Question

If \( A = \begin{bmatrix} 1 & 8 \\ 0 & 1 \end{bmatrix} \), then \( A^8 = \)

• A. \( \begin{bmatrix} 1 & 64 \\ 0 & 1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 1 & 32 \\ 0 & 1 \end{bmatrix} \)
• C. \( \begin{bmatrix} 1 & 16 \\ 0 & 1 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 8 \\ 0 & 1 \end{bmatrix} \)

Hint:

For matrices of this special form, raising them to a power simply involves multiplying the off-diagonal element by the power of the exponent.

Answer 1

The Correct Option is A

For the matrix \( A = \begin{bmatrix} 1 & 8 \\ 0 & 1 \end{bmatrix} \), notice that this is a special form of a matrix. We can calculate the power of the matrix by recognizing the structure of the matrix and multiplying it repeatedly. When we compute \( A^8 \), we get: \[ A^8 = \begin{bmatrix} 1 & 8 \\ 0 & 1 \end{bmatrix}^8 = \begin{bmatrix} 1 & 8 \times 8 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 64 \\ 0 & 1 \end{bmatrix}. \]