Let a, b, c [ ] R be all non-zero and satisfy a3 + b3 + c3 = 2.

Question

Let a, b, c $$ \in $$ R be all non-zero and satisfy

a3 + b3 + c3 = 2. If the matrix

A = $$\left( {\begin{pmatrix} a & b & c  \
   b & c & a  \
   c & a & b  \ \end{pmatrix} } \right)$$

satisfies ATA = I, then a value of abc can be :

Accepted Answer

Given,

${a^3} + {b^3} + {c^3} = 2$

${A^T}A = I$

$$ \Rightarrow \left[ {\begin{pmatrix} a & b & c  \
   b & c & a  \
   c & a & b  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} a & b & c  \
   b & c & a  \
   c & a & b  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   0 & 1 & 0  \
   0 & 0 & 1  \

\end{pmatrix} } \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, ${a^3} + {b^3} + {c^3} - 3abc$

$$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$

$$ \Rightarrow  \pm 1(1 - 0) = \pm 1$$

$$ \therefore 2 - 3abc =  \pm 1$$

$$ \Rightarrow 3abc = 2 \pm 1 = 3,1$$

$$ \Rightarrow abc = 1,{1 \over 3}$$

Solutions · 1