NIMCET 2026 — Mathematics PYQ

To find the value of the determinant, we can simplify the matrix using row operations.

Step 1: Simplify rows using subtraction

Let $R_1,R_2,R_3$ be the rows of the determinant. We apply the following row transformations:

• $R_2\to R_2-R_1$

• $R_3\to R_3-R_2$

Applying $R_2-R_1$:

• $(x+1)-x=1$

• $(x+3)-(x+1)=2$

• $(x+6)-(x+3)=3$

Applying $R_3-R_2$:

• $(x+3)-(x+1)=2$

• $(x+6)-(x+3)=3$

• $(x+10)-(x+6)=4$

The determinant becomes:

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

Step 2: Further simplification

Apply $R_3\to R_3-R_2$:

• $(2-1)=1$

• $(3-2)=1$

• $(4-3)=1$

The determinant becomes:

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

Step 3: Column operations

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

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

Step 4: Expand along the third row

Expanding along the third row ($1,0,0$):

$$\Delta =1\cdot \left|\begin{array}{cc}1& amp;2\\ 1& amp;1\end{array}\right|-0+0$$

$$\Delta =1\cdot (1-2)$$

$$\Delta =-1$$

Since the result is a constant independent of $x$, the value at $x=2026$ is also $-1$.

Correct Option: 1