Question

Let \( A = \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 3 & 2 & 1 \end{bmatrix} \). Find \( A^{100} \).

• A. Same as A
• B. Identity matrix
• C. \( \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 300 & 200 & 1 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 100 & 200 & 1 \end{bmatrix} \)

Hint:

If a matrix is of the form $I + B$ where $B^2 = 0$, then $(I+B)^n = I + nB$. Here, $B$ is the bottom row elements, and this rule applies!

Answer 1

The Correct Option is C

1. The given matrix \( A \) is: \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}\)
2. Notice that \( A \) is a lower triangular matrix, which means the diagonal elements (1, 1, and 1) are the main focus for determining the powers of the matrix.
3. To find \( A^{100} \), we need to analyze what happens when we multiply the matrix \( A \) by itself repeatedly. The key observation is: \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ x & y & 1 \end{bmatrix}^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ n \cdot x & n \cdot y & 1 \end{bmatrix}\)
4. This pattern arises because:
   • The multiplication of the upper-left 2x2 identity block with itself results in an identity block.
   • The bottom row, when multiplied with the matrix, adds multiples of the rows above it to its current values. If we expand this multiplication multiple times, the bottom-row elements accumulate by the number of multiplications.
5. Therefore, for \( A^{100} \), the matrix becomes: \(A^{100} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 100 \cdot 3 & 100 \cdot 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 300 & 200 & 1 \end{bmatrix}\)
6. Thus, the correct answer is:

\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 300 & 200 & 1 \end{bmatrix} \).