Contraction maps with $3\times 3$ matrices, choice for $x$? $f:\Bbb R^2\to \Bbb R^2$ given by  $$f(x)=\begin{bmatrix}\frac12&0\\0&\frac13\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$
We can determine if this is a contraction map by showing that $\|Ax\|\leq k \lt 1$ when $\|x\|=1$
Therefore we can take $x=(\cos t,\sin t)$
and hence $$Ax=\begin{pmatrix}\frac12 \cos t\\\frac13\sin t\end{pmatrix}$$
$$\|Ax\|^2=\frac14\cos^2t+\frac19\sin^2 t\leq \frac14$$
And hence $\|Ax\|\leq\frac12$, and thus this is a contraction.

What about in the case where we have a $3\times 3$ matrix?
$$f(x)=\begin{bmatrix}\frac12&0&\frac13\\0&\frac14&0\\-\frac12&0&\frac13\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$
$f:\Bbb R^3\to \Bbb R^3$, can we still just test for $\|x\|=1$ and what do we choose for $x$? We need three terms I guess?
 A: For your $3 \times 3$ problem: note that this is really just the $2 \times 2$ problem in disguise.  In particular, we note that
$$
\left\|\pmatrix{
\frac12&0&\frac13\\
0&\frac14&0\\
-\frac12&0&\frac13
}
\pmatrix{x_1\\x_2\\x_3}\right\|^2 = 
\left\|\pmatrix{
\frac12&\frac13\\
-\frac12&\frac13
}\pmatrix{x_1\\x_3}\right\|^2 + 
(x_2/4)^2
$$
From there, the technique you used last time applies.
I suspect that if they give you a $3 \times 3$ question on the exam, you can break it down like this.

Complete solution:
We see that
$$
\left\|\pmatrix{
\frac12&\frac13\\
-\frac12&\frac13
}\pmatrix{x_1\\x_3}\right\|^2 = 
\left\|
\pmatrix{(1/2) x_1 + (1/3)x_3\\ 
(-1/2)x_1 + (1/3)x_3}
\right\|^2 =\\
[(1/2) x_1 + (1/3)x_3]^2 + 
[(-1/2)x_1 + (1/3)x_3]^2 =\\
[x_1^2/4 + x_1x_3/3 + x_3^2/9] + 
[x_1^2/4 - x_1x_3/3 + x_3^2/9] =\\
(1/2)x_1^2 + (2/9)x_3^2 \leq\\
(1/2)x_1^2 + (1/2)x_3^2=\\
(1/2)\|(x_1,x_3)\|^2
$$
So, we have 
$$
\left\|\pmatrix{
\frac12&0&\frac13\\
0&\frac14&0\\
-\frac12&0&\frac13
}
\pmatrix{x_1\\x_2\\x_3}\right\|^2 = 
\left\|\pmatrix{
\frac12&\frac13\\
-\frac12&\frac13
}\pmatrix{x_1\\x_3}\right\|^2 + 
(x_2/4)^2 \leq\\
(1/2)\|(x_1,x_3)\|^2 + (1/4)x_2^2 \leq\\
(1/2)\|(x_1,x_3)\|^2 + (1/2)x_2^2 =\\
(1/2)(x_1^2 + x_2^2 + x_3^2) =\\
(1/2)\|(x_1,x_2,x_3)\|^2
$$
We conclude that $f$ is a contraction with $k = \frac{1}{\sqrt{2}}$.
A: Yes, you can still use $\|x\|=1$. 
Yes, you will need 3 terms. 
I would start by naively checking the norm of cos x, sin x, cos x
A: More generally, suppose you have a $3\times 3$ matrix $A$ whose column vectors are orthogonal vectors $\mathbf u$, $\mathbf v$, and $\mathbf w$. Then I claim that $\|A\mathbf x\| \le k\|\mathbf x\|$ for $k=\max(\|\mathbf u\|,\|\mathbf v\|,\|\mathbf w\|)$. For note that
$$A\mathbf x = x_1\mathbf u + x_2\mathbf v + x_3\mathbf w$$
and so
\begin{align*}
\|A\mathbf x\|^2 &= \|x_1\mathbf u + x_2\mathbf v + x_3\mathbf w\|^2 \\
&= \big(x_1\mathbf u + x_2\mathbf v + x_3\mathbf w\big)\cdot\big(x_1\mathbf u + x_2\mathbf v + x_3\mathbf w\big) \\
&= \|\mathbf u\|^2x_1^2 + \|\mathbf v\|^2x_2^2 + \|\mathbf w\|^2x_3^2 \\
&\le k^2(x_1^2+x_2^2+x_3^2) = k^2\|\mathbf x\|^2.
\end{align*}
