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

Step 1: Look at the matrix.
We have $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, an upper triangular matrix with ones on the diagonal. Such matrices have a very clean power pattern.
Step 2: Compute a couple of powers.
$A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $A^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}$. The top right entry just counts the power.
Step 3: Write the general power.
By this pattern $A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}$ for every positive integer $n$.
Step 4: Add the powers up to 12.
\[ S=\sum_{n=1}^{12}\begin{bmatrix}1&n\\0&1\end{bmatrix}=\begin{bmatrix}12&\sum_{n=1}^{12}n\\0&12\end{bmatrix} \]
Step 5: Evaluate the sum of the top right entries.
$\sum_{n=1}^{12}n=\dfrac{12\cdot 13}{2}=78$, so $S=\begin{bmatrix}12&78\\0&12\end{bmatrix}$.
Step 6: Add every element of S.
The total is $12+78+0+12=102$, which is option (3).
\[ \boxed{102} \]