Let $K_1 = \left\| I_1 \right\|$ and $K_2 = \left\| I_2 \right\|$. Suppose $I_2 = I_1 + \Delta I$. Therefore:
$$\Delta K = K_2 - K_1 = \left\| I_1 + \Delta I\right\| - \left\| I_1 \right\|$$
Is there a way I can calculate $\Delta K$ without having to compute the sum of $I_1 + \Delta I$ ?
Edit:
As an example:
$ I_1 = \begin{pmatrix} -0.4 & 0.2 & 0.1 \\ 0.3 & -0.4 & -0.1 \\ -0.2 & -0.4 & 0.2 \end{pmatrix} $ and $ \Delta I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0.1 & 0 \\ 0 & 0 & 0 \end{pmatrix} $
$K$ is calculated by taking the absolute values of a 2x2 submatrix of $I$. For example, $K_1$ is:
$ K_1 = \begin{pmatrix} 1.3 & 0.8 \\ 1.3 & 1.1 \end{pmatrix} $
For example, the element at row 1, col 1 of $K_1$ is $\left\| -0.4 \right\| + \left\| 0.2 \right\| + \left\| 0.3 \right\| + \left\| -0.4 \right\| = 0.4 + 0.2 + 0.3 + 0.4 = 1.3$.
The question now is how can I find $\Delta K$ given $I_1$ and $\Delta I$ without having to compute the sum of $I_1 + \Delta I$ ?
Yes, you can. When you add $\Delta I$ to $I_1$ the central element changes from $-.4$ to $-.3$, so its absolute value decreases by $.1$. All other elements are unchanged. Because the central element belongs to every $2\times2$ matrix, each element of $\left\|I_1+\Delta I\right\|$ will be $.1$ less than the corresponding element of $\left\|I_1\right\|$. That is $$\Delta K = \begin{pmatrix}-.1&-.1\\-.1&-.1\\\end{pmatrix}$$