# Prove that $a_1^2+a_2^2+a_3^2=b_1^2+b_2^2+b_3^2=c_1^2+c_1^2+c_3^2=1$, $a_1b_1+a_2b_2+a_3b_3=b_1c_1+b_2c_2+b_3c_3=c_1a_1+c_23a_2+c_3a_3=0$

Given $a_i, b_i, c_i (i=1,2,3)$ be all real and $a_1^2+b_1^2+c_1^2=a_2^2+b_2^2+c_2^2=a_3^2+b_3^2+c_3^2=1$, $a_1a_2+b_1b_2+c_1c_2=a_3a_2+b_3b_2+c_3c_2=a_3a_1+b_3b_1+c_3c_1=0$. Use matrix method to show that $$a_1^2+a_2^2+a_3^2=b_1^2+b_2^2+b_3^2=c_1^2+c_2^2+c_3^2=1,$$ $$a_1b_1+a_2b_2+a_3b_3=b_1c_1+b_2c_2+b_3c_3=c_1a_1+c_2a_2+c_3a_3= 0.$$

Attempt:

I understand that I have to start by assuming a matrix may be of the form $A=\left( \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right)$ and have to find its square which will be an identity matrix by using the results given. This A does not work. But what should the form of that A? How to solve the probelm?

• just FYI, $AA^T$ is the not the square of $A$. Aug 1, 2016 at 17:16
• @harvey Ok. Thanks a lot. Aug 1, 2016 at 17:16

Hint : By the assumption of the problem, your matrix became orthonormal matrix. i.e. $AA^{t}=I$. Then we also have $A^{t}A=I$.

• what if the product of A and $A^T$ is a multiple of I? i know the result is the same but how do i show that A$A^T$=cI=$A^T$A? May 17, 2021 at 22:55