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Consider the following matrix $$ A= \begin{bmatrix} -(k_1 + k_6)& k_2 & k_5 \\ k_1 & -(k_2+k_3) & k_4 \\ k_6 & k_3 & -(k_4+k_5) \end{bmatrix} $$

where all the parameters $k_j$, $j\in\{1, ..., 6\}$ are positive real numbers. Now if we want to solve the matrix-vector equation $Ax=0$ for $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$, how do we do so? I know that $(x_1,x_2,x_3)=(0,0,0)$ solves it but what other solutions are there and how do we get them?

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I'll write this linear systems of equations as augmented coefficient matrix: \begin{align} \left(\begin{array}{c c c | c} -(k_1 + k_6)& k_2 & k_5 & x_1\\ k_1 & -(k_2+k_3) & k_4 & x_2 \\ k_6 & k_3 & -(k_4+k_5) & x_3 \end{array}\right) & = \left(\begin{array}{c c c | c} k_1 & -(k_2+k_3) & k_4 & x_2 \\ k_6 & k_3 & -(k_4+k_5) & x_3 \\ -(k_1 + k_6)& k_2 & k_5 & x_1\\ \end{array}\right) \end{align} Now, by adding rows I and II onto the third column you get \begin{align} \left(\begin{array}{c c c | c} k_1 & -(k_2+k_3) & k_4 & x_2 \\ k_6 & k_3 & -(k_4+k_5) & x_3 \\ 0 & 0 & 0 & x_1 + x_2 + x_3\\ \end{array}\right) \\ \end{align} Here you can see that the matrix doesn't have full rank (either 1 or 2) and is therefore not invertible, as @Bernard pointed out. You can also see, that $x_1 + x_2 + x_3$ will have to be zero for this system to have a solution.

Now, subtract $\frac{k_6}{k_1}\cdot$I from the second to obtain \begin{align} \left(\begin{array}{c c c | c} k_1 & -(k_2+k_3) & k_4 & x_2 \\ 0 & k_2 + \frac{k_6}{k_1}(k_2 + k_3) & -(k_5 + \frac{k_6}{k_1} k_4) & x_3 + \frac{x_2(k_1 - k_6)}{k_6} \\ 0 & 0 & 0 & x_1 + x_2 + x_3 \\ \end{array}\right) \\ \end{align} Can you take it from here? To check your solutions, see here.

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