I'm trying to teach myself linear algebra for and one of the exercises of the book I'm reading is the following:
Find all solutions in $\vec x= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} ∈\mathbb R^3$ of the equation system $A\vec x= 12\vec x$, where $A= \begin{bmatrix} 6 & 4 & 3\\6 & 0 & 9\\0 & 8 & 0\end{bmatrix}$ and $\sum_{i=1}^3 x_i= 1$.
This is exercise 2.6 from here. I've realized that that last condition, $\sum_{i=1}^3 x_i= 1$, is another equation I can work with. I tried doing Gauss-Jordan with the following matrix:
$(A|b)= \left[\begin{array}{ccc|c} 6 & 4 & 3 &12x_1\\6 & 0 & 9 & 12x_2\\0 & 8 & 0 & 12x_3\\1&1&1&1\end{array}\right]$
and reached
$\left[\begin{array}{ccc|c} 1&0&0&5-4x_1-5x_2-\frac 92x_3\\ 0&1&0&\frac 32 x_3\\ 0&0&1&4x_2-x+3x_3\\ 0&0&0&4x_1-4+3x_3+5x_2 \end{array}\right]$
Now I don't know what to do, nor do I know if the approach I took was correct. How should I continue? And if I should not have done that, or if there's an easier way, what should I start with?
Your system is equivalent to $(A-12I)\vec{x} = 0$. If you solve this system you get $x_1 = t , x_2=t,x_3 = \frac{2}{3}t$ with $t \in R$. Now using the condition $x_1+x_2+x_3 =1$, which mean that $t+t+\frac{2}{3}t = \frac{8}{3}t = 1$, so $t=\frac{3}{8}$ and so $x_1 = \frac{3}{8}$, $x_2 = \frac{3}{8}$ and $x_3 = \frac{1}{4}$