How can I insert the value of a symbol in symbolic matrix and convert it into a numerical matrix?

syms t

PHI=[ 1, t, t/3 - (2*exp(-3*t))/9 + 2/9, (2*t)/3 + (2*exp(-3*t))/9 - 2/9;

0, 1, (5*exp(-3*t))/12 - (3*exp(t))/4 + 1/3, 2/3 - exp(t)/4 - (5*exp(-3*t))/12;

0, 0, exp(-3*t)/4 + (3*exp(t))/4, exp(t)/4 - exp(-3*t)/4;

0, 0, (3*exp(t))/4 - (3*exp(-3*t))/4, (3*exp(-3*t))/4 + exp(t)/4];

PHIT=transpose (PHI);

B=[0;1;2;1];

BT=transpose (B);

GRAM = PHI*B*BT*PHIT

%When t=1, Need to find the GRAM matrix

Answer 1

[David Hill]

t=1;
subs(GRAM);

[Shauvik Das]

It does not convert it to a numerical matrix.

Answer 2

[Star Strider]

Please see Plotting error: Need to plot the evolution of symbolic matrix? to convert it to a numerical matrix (as an anonymous function).

[Shauvik Das]

Well, here i need to substitute t=1 at the end and get a numerical matrix.

[Star Strider]

Try this:

PHI = @(t) [ 1, t, t/3-(2*exp(-3*t))/9+2/9, (2*t)/3+(2*exp(-3*t))/9-2/9;
    0, 1, (5*exp(-3*t))/12-(3*exp(t))/4+1/3, 2/3-exp(t)/4-(5*exp(-3*t))/12;
    0, 0, exp(-3*t)/4+(3*exp(t))/4, exp(t)/4-exp(-3*t)/4;
    0, 0, (3*exp(t))/4-(3*exp(-3*t))/4, (3*exp(-3*t))/4+exp(t)/4];
N = @(t) norm(PHI(t));

t = 1
Matrix = PHI(t)
Norm = N(t)

producing:

t =
     1

Matrix =
            1            1      0.54449      0.45551
            0            1      -1.6846    -0.033648
            0            0       2.0512      0.66712
            0            0       2.0014      0.71691

Norm =
       3.5131

Different values of ‘t’ will of course return different results.

[Shauvik Das]

It does not help.

Code:

syms t

PHI=[ 1, t, t/3 - (2*exp(-3*t))/9 + 2/9, (2*t)/3 + (2*exp(-3*t))/9 - 2/9;

0, 1, (5*exp(-3*t))/12 - (3*exp(t))/4 + 1/3, 2/3 - exp(t)/4 - (5*exp(-3*t))/12;

0, 0, exp(-3*t)/4 + (3*exp(t))/4, exp(t)/4 - exp(-3*t)/4;

0, 0, (3*exp(t))/4 - (3*exp(-3*t))/4, (3*exp(-3*t))/4 + exp(t)/4];

PHIT=transpose (PHI);

B=[0;1;2;1];

BT=transpose (B);

GRAM = PHI*B*BT*PHIT

%% Now I want replace t=1 for the matrix Gram??

[Star Strider]

Try this:

PHI = @(t) [ 1, t, t/3-(2*exp(-3*t))/9+2/9, (2*t)/3+(2*exp(-3*t))/9-2/9;
    0, 1, (5*exp(-3*t))/12-(3*exp(t))/4+1/3, 2/3-exp(t)/4-(5*exp(-3*t))/12;
    0, 0, exp(-3*t)/4+(3*exp(t))/4, exp(t)/4-exp(-3*t)/4;
    0, 0, (3*exp(t))/4-(3*exp(-3*t))/4, (3*exp(-3*t))/4+exp(t)/4];

PHIT = @(t) transpose(PHI(t));
B=[0;1;2;1];
BT=transpose(B);
GRAM = @(t) PHI(t)*B*BT*PHIT(t);

calling:

t = 1
GRAM(t)

produces:

t =
     1

ans =
       6.4744      -6.1142       12.136       12.009
      -6.1142        5.774      -11.461      -11.341
       12.136      -11.461       22.748        22.51

Make appropriate changes to get different results.