IPMAT Indore 2026 · Quantitative Ability · Question 41

Step 1 — find the pattern in the powers.

{A}^{2}=\left[ \begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right],{A}^{3}=\left[ \begin{matrix}1 & 3 \\ 0 & 1\end{matrix}\right], \ldots{A}^{n}=\left[ \begin{matrix}1 & n \\ 0 & 1\end{matrix}\right]

(Each multiplication by A simply adds 1 to the top-right entry.)

Step 2 — add the first 13 powers.

\sum_{n=1}^{13}{{A}^{n}}=\left[ \begin{matrix}13 & 1+2+\cdots +13 \\ 0 & 13\end{matrix}\right]=\left[ \begin{matrix}13 & 91 \\ 0 & 13\end{matrix}\right]

using 1+2+\cdots +13=\frac{13\times 14}{2}=91.

Step 3 — take the determinant. For an upper-triangular matrix it is the product of the diagonal:

13\times 13=169

Answer: Option 3 — 169

Shortcut. The matrix is upper triangular with 1s on the diagonal, so the sum is upper triangular with 13s on the diagonal and the determinant is {13}^{2} — the top-right entry 91 never matters. Option (a) 91 is there to catch anyone who computes the sum and stops.