how to calculate scalar with matrix

I need to code the following: I got to part b and I am unsure how to get a scalar in the code.

Goal: (b)

%  chapter 2-1
clear; clc; close all

%% normal stress

fprintf('\n ====== Exercise 2.1 a=======\n\n')
    
tau = [-30 -20; -20 -40];   %2D stress tensor (Mpa)

theta = 10;

fhat = [sind(theta) , cosd(theta)];

nhat = [ cosd(theta) , -sind(theta)];

tnhat = tau * nhat.';

tn = nhat * tnhat %normal stress

%% shear stress

ts = fhat * tnhat

fprintf('\n ====== Exercise 2.1 a end=======\n\n')

%%

fprintf('\n ====== Exercise 2.1 b =======\n\n')

I = [1 0; 0 1];

det[-30-x -20; -20 -40-x] = 0

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

fprintf('\n ====== Exercise 2.1 b end=======\n\n')




 

[VBBV]

Use the symbolic toolbox to declare the unknown variable. Then solve the determinant using solve function for the unknown

syms x
D = [-30-x -20; -20 -40-x];
S = det(D) % determinant
sol = solve(S==0,x)

[VBBV]

syms lambda  % define lambda as symbolic variable (eigen value)
tau = [-30 -20; -20 -40];  % shear stress

I = [1 0; 0 1];  % identity matrix
S = det(tau - I*lambda) % determinant of characteristic equation

S = 

sol = solve(S==0,lambda) % solve for eigen values 

sol = 

double(vpa(sol))

ans = 2×1

-55.6155 -14.3845

Answer 1

[Matt J]

Don't see any reason why you'd have to reinvent eigendecomposition:

lambda=eig([-30,-20; -20 -40])

lambda = 2×1

-55.6155 -14.3845