I am trying to prove that the linear system $x = Ax + b$ has a unique solution using the contraction mapping principle, where $$
A =
\begin{pmatrix}
 \frac{1}{4} & -\frac{1}{4} & \frac{2}{15} \\
 \frac{1}{4} & \frac{1}{5} & \frac{1}{2} \\
 -\frac{1}{4} & \frac{1}{3} & -\frac{1}{3}
\end{pmatrix}
~~\text{ and }~~
b = \begin{pmatrix}
 3 \\
 -1 \\
 2 
\end{pmatrix}.
$$

Toward this end, I've defined the map $f(x) = Ax + b$ with the next logical step being to prove that $\forall x,y \in R^3 \times R^3$ that $$
\|f(x) - f(y)\| = \|A(x - y)\| \le \|A\|\|x-y\|
$$
and $\|A\| < 1$. But I'm not sure how to prove that $\|A\| < 1$.

One thing I notice here is that the L2-norms of the rows of $A$ are all less than one, and the L2-norms of the columns of $A$ are all less than one. Is this sufficient to prove the claim?