Question

Given a matrix $A=\left[ \begin{matrix}
   a & b & c \\
   b & c & a \\
   c & a & b \\
\end{matrix} \right]$ , where a, b, c are real positive numbers, abc = 1and ${{A}^{T}}A=I$ then find the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}.$

Answer 1

Hint:Find${{A}^{T}}$ by changing the row elements to column. Now, use the relation ${{A}^{T}}A=I$ and $abc=1$to get the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$. Use the relation
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$

Complete step by step answer:
Here, matrix A is given as
$A=\left[ \begin{matrix}
   a & b & c \\
   b & c & a \\
   c & a & b \\
\end{matrix} \right]$……….. (i)
Where a, b, c are real numbers, with relations
abc = 1 ……………….. (ii)
${{A}^{T}}A=I$ ……………(iii)
We need to determine the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$ from the above equations.
Now, let us find the value of ${{A}^{T}}$ and put values of A and ${{A}^{T}}$in equation (iii) to relation in a, b, c.
So, ${{A}^{T}}$ can be written using the equation (i) i.e. from ‘A’:
${{A}^{T}}=\left[ \begin{matrix}
   a & b & c \\
   b & c & a \\
   c & a & b \\
\end{matrix} \right]$………. (iv)
So, putting values of A and ${{A}^{T}}$ in equation (iii), we get
A.${{A}^{T}}$ = I
Where I can be given as
$I=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
Hence,
$\left[ \begin{matrix}
   a & b & c \\
   b & c & a \\
   c & a & b \\
\end{matrix} \right]\left[ \begin{matrix}
   a & b & c \\
   b & c & a \\
   c & a & b \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
Now, we can get A.${{A}^{T}}$using above equation by multiplying the matrices A and ${{A}^{T}}$in the above equation; we get,
\[\begin{align}
  & \\
 & \left[ \begin{matrix}
   {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} & \text{ab+bc+ca} & \text{ac+ba+bc} \\
   \text{ab+bc+ac} & {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} & \text{bc+ac+ab} \\
   \text{ac+ab+bc} & \text{bc+ac+ab} & {{\text{c}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right] \\
\end{align}\]

Now, we can compare the above two matrices and equate the elements to get relations in (a, b, c). Hence, we can further write the above matrix as:
\[\left[ \begin{matrix}
   {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} & \text{ab+bc+ca} & \text{ab+bc+ca} \\
   \text{ab+bc+ca} & {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} & \text{ab+bc+ca} \\
   \text{ab+bc+ca} & \text{ab+bc+ca} & {{a}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\]
Hence, we get two relations from above equations as
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ ………………. (v)
And
$ab+bc+ca=0$……………… (vi)
Now, we know the algebraic identity of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ can be given as
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$
Or
$\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-\left( ab+bc+ca \right) \right)$…………..(vii)
Now, put the values of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}},abc,ab+bc+ca$ from equation (v),(ii) and (vi) respectively in the equation (vii), hence, we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3\left( 1 \right)=\left( a+b+c \right)\left( 1-0 \right)$
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3+\left( a+b+c \right)$………..(viii)
Now, we can use value of a + b + c by using the algebraic identity
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$
Now, using relation (v) and (vi), we can get value of $\left( a+b+c \right)$ as
$\begin{align}
  & {{\left( a+b+c \right)}^{2}}=1+2\left( 0 \right) \\
 & {{\left( a+b+c \right)}^{2}}=1 \\
\end{align}$
And hence,
$a+b+c=\pm 1$
As we already know the values of (a, b, c) are real and positive from the question, so, $a+b+c$ can never be ‘-1’. Hence, the value of (a + b + c) be ‘1’. Hence,
a + b + c = 1 ………….. (ix).
So, putting value of (a + b + c) in equation
(viii) to get the value of${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$. Hence,we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3+1=4$
So,
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=4$

Note: One can go wrong with the algebraic identities ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ and ${{\left( a+b+c \right)}^{2}}$. So, identities should be very clear for solving these kinds of problems.
One may try calculate exact values of given relations i.e. abc = 1, ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$and $ab+bc+ca=0$. So, one can think, we have three equations and three variables so, we can calculate exact values of a,b,c which is the wrong approach for getting value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$. As, we don’t need exact values of a, b, c we need value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$, so try to relate the problem with the algebraic identities. Calculating values of (a, b, c) is the complex approach for these kinds of problems.