Question:
The value of
\(\begin{vmatrix}(a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1\end{vmatrix}\)
is :
The value of
\(\begin{vmatrix}(a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1\end{vmatrix}\)
is :
\(\begin{vmatrix}(a+1)(a+2) & a+2 & 1\\ (a+2)(a+3) & a+3 & 1\\ (a+3)(a+4) & a+4 & 1\end{vmatrix}\)
is :
Show Hint
Subtract row 2 from row 3 and row 1 from row 2 to clear the last column.
Updated On: Nov 4, 2026
- \(-2\)
- 0
- (a + 2)(a + 3)(a + 4)
- (a + 1)(a + 2)(a + 3)
Show Solution
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The Correct Option is A
Solution and Explanation
Put \(u=a+2\). Rows: \(R_1=(u^2-u,\ u,\ 1)\), \(R_2=(u^2+u,\ u+1,\ 1)\), \(R_3=(u^2+3u+2,\ u+2,\ 1)\).
\(R_3\to R_3-R_2=(2u+2,\ 1,\ 0)\); then \(R_2\to R_2-R_1=(2u,\ 1,\ 0)\).
Expanding along the third column (only \(R_1\) has a non-zero entry, 1, with sign \(+\)): \(\begin{vmatrix}2u&1\\2u+2&1\end{vmatrix}=2u-(2u+2)=-2\).
Answer: (1) \(-2\)
\(R_3\to R_3-R_2=(2u+2,\ 1,\ 0)\); then \(R_2\to R_2-R_1=(2u,\ 1,\ 0)\).
Expanding along the third column (only \(R_1\) has a non-zero entry, 1, with sign \(+\)): \(\begin{vmatrix}2u&1\\2u+2&1\end{vmatrix}=2u-(2u+2)=-2\).
Answer: (1) \(-2\)
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