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

Step 1: Understanding the Concept
To find high powers of a matrix, look for a pattern by calculating $A^2$ and $A^3$.

Step 2: Key Formula or Approach
Matrix multiplication $A^2 = A \cdot A$.

Step 3: Detailed Explanation
1. Calculate $A^2$: \[ \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 3(1)+3(1) & 2(1)+2(1) & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 6 & 4 & 1 \end{bmatrix} \] 2. Observe the pattern: For $A^n$, the elements $a_{31}$ and $a_{32}$ become $n \times a_{31}$ and $n \times a_{32}$ respectively, while the identity structure remains. 3. For $n = 100$: \[ a_{31} = 100 \times 3 = 300, \quad a_{32} = 100 \times 2 = 200 \]

Step 4: Final Answer
The matrix \( A^{100} \) is \( \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 300 & 200 & 1 \end{bmatrix} \).