Then, you want the values for all i == {1,2,3}. The question is: why your result for i == 1 is 2? U(0) is not defined. Are you imposing that U(0) == 1?

[jaah navi]

i is not equal to 1.

i can be equal to 2,3 and so on.

[Alex Mcaulley]

Then, I don't understand your result. With your data you only can compute the values for i == 2 and i == 3. To compute the value for i == 4 you need V(4). Then, how do you obtain 3 numbers in your solution?

i == 2 -> U(1)*(V(2)-V(1)) = 15*(3-2)
i == 3 -> U(2)*(V(3)-V(1)-V(2)) = 14*(4-3-2)

[jaah navi]

the first value remain itself only for second and third value the loop needs to be performed

[Alex Mcaulley]

ok, then using a for loop:

U=[15;
14;
13];
V=[2;
3;
4];

sol = V;
for a = 2:numel(V)
    sol(a) = U(a-1)*(V(a)-sum(V(1:a-1)));
end
%Where sol is:
sol =
     2
    15
   -14

[jaah navi]

The solution should be [2;

(3-2)*15;

4*15-(2+3)*15]

which will be[2;

15;

-15]

but the value is -14 instead of 15.

[Alex Mcaulley]

With your data the solution for i == 3 is -14:

i == 3 -> U(2)*(V(3)-V(1)-V(2)) = 14*(4-3-2) = -14

[jaah navi]

But i want to have the third row value to be -15.

so the command needs to perform in such a way that the third value needs to be (4*15-(2+3)*15)=-15.

[Alex Mcaulley]

It doesn't follow the equation you shared. Anyway, a possible trick:

sol = V;
for a = 2:numel(V)
    sol(a) = U(1)*(V(a)-sum(V(1:a-1)));
end