Question

If a matrix \( A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \) satisfies \( A^6 = kA' \), then the value of \( k \) is:

• A. 1
• B. \( \frac{1}{32} \)
• C. 6
• D. 32

Hint:

For powers of matrices, you can often use the properties of the matrix to simplify calculations. Here, matrix \( A \) had a repeating structure, making it easier to calculate successive powers.

Answer 1

The Correct Option is D

If a matrix \( A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \) satisfies \( A^6 = kA' \), then the value of \( k \) is:

Step 1: Compute \( A^2 \)
We first compute the square of the matrix \( A \) (i.e., \( A^2 \)): \[ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \] Multiply \( A \) by itself: \[ A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \]

Step 2: Compute \( A^3 \)
Next, we compute the cube of the matrix \( A \), which is \( A^3 \): \[ A^3 = A^2 \times A = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \]

Step 3: Generalize the power of \( A \)
From the calculations above, we see that \( A^n \) for any power \( n \) will always result in a matrix of the form: \[ A^n = 2^{n-1} A \] So, for \( A^6 \), we have: \[ A^6 = 2^5 A = 32A \]

Step 4: Compute the transpose of \( A \)
The transpose of the matrix \( A \), denoted \( A' \), is given by swapping its rows and columns. Since \( A \) is a symmetric matrix: \[ A' = A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]

Step 5: Use the equation \( A^6 = kA' \)
We are given that \( A^6 = kA' \), and since \( A' = A \), we can substitute: \[ 32A = kA \] Canceling \( A \) (since \( A \neq 0 \)): \[ k = 32 \]

Conclusion:
The value of \( k \) is \( \boxed{32} \).