Question:medium

Find the value of determinant at \(x=2026\): \[ \begin{vmatrix} x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10 \end{vmatrix} \]

Show Hint

When consecutive entries differ by fixed amounts, use row or column differences first. Such determinants often become constant and independent of the variable.
Updated On: Jun 11, 2026
  • \(2026\)
  • \(-1\)
  • \(0\)
  • \(1\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Note the variable cancels with row operations.
Every entry contains $x$, so subtracting rows should wipe out $x$ and leave a constant. That is why the answer does not depend on $x=2026$.
Step 2: Subtract the first row from the others.
Do $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$. Row 2 becomes $(1,2,3)$ and row 3 becomes $(3,5,7)$, both free of $x$.
Step 3: Expand along the first row.
$D=x\begin{vmatrix}2&3\\5&7\end{vmatrix}-(x+1)\begin{vmatrix}1&3\\3&7\end{vmatrix}+(x+3)\begin{vmatrix}1&2\\3&5\end{vmatrix}$.
Step 4: Compute the small minors.
$14-15=-1$; $7-9=-2$; $5-6=-1$.
Step 5: Substitute the minors.
$D=x(-1)-(x+1)(-2)+(x+3)(-1)=-x+2x+2-x-3$.
Step 6: Collect terms.
The $x$ terms cancel: $-x+2x-x=0$, leaving $2-3=-1$. So $D=-1$ for any $x$, including $2026$, which is option 2.
\[ \boxed{-1} \]
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