Question

If \( A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \), then the value of \( A^{2017} \) is equal to:

• A. \( 2^{2015} A \)
• B. \( 2^{2016} A \)
• C. \( 2^{2014} A \)
• D. \( 2^{2017} A \)
• E. \( 2^{2020} A \)

Hint:

For any matrix \( A \) where all elements are \( 1 \) and the order is \( n \), the identity \( A^k = n^{k-1}A \) holds true. This is a very frequent pattern in linear algebra assessments.

Answer 1

The Correct Option is B

Concept: For a matrix where every entry is \( 1 \), higher powers follow a geometric progression based on the size of the matrix. Specifically, if \( A \) is a \( n \times n \) matrix of ones, then \( A^2 = nA \), and by induction, \( A^k = n^{k-1}A \). This occurs because the dot product of any row and column will always result in the value \( n \).

Step 1: Establishing the pattern through direct multiplication.
Let's calculate the first few powers of \( A \) to identify the growth factor: \[ A^2 = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = 2A \] Now, calculate \( A^3 \): \[ A^3 = A^2 \cdot A = (2A) \cdot A = 2A^2 = 2(2A) = 4A = 2^2 A \] Similarly, calculate \( A^4 \): \[ A^4 = A^3 \cdot A = (2^2 A) \cdot A = 2^2 A^2 = 2^2(2A) = 2^3 A \]

Step 2: Generalizing to the power of 2017.
From the observed pattern, the power of \( 2 \) is always one less than the power of the matrix \( A \): \[ A^n = 2^{n-1} A \] Substituting \( n = 2017 \): \[ A^{2017} = 2^{2017-1} A = 2^{2016} A \]