my first variable (y1) is coming us as undefined

var y1
var t1
var y2
var t2
var y3
var t3
var y4
var t4
var y5
var t5 
var t6 
var y7
var t7 
var y8
var t8 
var y9 
var t9 
var t10


A = [4224.3 7392.5 -4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
     7392.5 17249.14 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    -4224.3 -7392.5 8448.6 0 -4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    7392.5 8624.57 0 34498.28 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 -4224.3 -7392.5 8448.6 0 4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 7392.5 8624.57 0 34498.28 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 -4224.3 -7392.5 447045.3176 158665.3816 -442821.0176 166057.8816 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 7392.5 8624.57 158665.3816 100278.081 -166057.8816 41514.47 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 -442821.0176 -166057.8816 608878.9 0 -442821.0176 166057.8816 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 166057.8816 41514.47 0 166057.822 -166057.8816 41514.47 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 -442821.0176 -166057.8816 453199.6376 -155679.2816  -6919.08 10378.6 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 166057.8816 41514.47 -155679.2816 103786.181 -10378.6 10378.62 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 10387.6 10378.6 0 41514.48 -10378.6 10378.6 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.62 0 41514.48 -10378.6 10378.6 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.62 0 41514.48 -10378.6 10378.6;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 6919.08 -10378.6;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.6 -10378.6 20757.24];

B = [  y1;t1; y2;t2; y3;t3; y4;t4; y5;t5; 0;t6; y7;t7; y8;t8; y9;t9; 0;t10];

X = [150;0; 0;0; 0;0; 0;0; 0;0; 300;0; 0;0; 0;0; 0;0; 150;0];

 solve(X.*A==B);

Answer 1

[Walter Roberson]

var y1

uses "Command/function duality", and means the same thing as

var( 'y1' )

which requestions that the variance be calculated of the vector of values ['y', '1'] . Because there is no semi-colon at the end of the line, the result of the variance calculation would be displayed. No symbolic variable would be defined.

Perhaps you wanted to use syms instead of var

[Kevin Krone]

Now nothing is being displayed

syms y1
syms t1
syms y2
syms t2
syms y3
syms t3
syms y4
syms t4
syms y5
syms t5 
syms t6 
syms y7
syms t7 
syms y8
syms t8 
syms y9 
syms t9 
syms t10


A = [4224.3 7392.5 -4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
     7392.5 17249.14 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    -4224.3 -7392.5 8448.6 0 -4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    7392.5 8624.57 0 34498.28 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 -4224.3 -7392.5 8448.6 0 4224.3 7392.5 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 7392.5 8624.57 0 34498.28 -7392.5 8624.57 0 0 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 -4224.3 -7392.5 447045.3176 158665.3816 -442821.0176 166057.8816 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 7392.5 8624.57 158665.3816 100278.081 -166057.8816 41514.47 0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 -442821.0176 -166057.8816 608878.9 0 -442821.0176 166057.8816 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 166057.8816 41514.47 0 166057.822 -166057.8816 41514.47 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 166057.8816 41514.47 -155679.2816 103786.181 -10378.6 10378.62 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 10387.6 10378.6 0 41514.48 -10378.6 10378.6 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.62 0 41514.48 -10378.6 10378.6 0 0;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 -6919.08 -10378.6 17297.68 0 -6919.08 10378.6;
    0 0 0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.62 0 41514.48 -10378.6 10378.6;

    0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10378.6 10378.6 -10378.6 20757.24];

B = [y1; t1; y2; t2; y3; t3; y4; t4; y5; t5; t6; y7; t7; y8; t8; y9; t9;t10];

X = [150;0; 0;0; 0;0; 0;0; 0;0; 0; 0;0; 0;0; 0;0; 0];

 solve(A.*X==B);

[Walter Roberson]

Your A is 18 * 20. Your X is 18 x 1. In releases up to R2016a, A.*X would be invalid. In R2016b and later, A.*X is treated similar to bsxfun(@times, A, X), which is effectively the same as A.*repmat(X, 1, size(A,2)) . The result will be the same size as A, which is 18*20 . You then want to solve for an 18*20 matrix to equal an 18 x 1 matrix

Typically when people use notations with A, x, and b, they are trying to solve A*x == b with known A and known b, and x having as many rows as A has, and as many columns as b has. Notice this is the * operator, not the .* operator. The * operator is matrix multiplication.

Perhaps what you want is

temp = num2cell( A\X );
[y1, t1, y2, t2, y3, t3, y4, t4, y5, t5, t6, y7, t7, y8, t8, y9, t9, t10] = deal(temp{:});