If \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1&3\\0&1\end{bmatrix}\cdots\begin{bmatrix}1&12\\0&1\end{bmatrix} = \begin{bmatrix}1&p\\0&1\end{bmatrix}\), then the value of p is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 78
For any numbers a and b, multiplying row by column:
\( \displaystyle \begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix} = \begin{bmatrix}1&b + a\\0&1\end{bmatrix} \)
So each new factor keeps the same form and adds its upper-right element to the running total. After all twelve factors:
\(p = 1 + 2 + 3 + \cdots + 12\)
\( \displaystyle p = \frac{12 \times 13}{2} = 78 \)