**Step 1: Recognize the Infinite Geometric Series** The given expression \(I - A + A^2 - A^3 + ...\) is an infinite geometric series with the first term \(I\) (identity matrix) and common ratio \(-A\).

**Step 2: Apply the Formula for the Sum of an Infinite Geometric Series** The sum of an infinite geometric series \(a + ar + ar^2 + ar^3 + ...\) is given by \(\frac{a}{1 - r}\), provided \(|r| < 1\). In this case, \(a = I\) and \(r = -A\). Therefore, the sum is \((I + A)^{-1}\), assuming the series converges.

**Step 3: Check if the Series Converges** For the series to converge, the eigenvalues of \(A\) must have absolute values less than 1. Let's find the eigenvalues of \(A\). The characteristic equation is given by \(det(A - \lambda I) = 0\), where \(\lambda\) are the eigenvalues. \(det([\begin{matrix}2-\lambda&1\\ -4&-2-\lambda\end{matrix}]) = (2-\lambda)(-2-\lambda) - (1)(-4) = 0\) \(-4 - 2\lambda + 2\lambda + \lambda^2 + 4 = \lambda^2 = 0\) So, \(\lambda = 0\) (repeated eigenvalue). Since the eigenvalue is 0, the condition for convergence is satisfied.

**Step 4: Calculate \(I + A\)** \(I + A = [\begin{matrix}1&0\\ 0&1\end{matrix}] + [\begin{matrix}2&1\\ -4&-2\end{matrix}] = [\begin{matrix}3&1\\ -4&-1\end{matrix}]\)

**Step 5: Calculate the Inverse of \(I + A\)** Let \(B = I + A = [\begin{matrix}3&1\\ -4&-1\end{matrix}]\). The determinant of \(B\) is \(det(B) = (3)(-1) - (1)(-4) = -3 + 4 = 1\). The inverse of \(B\) is given by \(B^{-1} = \frac{1}{det(B)} [\begin{matrix}-1&-1\\ 4&3\end{matrix}] = \frac{1}{1} [\begin{matrix}-1&-1\\ 4&3\end{matrix}] = [\begin{matrix}-1&-1\\ 4&3\end{matrix}]\)

**Step 6: State the Final Answer** Therefore, \(I - A + A^2 - A^3 + ... = (I + A)^{-1} = [\begin{matrix}-1&-1\\ 4&3\end{matrix}]\)