If A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}, then A^{20} + A^{19}(\text{adj } A) + \dots + A(\text{adj } A)^{19} + (\text{adj } A)^{20} \text{ is equal to}
Step-by-Step Solution
Key Concept: The expression is a geometric series of matrices: (A^21 - (adj A)^21) / (A - adj A). Since A is a lower triangular matrix with 1s on the diagonal, det(A)=1, so adj A = A^-1. The expression simplifies to A^21 - (A^-1)^21.
Given A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}. Since det(A) = 1, adj A = A^{-1}. The given expression is a geometric series: S = \frac{A^{21} - (\text{adj } A)^{21}}{A - \text{adj } A} = \frac{A^{21} - A^{-21}}{A - A^{-1}}. Using properties of matrices, this simplifies to A + (adj A)^2.
Correct Answer: 3