Question

If \[ A= \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} \] and \[ S=A+A^{2}+A^{3}+...+A^{12} \] then sum of all elements of matrix S is

• A. 104
• B. 96
• C. 102
• D. 81

Hint:

For matrices of form \[ \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} \] remember shortcut: \[ A^n= \begin{bmatrix} 1&n\\ 0&1 \end{bmatrix} \]

Answer 1

The Correct Option is C

Concept: For triangular matrix \[ A= \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} \] powers follow standard pattern.

Step 1: Find general power.
Observe \[ A^2= \begin{bmatrix} 1&2\\ 0&1 \end{bmatrix} \] \[ A^3= \begin{bmatrix} 1&3\\ 0&1 \end{bmatrix} \] Thus \[ A^n= \begin{bmatrix} 1&n\\ 0&1 \end{bmatrix} \]

Step 2: Find summation.
\[ S= \sum_{n=1}^{12} \begin{bmatrix} 1&n\\ 0&1 \end{bmatrix} \] \[ = \begin{bmatrix} 12&\sum n\\ 0&12 \end{bmatrix} \] Now \[ \sum_{n=1}^{12}n=\frac{12(13)}2=78 \] Hence \[ S= \begin{bmatrix} 12&78\\ 0&12 \end{bmatrix} \]

Step 3: Add all entries.
Total sum \[ 12+78+0+12 = 102 \] Thus \[ \boxed{102} \]