3 and 4 .Determinants and Matrices
medium

माना $a, b, c$ के लिए $b(a+c) \neq 0$ । यदि

$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$

तो $n$ का मान है

A

$0$

B

कोई भी सम पूर्णांक

C

कोई भी विषम पूर्णांक

D

कोई भी पूणांक

(AIEEE-2009)

Solution

$\left| {\begin{array}{*{20}{c}}
a&{a + 1}&{a – 1}\\
{ – b}&{b + 1}&{b – 1}\\
c&{c – 1}&{c + 1}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
{{{\left( { – 1} \right)}^{n + 2}}a}&{a + 1}&{a – 1}\\
{{{\left( { – 1} \right)}^{n + 1}}b}&{b + 1}&{b – 1}\\
{{{\left( { – 1} \right)}^n}c}&{c – 1}&{c + 1}
\end{array}} \right|$

$ = \left| {\begin{array}{*{20}{c}}
{a + {{\left( { – 1} \right)}^{n + 2}}a}&{a + 1}&{a – 1}\\
{ – b + {{\left( { – 1} \right)}^{n + 1}}b}&{b + 1}&{b – 1}\\
{c + {{\left( { – 1} \right)}^n}c}&{c – 1}&{c + 1}
\end{array}} \right|$

$=0$ if $n$ is an obb integer.

Standard 12
Mathematics

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