A system of $4$ linear equations in $3$ unknowns

Question

I am a having hard times solving this. Any help will be appreciated.

$0.4a +0.8b+0.1c=a$,
$0.3a+0.2b+0.3c=b$,
$0.3a+0.6c=c$,

and we have 1 more info

$a+b+c=1$

Answer 1

Your first three equations can be written as a matrix equation.

Let $$ A = \begin{bmatrix} 0.4 & 0.8 & 0.1 \\ 0.3 & 0.2 & 0.3 \\ 0.3 & 0 & 0.6 \end{bmatrix}, \mathbf{v} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}. $$

You want to find a $\mathbf{v}$ such that $A\mathbf{v} = \mathbf{v}$. This should remind you of eigenvalues and eigenvectors, where in your case, the eigenvector you're looking for has a corresponding eigenvalue $1$.

Once you have found a $\mathbf{v}$ that solves this equation, you have to deal with your additional constraint, being that $a + b + c = 1$. This constraint shouldn't be hard to deal with, since for any eigenvector $\mathbf{u}$, the vector $c\mathbf{u}$, where $c \in \mathbb{R}$, is also an eigenvector.

If you have limited knowledge of eigenvalues and eigenvectors, there is tons of material on this topic, e.g. https://math.mit.edu/~gs/linearalgebra/ila0601.pdf.