solving 4 equations with 4 unknowns
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why do matlab gives two values for each unknown in the following system?
syms a b d e
eqn1 = a+b-4 == 0;
eqn2 = d+e-5 == 0;
eqn3 = a + b/2 + d/2 - 6.5/1.3 == 0;
eqn4 = d*b - exp(1.51)*e*a == 0;
[a,b,d,e]=solve(eqn1,eqn2,eqn3,eqn4)
采纳的回答
Because there are two solutions.
For example, how many solutions would you expect to see for the problem
syms x
xsol = solve(x^2 + x + 2,x)
Two, of course, this is just a quadratic polynomial. Would you have expected only one solution? If then, which one?
And, while you may not think of your problem as being implicitly a quadratic in a sense it is one. That is, look at the products of variables in eqn4. If you eliminate some of the variables using substitution, at the end, you would find one quadratic polynomial in one unknown. Ergo, two solutions. And once you then have two solutions for that final variable, substituting back in to recover the other variables yields EXACTLY two solutions for each of them.
更多回答（1 个）
William Rose
2023-3-7
编辑：William Rose
2023-3-7
0 个投票
@Hamza Harbi, because the equation has two solutions. The first values of a,b,d,e are one solution; the second values of a,b,d,e are another.
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