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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
If $x$ is a positive integer, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$ is equal to
  • A
    $2(x!)(x + 1)!$
  • ✓
    $2(x!)(x + 1)!(x + 2)!$
  • C
    $2(x!)(x + 3)!$
  • D
    None of these
✓

Answer

Correct option: B.
$2(x!)(x + 1)!(x + 2)!$
b
(b) $\Delta = \left| {\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)\,!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)\,!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)\,!}\end{array}} \right|$

= $x\,!\,(x + 1)\,!\,(x + 2)\,!\,\left| {\,\begin{array}{*{20}{c}}1&{(x + 1)}&{(x + 2)\,(x + 1)}\\1&{(x + 2)}&{(x + 3)\,(x + 2)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

Applying ${R_1} \to {R_2} - {R_1},{R_2} \to ({R_3} - {R_2})$ we get

$ = x\,!(x + 1)\,!(x + 2)\,!$ $\,\left| {\,\begin{array}{*{20}{c}}0&1&{2(x + 2)}\\0&1&{2(x + 3)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

$ = 2\,x!(x + 1)!(x + 2)!$,(on simplification).

Trick: Put $x = 1$ and then match the alternate.

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