Question

Let a, b, c \(\in \mathbb{R}\) be all non-zero and satisfy \(a^3+b^3+c^3=2\). If the matrix \(A=\begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}\) satisfies \(A^TA = I\), then a value of abc can be :

• A. 3
• B. \(\frac{1}{3}\)
• C. \(-\frac{1}{3}\)
• D. \(\frac{2}{3}\)

Hint:

\(A^TA=I\) gives \(a^2+b^2+c^2=1\) and \(ab+bc+ca=0\), so \((a+b+c)^2=1\). Then use \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

Answer 1

The Correct Option is B

Diagonal entries of \(A^TA\): \(a^2+b^2+c^2=1\). Off-diagonal entries: \(ab+bc+ca=0\).
Then \((a+b+c)^2 = 1+0=1\Rightarrow a+b+c=\pm1\).
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)(1)=\pm1\).
With \(a^3+b^3+c^3=2\): \(2-3abc=\pm1\Rightarrow abc=\frac13\) (for +1) or \(abc=1\) (for -1).
Only \(\frac13\) is among the options.
Answer: (2) \(\frac{1}{3}\)