how to get the vector solution with a system of 3 variables and 4 equations

Question

I'm doing an exercise of solid mechanics in which I got these 4 equations

Own vector associated with eI = 7,391 (i) System of equations:

$1.39116499156263*nx + 0.5*ny + 3.0*nz = 0$

$0.5*nx - 7.39116499156263*ny + 1.5*nz = 0$

$3.0*nx + 1.5*ny - 7.39116499156263*nz = 0$

$nx^2 + ny^2 + nz^2 - 1 = 0$

The solutions is $ \vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3.$

I put the system in a Linear Equation Systems Calculator and I got system impossible (and I agree with the solution)

However I can't understand why the vector solution is $\vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3$ ... Could someone explain it to me? Shouldn't the solution be the vector $(0,0,0)$?

Answer 1

The first three homogeneous equations have a non trivial solution (i.e. not all zeros) if the determinant of its coefficients is zero.

If the first equation has the first coefficient $-1.39116499156263$ then the matrix of coefficients has determinant almost zero $$\left| \begin{array}{rrr} -1.39116 & 0.5 & 3 \\ 0.5 & -7.39116 & 1.5 \\ 3 & 1.5 & -7.39116 \\ \end{array} \right|\approx 0 $$ Matrix rank is $2$ so we can solve the system formed by the first two equations $$ \begin{cases} 0.5ny + 3.0nz &= 1.39116499156263nx \\ - 7.39116499156263ny + 1.5nz & = -0.5nx\\ \end{cases} $$ we get the non trivial solutions: $$ny= 0.156466nx,nz= 0.437644nx$$ Plugging in the third equation we get $$nx^2+(0.156466nx)^2+(0.437644nx)^2=1$$ therefore the results are (I suppose that $nx>0$) $$nx= 0.907;\;ny=0.142;\; nz= 0.397$$

Answer 2

As in comments and answers, admitting a sign error in the first equation, using whole numbers

$$M=\left( \begin{array}{ccc} -\frac{139116499156263}{100000000000000} & \frac{1}{2} & 3 \\ \frac{1}{2} & -\frac{739116499156263}{100000000000000} & \frac{3}{2} \\ 3 & \frac{3}{2} & -\frac{739116499156263}{100000000000000} \end{array} \right)$$

$$|M|=259204707122654630160513716553\times 10^{-42}$$

Using the second and third equations $$y=\frac{239116499156263 }{1528232998312526}x \qquad z=\frac{100323299831252323990892125169 }{229234949746878900000000000000}x$$ Using the fourth equation $$x^2=\frac{52548662185454094659180291265210000000000000000000000000000}{638999024282822 57528322078121921616822143238895937163278561}$$

All of that gives in decimals $$x=0.9068401383\qquad y=0.1418896460 \qquad z=0.3968731434 $$