Question:
Find the value of determinant at \(x=2026\):
\[
\begin{vmatrix}
x & x+1 & x+3\\
x+1 & x+3 & x+6\\
x+3 & x+6 & x+10
\end{vmatrix}
\]
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
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The Correct Option is B
Solution and Explanation
Concept:
For determinants having a common variable in all entries, row and column operations often simplify the determinant dramatically.
Step 1: Apply row operations.
Let \[ D= \begin{vmatrix} x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10 \end{vmatrix}. \] Perform \[ R_2\to R_2-R_1, \qquad R_3\to R_3-R_1. \] Then \[ D= \begin{vmatrix} x & x+1 & x+3\\ 1 & 2 & 3\\ 3 & 5 & 7 \end{vmatrix}. \]
Step 2: Expand along the first row.
\[ D=x \begin{vmatrix} 2&3\\ 5&7 \end{vmatrix} -(x+1) \begin{vmatrix} 1&3\\ 3&7 \end{vmatrix} +(x+3) \begin{vmatrix} 1&2\\ 3&5 \end{vmatrix}. \] Compute minors: \[ \begin{vmatrix} 2&3\\ 5&7 \end{vmatrix} =14-15=-1, \] \[ \begin{vmatrix} 1&3\\ 3&7 \end{vmatrix} =7-9=-2, \] \[ \begin{vmatrix} 1&2\\ 3&5 \end{vmatrix} =5-6=-1. \] Thus \[ D=x(-1)-(x+1)(-2)+(x+3)(-1). \] \[ D=-x+2x+2-x-3. \] \[ D=-1. \]
Step 1: Apply row operations.
Let \[ D= \begin{vmatrix} x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10 \end{vmatrix}. \] Perform \[ R_2\to R_2-R_1, \qquad R_3\to R_3-R_1. \] Then \[ D= \begin{vmatrix} x & x+1 & x+3\\ 1 & 2 & 3\\ 3 & 5 & 7 \end{vmatrix}. \]
Step 2: Expand along the first row.
\[ D=x \begin{vmatrix} 2&3\\ 5&7 \end{vmatrix} -(x+1) \begin{vmatrix} 1&3\\ 3&7 \end{vmatrix} +(x+3) \begin{vmatrix} 1&2\\ 3&5 \end{vmatrix}. \] Compute minors: \[ \begin{vmatrix} 2&3\\ 5&7 \end{vmatrix} =14-15=-1, \] \[ \begin{vmatrix} 1&3\\ 3&7 \end{vmatrix} =7-9=-2, \] \[ \begin{vmatrix} 1&2\\ 3&5 \end{vmatrix} =5-6=-1. \] Thus \[ D=x(-1)-(x+1)(-2)+(x+3)(-1). \] \[ D=-x+2x+2-x-3. \] \[ D=-1. \]
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