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: \([ A^8 = \begin{bmatrix} 1 & 8  0 & 1 \end{bmatrix}^8\)
= \(\begin{bmatrix} 1 & 8 \times 8  0 & 1 \end{bmatrix}\)
=\( \begin{bmatrix} 1 & 64  0 & 1 \end{bmatrix}\)