Question

If \( A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \), then \( A^n + nI \) is equal to:

• A. \( I \)
• B. \( nA \)
• C. \( I + nA \)
• D. \( I - nA \)
• E. \( nA - I \)

Hint:

For matrices of the form \( \begin{pmatrix} 1 & 0 k & 1 \end{pmatrix} \), the \( n \)-th power is simply \( \begin{pmatrix} 1 & 0 nk & 1 \end{pmatrix} \). This is a very common shortcut in entrance exams.

Answer 1

The Correct Option is C

Concept: For a specific type of matrix where the diagonal entries are 1 and there is a single non-zero entry off-diagonal, the power \( A^n \) follows a predictable linear pattern.

Step 1: Finding a pattern for \( A^n \).
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 2 & 1 \end{pmatrix} \] \[ A^3 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \] By induction, \( A^n = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \).

Step 2: Evaluating the expression \( A^n + nI \).
\[ A^n + nI = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} + \begin{pmatrix} n & 0 0 & n \end{pmatrix} = \begin{pmatrix} 1+n & 0 \\ n & 1+n \end{pmatrix} \] Now, let's check Option (C): \[ I + nA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} n & 0 \\ n & n \end{pmatrix} = \begin{pmatrix} 1+n & 0 \\ n & 1+n \end{pmatrix} \] The results match.