Matrix finds the cube

Algebra Level 4

Let $A = \left( \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} \right)$ be a matrix where $a, b, c$ are complex numbers. If $abc = 1$ and $A^T A = I$ , what is the value of $a^3 + b^3 + c^3$ ?

Clarification : $A^T$ is the transpose of $A$ . $I$ is the identity matrix of order $3 \times 3$ .

3 9 1 4

2 solutions

Aniket Sanghi
Feb 17, 2017

It can be solved in a much better way

Take mod of |AA'| = |I| which implies |A|= 1 or -1...

Now $det(A) = 3abc - a^3 - b^3 - c^3$ which is always negative for positive a,b,c as can be easily proved

Hence det(A) = -1.

So $a^3 + b^3 + c^3 = 3abc - (-1) = 4$

Easy one .... @Despicable Tamim you must have included 2 also in options to make it difficult! As determinant has 2 possible values :)

Yea! Adding 2 to the options would have tricked many people!

Sumanth R Hegde - 4 years, 3 months ago

@Arpit Dhimanj This setup is incorrect. There is no system of (positive) real solutions to $abc = 1, a^2 + b^2 + c^2 = 1, ab + bc + ca = 0$ .

There are complex roots, which does yield $a^3 + b^3 + c^3 = 4$ , but this particular solution would no longer apply (since it uses the assumption of real values).

I've edited that the problem is for complex numbers.

Calvin Lin Staff - 4 years, 3 months ago

Despicable Tamim
Apr 25, 2015

ANSWER IS 4

Matrix finds the cube

2 solutions

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