To solve the nonlinear system of equations

using the problem-based approach, first define x as a two-element optimization variable.

x = optimvar('x',2);

Create the first equation as an optimization equality expression.

eq1 = exp(-exp(-(x(1) + x(2)))) == x(2)*(1 + x(1)^2);

Similarly, create the second equation as an optimization equality expression.

eq2 = x(1)*cos(x(2)) + x(2)*sin(x(1)) == 1/2;

Create an equation problem, and place the equations in the problem.

prob = eqnproblem;
prob.Equations.eq1 = eq1;
prob.Equations.eq2 = eq2;

Review the problem.

show(prob)

  EquationProblem : 

	Solve for:
       x


 eq1:
       exp(-exp(-(x(1) + x(2)))) == (x(2) .* (1 + x(1).^2))

 eq2:
       ((x(1) .* cos(x(2))) + (x(2) .* sin(x(1)))) == 0.5

Solve the problem starting from the point [0,0]. For the problem-based approach, specify the initial point as a structure, with the variable names as the fields of the structure. For this problem, there is only one variable, x.

x0.x = [0 0];
[sol,fval,exitflag] = solve(prob,x0)

Solving problem using fsolve.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

sol = struct with fields:
    x: [2x1 double]

fval = struct with fields:
    eq1: -2.4069e-07
    eq2: -3.8255e-08

exitflag = 
    EquationSolved

View the solution point.

disp(sol.x)

    0.3532
    0.6061

ls1 = fcn2optimexpr(@(x)exp(-exp(-(x(1)+x(2)))),x);
eq1 = ls1 == x(2)*(1 + x(1)^2);
ls2 = fcn2optimexpr(@(x)x(1)*cos(x(2))+x(2)*sin(x(1)),x);
eq2 = ls2 == 1/2;

When x is a 2-by-2 matrix, the equation

is a system of polynomial equations. Here, $$x^3$$ means $$x*x*x$$ using matrix multiplication. You can easily formulate and solve this system using the problem-based approach.

First, define the variable x as a 2-by-2 matrix variable.

x = optimvar('x',2,2);

Define the equation to be solved in terms of x.

eqn = x^3 == [1 2;3 4];

Create an equation problem with this equation.

prob = eqnproblem('Equations',eqn);

Solve the problem starting from the point [1 1;1 1].

x0.x = ones(2);
sol = solve(prob,x0)

Solving problem using fsolve.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

sol = struct with fields:
    x: [2x2 double]

Examine the solution.

disp(sol.x)

   -0.1291    0.8602
    1.2903    1.1612

Display the cube of the solution.

sol.x^3

ans = 2×2

    1.0000    2.0000
    3.0000    4.0000