Question

Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, then $A^{10}$ is equal to:

• A. $2^8 A$
• B. $2^9 A$
• C. $2^{10} A$
• D. $2^{11} A$

Answer 1

The Correct Option is B

First, compute $A^2$: \[ A^2 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2A \] Notice that: \[ A^n = 2^{n-1} A \quad \text{for } n \geq 1. \] Therefore: \[ A^{10} = 2^{10-1} A = 2^9 A. \] Hence, the correct answer is $2^9 A$.

Answer 2

1. Understand the problem:

Given matrix \( A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \), we need to find \( A^{10} \).

2. Observe the pattern for powers of A:

Compute lower powers to identify a pattern:

\[ A^2 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2A \]

\[ A^3 = A^2 \cdot A = 2A \cdot A = 2A^2 = 4A \]

Thus, \( A^n = 2^{n-1} A \).

3. Compute \( A^{10} \):

Using the pattern:

\[ A^{10} = 2^{10-1} A = 2^9 A \]

Correct Answer: (B) \( 2^9 A \)