If A=a2abacabb2bcacbcc2 and a2+b2+c2=1, then A2 =

A
2A
B
A
C
$A^{- 1}$
D
none

Solution:

$A^{2} = [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] = [\begin{matrix} a^{4} + a^{2} b^{2} + a^{2} c^{2} & a^{3} b + a b^{3} + a b c^{2} & a^{3} c + a b^{2} c + a c^{3} \\ a^{3} b + a b^{3} + a b c^{2} & a^{2} b^{2} + b^{4} + b^{2} c^{2} & a^{2} b c + b^{3} c + b c^{3} \\ a^{3} c + a b^{2} c + a c^{3} & a^{2} b c + b^{3} c + b c^{3} & a^{2} c^{2} + b^{2} c^{2} + c^{4} \end{matrix}]$

$= [\begin{matrix} a^{2} (a^{2} + b^{2} + c^{2}) & a b (a^{2} + b^{2} + c^{2}) & a c (a^{2} + b^{2} + c^{2}) \\ a b (a^{2} + b^{2} + c^{2}) & b^{2} (a^{2} + b^{2} + c^{2}) & b c (a^{2} + b^{2} + c^{2}) \\ a c (a^{2} + b^{2} + c^{2}) & b c (a^{2} + b^{2} + c^{2}) & c^{2} (a^{2} + b^{2} + c^{2}) \end{matrix}] = [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] = A$

If $A = [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] a n d a^{2} + b^{2} + c^{2} = 1, t h e n A^{2} =$

Solution:

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