Question

Let the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 1 \\0 & 1 & 0 \end{pmatrix} $ satisfy $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $. Then the sum of all the elements of $ A^{50} $ is:

• A. 53
• B. 52
• C. 39
• D. 44

Hint:

Use matrix recurrence relations and properties to compute higher powers of matrices efficiently.

Answer 1

The Correct Option is A

Using the recurrence relation \( A^n = A^{n-2} + A^2 - I \) for \( n \geq 3 \), we can compute higher powers of the matrix \( A \). By using matrix algebra, we find that the sum of the elements of \( A^{50} \) is 53.
Thus, the correct answer is \( 53 \).

Answer 2

Step 1: We are given the matrix equation: 

\[ A^n - A^{n-2} = A^2 - I \] and for higher powers, this holds: \[ A^{50} - A^{48} = A^2 - I, \] \[ A^{48} - A^{46} = A^2 - I. \]

Step 2: Now, we calculate \( A^2 \):

\[ A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \]

So, \( A^2 = I \), the identity matrix.

Step 3: Using the result to compute \( A^{50} \):

\[ A^{50} - A^2 = 24(A^2 - I) \] Thus: \[ A^{50} = 25A^2 - 24I. \]

Step 4: Final Matrix Calculation:

Substituting \( A^2 = I \) into the expression: \[ 25A^2 - 24I = 25 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 24 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 25 & 0 \\ 0 & 25 \end{pmatrix} - \begin{pmatrix} 24 & 0 \\ 0 & 24 \end{pmatrix}. \] This simplifies to: \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \]

Step 5: Summing All Elements:

The sum of all the elements is: \[ 1 + 25 + 25 + 1 + 1 + 1 = 53. \]