Solving System of 3 equations non trivial

Px = 8.08*10^5
Py = 4.63*10^5
P_A = 9.54*10^4
pCr = .89*10^5
m = [0 (pCr-Px) (pCr*.076);(pCr-Py) 0 0; 0 (.076*pCr) (.0089*(pCr-P_A))]
B = [0; 0; 0];
for pCr = .88*10^5:.1:.9*10^5
    m = [0 (pCr-Px) (pCr*.076);(pCr-Py) 0 0; 0 (.076*pCr) (.0089*(pCr-P_A))]
    X = linsolve(m,B)
    if X ~= [0;0;0]
        PCrit = pcr;
        break;
    end 
end

I already know one critical values and my current proccess odviously does not work by testing a lot of Pcr values until I do not have a trivial I think there is a way to do it through Linear Algebra but I am brain farting right now.

Answer 1

[Walter Roberson]

Px = 8.08*10^5

Px = 808000

Py = 4.63*10^5

Py = 463000

P_A = 9.54*10^4

P_A = 9.5400e+04

syms pCr;
m = [0 (pCr-Px) (pCr*.076);(pCr-Py) 0 0; 0 (.076*pCr) (.0089*(pCr-P_A))]

m = 

%are there general X such that m*x = [0;0;0] ? 
%if so that would be the null space
null(m)

 
ans =
 
Empty sym: 3-by-0
 

%nope, it is generally empty.

%what happens if we substitute in a particular value for pCr

m123 = subs(m, pCr, 123);
rank(m123)

ans = 3

null(m123)

 
ans =
 
Empty sym: 3-by-0
 

%nope, full rank, no non-trival null space.

%but are there specific values of pCr that lead to lower rank?

pCrit = solve(det(m))

pCrit = 

vpa(pCrit)

ans = 

for K = 1 : size(pCrit,1)
    disp('value being substituted for pCr:')
    disp(pCrit(K))
    disp('substituted matrix')
    M = subs(m, pCr, pCrit(K))
    disp('rank of substituted')
    rank(M)
    disp('null space of substituted')
    null(M)
end

value being substituted for pCr:

463000

substituted matrix

M = 

rank of substituted

ans = 2

null space of substituted

ans = 

value being substituted for pCr:

substituted matrix

M = 

rank of substituted

ans = 2

null space of substituted

ans = 

value being substituted for pCr:

substituted matrix

M = 

rank of substituted

ans = 2

null space of substituted

ans = 

[Jonathan Leutz]

Perfect thank you!!