MATLAB: The first variable (y1) is coming us as undefined

matrix manipulationvariables

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);

Best Answer

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
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