Question

Suppose that a, b, c are all real numbers such that $a+b+c=1$. If the matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ is an orthogonal matrix, then which of the following is correct?

• A. A must be a singular matrix.
• B. $|A|=1$
• C. $a^2+b^3+c^3-3abc=1$
• D. atleast one of a,b,c must be zero.

Answer 1

The correct option is C $a^2+b^3+c^3-3abc=1$

As A is an orthogonal matrix, $\Rightarrow A{A}^T=I\Rightarrow |A|=\pm 1$
Therefore, A is a non-singular matrix.
$\Rightarrow |A|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$\Rightarrow |A|=3abc-a^3-b^3-c^3\Rightarrow |A|=-\frac{(a+b+c)}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

$\Rightarrow |A|=\frac{-1}{2}\left(\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)\le 0$
$\therefore |A|\ne 1$
$So,|A|=-1\Rightarrow a^3+b^3+c^3-3abc=1$