If A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}, then A^{20} + A^{19}(A-I) + A^{18}(A-I)^2 + \dots + (A-I)^{20} is equal to
Step-by-Step Solution
Key Concept: The expression is a geometric series of the form (x^n - y^n)/(x - y) where x = A and y = A-I. Since x - y = I, the sum simplifies to A^21 - (A-I)^21.
Let S = A^{20} + A^{19}(A-I) + A^{18}(A-I)^2 + \dots + (A-I)^{20}. This is a geometric series with first term a = A^{20}, common ratio r = (A-I)/A, and number of terms n = 21. Alternatively, note that this is of the form \frac{x^n - y^n}{x - y} where x = A and y = A-I. Since x - y = A - (A-I) = I, the sum is \frac{A^{21} - (A-I)^{21}}{A - (A-I)} = \frac{A^{21} - (A-I)^{21}}{I} = A^{21} - (A-I)^{21}.
Correct Answer: 3