JEE MAIN - Mathematics (2009 - No. 10)
$$\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + \left| {\matrix{
{a + 1} & {b + 1} & {c - 1} \cr
{a - 1} & {b - 1} & {c + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr
} } \right| = 0$$
then the value of $$n$$ :
then the value of $$n$$ :
any even integer
any odd integer
any integer
zero
Explanation
$$\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + \left| {\matrix{
{a + 1} & {b + 1} & {c - 1} \cr
{a - 1} & {b - 1} & {c + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr
} } \right| = 0$$
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$$
(Taking transpose of second determinant)
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$$
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$$
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$
$${R_1} + {R_3}$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$$
$$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$$
$$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$$ $$\,\,\,\,\,$$ as $$\,\,\,\,\,b\left( {a + c} \right) \ne 0$$
$$ \Rightarrow n$$ should be an odd integer.
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$$
(Taking transpose of second determinant)
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$$
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$$
$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$
$${R_1} + {R_3}$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$
$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$$
$$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$$
$$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$$ $$\,\,\,\,\,$$ as $$\,\,\,\,\,b\left( {a + c} \right) \ne 0$$
$$ \Rightarrow n$$ should be an odd integer.
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