Q:

Infinity matrix norm example

I have a brief question regarding the infinity matrix norm.  
The subordinate matrix infinity norm is defined as:
$$\|A\|_{\infty} =\max_{1 \leq i \leq n}\sum_{j=1}^{n}|a_{ij}|.$$
This is derived from the general definition of a subordinate matrix norm which is defined as:
$$\|A\| = \max \left\{\frac{\|Ax\|}{\|x\|} : x \in K^{n}, x \neq 0\right\}.$$
I wanted to try this out in an example.  So say we define the matrix:
$$A = \begin{bmatrix} 1 & 4 & 2 \\ 3 & 1 & 2 \\ 4 & 4 & 3 \end{bmatrix}$$
and
$$x = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$
Now if we use the first definition, it is easy to see that $\|A\|_{\infty} = 11$
But if we use the general definition, we get:
$$\|A\|_{\infty} =\max \left\{\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} : x \in K^{n}, x \neq 0\right\}.$$
Now, we have:
$$Ax = \begin{bmatrix} 15 \\ 11 \\ 21 \end{bmatrix}.$$
Since the infinity vector norm is defined as:
$$\|x\|_{\infty} =\max_{1 \leq i \leq n} |x_i|$$
it follows that:
$$\|Ax\|_{\infty} = 21$$
and:
$$\|x\|_{\infty} = 3$$
But then we have:
$$\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} = \frac{21}3 = 7$$
that does not correlate with the fact that we previously found that $\|A\|_{\infty} = 11$.
If anyone can explain to me what is wrong with my reasoning here, I would appreciate it!

A:

The supremum occurs at $$x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$

