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The question is formulated as follows.

Find all solutions in $x$=$\begin{bmatrix}x_1\\x_2\\x_3\\\end{bmatrix}$ $\in R^3$ of the equation system $Ax=12x$

$A$=$\begin{bmatrix}6 & 4 & 3\\6 & 0 & 9\\0 & 8 & 0\\\end{bmatrix}$

and $\sum\limits_{i=1}^3 x_i = 1$.

I have started with finding all of the vectors of $A$ in the nullspace of $Ax=12x$ which means solving for $A-12I=0$ and converting the resultant matrix into a reduced row-echelon form which is given as follows.

$rref(A)$=$\begin{bmatrix}1 & 0 & -3/2\\0 & 1 & 3/2\\0 & 0 & 0\\\end{bmatrix}$

Now this says that $x_1=(3/2) x_3$ and $x_2= -(3/2) x_3$ but this does not satisfy the summation constraint what am i doing wrong here?.

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    $\begingroup$ You have a line of solutions parametrized by $x_3$; take $x_3=1$ to satisfy the summation constraint, though I think you should get $x_2=+\frac32x_3$ $\endgroup$ May 3, 2020 at 14:53
  • $\begingroup$ @J.W.Tanner that would give me (3/2,-3/2,1) which does satisfy the constraint. Thanks $\endgroup$ May 3, 2020 at 14:57
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    $\begingroup$ $(3/2,-3/2,1)$ does satisfy the summation constraint, but as I said in my revised comment I think you made a sign error on one component of rref$(A)$ $\endgroup$ May 3, 2020 at 14:59
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    $\begingroup$ Note: you're looking for eigenvectors with eigenvalue $12$ $\endgroup$ May 3, 2020 at 15:07

4 Answers 4

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It should be $rref(A)$=$\begin{bmatrix}1 & 0 & -3/2\\0 & 1 & \color{red}-3/2\\0 & 0 & 0\\\end{bmatrix}$.

So $x_1=x_2=\frac32x_3$.

To get $x_1+x_2+x_3=\frac32x_3+\frac32x_3+x_3=1$, take $x_3=\frac14$.

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$$1=x_1+x_2+x_3=\dfrac{3}{2}x_3\color{red}+\dfrac{3}{2}x_3+x_3=\color{red}4x_3 \implies x_3=\color{red}{\dfrac{1}{4}}.$$

So the only solution satisfying the constraints is $\Big[\dfrac{3}{2},\dfrac{3}{2},\dfrac{1}{4}\Big].$


$\color{red}{\text{Edited}}$ thanks to @amd and @J.W.Tanner.

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  • $\begingroup$ Exactly, so stupid of me to not think this way $\endgroup$ May 3, 2020 at 14:59
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Ax=12x gives.

$6x_1+4x_2+3x_3=12x_1$ (1)

$6x_1+9x_3=12x_2$ (2)

$8x_2=12x_3$ (3)

Which gives us.

$x_2=1.5x_3$ subbed into (2) gives

$6x_1=6x_2$

$x_1=x_2=1.5x_3$

$x_1+x_2+x_3=1$

$1.5x_3+1.5x_2+x_3=1$

$x_3=1/4$

$x_1=x_2=3/8$

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The constraint on the components of the solution is just another linear equation to include in the system. That is, instead of first solving $Ax=12x$ via row-reduction and then trying to restrict the solution space so that it satisfies the constraint, you can row reduce the matrix $$\begin{bmatrix}1&1&1&1\\-6&4&3&0\\6&-12&9&0\\0&8&-12&0\end{bmatrix} \to \begin{bmatrix}1&0&0&\frac38 \\ 0&1&0&\frac38 \\ 0&0&1&\frac14 \\ 0&0&0&0\end{bmatrix},$$ which gives you the solution directly.

Continuing with your method from where you left off, on the other hand, since the solution space is one-dimensional (we expect an infinite number of solutions since this is really an eigenvalue problem in disguise), take any of the solutions and divide it by the sum of its elements. So, take your general solution to $Ax=12x$ (with sign error corrected), namely $\left(\frac32x_3,\frac32x_3,x_3\right)$ and divide it by $\frac32x_3+\frac32x_3+x_3=4x_3$ to get $\left(\frac38,\frac38,\frac14\right)$. We can of course assume that $x_3\ne0$ since there’s no way to scale the zero vector so that its elements sum to $1$.

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