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- 03:58codecov[bot] commented #17753
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- Oct 22 22:05oscarbenjamin commented #17696
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the example in maple is shown above

just to make clear, the float points in my PI matrix are not exactly 3 digits precision, I'm just showing the first 3 most significant digits using:

`np.set_printoptions(precision=3)`

for sake of this test it was created using the ratios, as in

```
PI = np.array([
[-11/12, 1/24, 0, 0, -1],
[ 17/24, -9/8, 1/24, 0 ,5],
[3/8, 9/8, -9/8, 1/24, -10],
[-5/24, -1/24, 9/8, -9/8, 10],
[1/24, 0, -1/24, 9/8, -5],
[0, 0, 0, -1/24, 1]
])
```

Ideas, are very likely welcome.

It seems that your system is over-determined. It has 6 equations in 5 variables. It can have solutions only if the determinant of the extended system is 0.

```
>>> P = Matrix([[ -0.917, 0.042, 0. , 0. , -1. ],
... [ 0.708, -1.125, 0.042, 0. , 5. ],
... [ 0.375, 1.125, -1.125, 0.042, -10. ],
... [ -0.208, -0.042, 1.125, -1.125, 10. ],
... [ 0.042, 0. , -0.042, 1.125, -5. ],
... [ 0. , 0. , 0. , -0.042, 1. ]])
>>> c = Matrix([-1. , 0. , 0. , -0.042, 1.083, -0.042])
>>> Pc = P.row_join(c)
>>> Pc.det()
0.000998125750125132
```

In this example, the determinant not 0 because of the inaccuracy of floating point numbers. Hence there is no solution.

i am learning python megacourse on udemy and it is also teaching numpy, so i was wondering if it might be useful

A potential solution could be to define

`rationalize = lambda x: [nsimplify(e) for e in x]`

(this will turn numbers from a list into rationals). And then do

```
PI = Matrix(map(rationalize, [
[-11/12, 1/24, 0, 0, -1],
[ 17/24, -9/8, 1/24, 0 ,5],
[3/8, 9/8, -9/8, 1/24, -10],
[-5/24, -1/24, 9/8, -9/8, 10],
[1/24, 0, -1/24, 9/8, -5],
[0, 0, 0, -1/24, 1]
]))
```

and similarly for `c`

. I've just checked and this returns the right solutions

@valglad Amazing! I will test.

There are 6 rows and only 5 columns. One row could be omitted.

I have created

`mrationalize = lambda x: [nsimplify(e) for e in [r for r in x]]`

that I do not like much the fell.
I have a problem with calculating eigenvectors for a simple 3x3 matrix in sympy. Trying to execute

```
q = symbols("q", positive=True)
m = Matrix([[-2, exp(-q), 1], [exp(q), -2, 1], [1, 1, -2]])
#m.eigenvects(simplify=True)
m.eigenvals()
```

results in very complicated expressions for the eigenvalues, trying to get the eigenvectors fails with a NotImplementedError. I expect the latter to be due to the former. However, when trying to do the same thing in Mathematica, I get much simpler expressions for everything. Is there some option / flag that I can set to have sympy compute these eigenvalues and eigenvectors? Is there currently some form of limitation within sympy that prevents me from doing this?

@adoa does Mathematica give a simpler expression for the eigenvalues?

Oh apparently they can be simplified if you call

`simplify`

@adoa it seems there is a

`simplify`

flag, but it doesn't work
This may help you

You should use whichever you feel more comfortable with

(sorry for these silly questions... I am trying my best ,i just need some guidance )

I guess it's because

`Add`

does the smart substitution but `Symbol`

does not
The cosine is irrelevant. It's the same with

`x.subs(x - 1, y)`

vs. `(x + 1).subs(x - 1, y)`