Question

MATRICES CHALLENGE!

Answer 1

Let\[A = \left[\begin{matrix} a & b & c \\ b& c& a \\ c&a&b\end{matrix}\right]\]such that \(abc = 1\) and \(A^T A = I\).
Find the value of \(a^3 + b^3 + c^3\).

Answer 2

\(a, b, c\) are real and positive.

Answer 3

Why do you stipulate \(A^T A=I\) why not \(A^Q\) where Q=T+1?

Answer 4

ohhhhh cause that means transpose doesn't it....sorry, ode mind

Answer 5

took determinant on both sides |A|^2= 1...solved the determinant a^3+b^3+c^3-3abc=1

Answer 6

Yup.

Answer 7

Nice!

Answer 8

\[[\det (A)]^2 = 1\]\[\Rightarrow (-(a^3 + b^3 + c^3 - 3abc))^2 = 1\]\[\Rightarrow (a^3 + b^3 + c^3 - 3)^2 = 1\]\[\Rightarrow a^3 + b^3 + c^3 = 2~\rm or ~ 4\]Which one of these?

Answer 9

it's 2
so \[A = \left[\begin{matrix} a & b & c \\ b& c& a \\ c&a&b\end{matrix}\right]\]
\[A^T = \left[\begin{matrix} a & b & c \\ b& c& a \\ c&a&b\end{matrix}\right]\]
then \[A^T \times A=A^2\]\[A^2=I\]\[det(A^2)=det(I)\]\[a(cb-a^2)-b(b^2-ac)+c(ba-c^2)=1\] \[abc-a^3-b^3+abc-c^3+abc=1\]
\[a^3+b^3+c^3=-1+3abc\]
\[a^3+b^3+c^3=-1+3=2\]

Answer 10

oh missed a step ugh

Answer 11

AM>GM so i would choose 4

Answer 12

4

Answer 13

Yup.

Answer 14

yea, I didn't square A, I just found it's det.