JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 1)
Let a, b, c $$ \in $$ R be all non-zero and satisfy
a3 + b3 + c3 = 2. If the matrix
A = $$\left( {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right)$$
satisfies ATA = I, then a value of abc can be :
a3 + b3 + c3 = 2. If the matrix
A = $$\left( {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right)$$
satisfies ATA = I, then a value of abc can be :
3
$${1 \over 3}$$
-$${1 \over 3}$$
$${2 \over 3}$$
Explanation
Given,
$${a^3} + {b^3} + {c^3} = 2$$
$${A^T}A = I$$
$$ \Rightarrow \left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right]\left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$$
and ab + bc + ca = 0
Now (a + b + c)2 = 1
$$ \Rightarrow (a + b + c) = \pm 1$$
So, $${a^3} + {b^3} + {c^3} - 3abc$$
$$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$ \Rightarrow \pm 1(1 - 0) = \pm 1$$
$$ \therefore 2 - 3abc = \pm 1$$
$$ \Rightarrow 3abc = 2 \pm 1 = 3,1$$
$$ \Rightarrow abc = 1,{1 \over 3}$$
$${a^3} + {b^3} + {c^3} = 2$$
$${A^T}A = I$$
$$ \Rightarrow \left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right]\left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$$
and ab + bc + ca = 0
Now (a + b + c)2 = 1
$$ \Rightarrow (a + b + c) = \pm 1$$
So, $${a^3} + {b^3} + {c^3} - 3abc$$
$$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$ \Rightarrow \pm 1(1 - 0) = \pm 1$$
$$ \therefore 2 - 3abc = \pm 1$$
$$ \Rightarrow 3abc = 2 \pm 1 = 3,1$$
$$ \Rightarrow abc = 1,{1 \over 3}$$
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