The problem is to use Banach's fixed point theorem to prove that $\vec{x} = A\vec{x} + \vec{b}$ has a unique solution, where $b = [3, -1, 2]^T$ and $$ A = \begin{bmatrix} 1/4 & -1/4 & 2/15 \\ 1/4 & 1/5 & 1/2 \\ -1/4 & 1/3 & -1/3 \end{bmatrix} $$
My attempt: Arrows are abbreviated. Let $T({x}) = A{x} + {b}$. Then for ${x}$, ${y}$, ${v} = {x} - {y}$, we have
$$ \begin{align*}
||T(x)-T(y)|| &= ||A(x-y)|| \quad \text{(By Linearity)} \\
&= ||Av||
\end{align*} $$
Choose the norm $||v||$ as the sum of the absolute values of each components. Then, we have
$$ \begin{align*} ||Av|| &= \left| \frac{1}{4}v_1 - \frac{1}{4}v_2 + \frac{2}{15}v_3 \right| + \left| \frac{1}{4}v_1 + \frac{1}{5}v_2 + \frac{1}{3}v_3 \right| + \left| -\frac{1}{4}v_1 + \frac{1}{3}v_2 - \frac{1}{3}v_3 \right| \\ & \leq \frac{3}{4}|v_1| + \frac{47}{60}|v_2| + \frac{29}{30}|v_3| \quad \text{(By triangular inequality)}\\ & \leq \frac{29}{30}||v||\end{align*}
$$
Thus, by Banach's fixed point theorem, $T$ has a unique fixed point theorem, which is the solution to the given equation.
I am quite new to this 'generalized norm' and 'fixed point' concepts, so I'm not sure if what I've done is right. Is this solve valid?