NIMCET 2026 — Mathematics PYQ

To simplify the determinant, we apply row and column transformations.

1. Apply row operations:

Let $R_1,R_2,$ and $R_3$ be the rows of the determinant. Perform the following operations to simplify:

• $R_2\to R_2-R_1$

• $R_3\to R_3-R_2$

$$\Delta =\left|\begin{array}{ccc}x& amp;x+1& amp;x+3\\ (x+1)-x& amp;(x+3)-(x+1)& amp;(x+6)-(x+3)\\ (x+3)-(x+1)& amp;(x+6)-(x+3)& amp;(x+10)-(x+6)\end{array}\right|$$

$$\Delta =\left|\begin{array}{ccc}x& amp;x+1& amp;x+3\\ 1& amp;2& amp;3\\ 2& amp;3& amp;4\end{array}\right|$$

2. Apply another row operation:

Perform $R_3\to R_3-R_2$:

$$\Delta =\left|\begin{array}{ccc}x& amp;x+1& amp;x+3\\ 1& amp;2& amp;3\\ 2-1& amp;3-2& amp;4-3\end{array}\right|$$

$$\Delta =\left|\begin{array}{ccc}x& amp;x+1& amp;x+3\\ 1& amp;2& amp;3\\ 1& amp;1& amp;1\end{array}\right|$$

3. Apply column operations:

Perform $C_2\to C_2-C_1$ and $C_3\to C_3-C_2$:

$$\Delta =\left|\begin{array}{ccc}x& amp;1& amp;2\\ 1& amp;1& amp;1\\ 1& amp;0& amp;0\end{array}\right|$$

Expanding along the third row:

$$\Delta =1\cdot \left|\begin{array}{cc}1& amp;2\\ 1& amp;1\end{array}\right|=1(1-2)=-1$$

Since the simplified determinant is $-1$ and is independent of $x$, the value at $x=2026$ is $-1$.

Correct Option: (b)