Question

If matrix A is a circulant matrix whose elements of first row are a, b, c and all the given elements are positive such that $abc=1$ and ${{A}^{T}}A=I$, then find the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$?
(a) 0
(b) 3
(c) 1
(d) 4

Answer 1

Hint: We start solving the solving by recalling the definition of circulant matrix. We write the matrix A by using this definition and find the transpose of it. We now multiply the given matrix and its transpose using the given ${{A}^{T}}A=I$. We then compare both sides and get the required values which will be useful to find the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$.

Complete step by step answer:
Given that we have matrix A which is a circulant matrix. We have elements of the first row as a, b, c which are positive and the value of $abc$ is 1. We also have that the matrix satisfying ${{A}^{T}}A=I$ and we need to find the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$.
We know that the circulant matrix is a square matrix in which the elements of the next row are rotated one element to the right relative to the previous row. Here, the last element of the previous row will be the first element of the next row. First element of the previous row will be the second element off the next row and so on.
Now, we write the circulant matrix with elements of the first row as a, b and c. Since we have only 3 elements in the first row, the order of the matrix will be $3\times 3$ as the circulant matrix is a square matrix.
We get the circulant matrix A as $A=\left[ \begin{matrix}
   a & b & c \\
   c & a & b \\
   b & c & a \\
\end{matrix} \right]$.
We know that the transpose of the matrix is formed by interchanging rows and columns of a matrix.
So, we get transpose of matrix A as ${{A}^{T}}=\left[ \begin{matrix}
   a & c & b \\
   b & a & c \\
   c & b & a \\
\end{matrix} \right]$.
According to the problem, we have ${{A}^{T}}A=I$. We know that is an identity matrix which has the value of all elements of principal diagonal as 1 and others as 0.
$\Rightarrow {{A}^{T}}A=I$.
$\Rightarrow \left[ \begin{matrix}
   a & b & c \\
   c & a & b \\
   b & c & a \\
\end{matrix} \right]\times \left[ \begin{matrix}
   a & c & b \\
   b & a & c \\
   c & b & a \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   \left( a\times a \right)+\left( b\times b \right)+\left( c\times c \right) & \left( a\times c \right)+\left( b\times a \right)+\left( c\times b \right) & \left( a\times b \right)+\left( b\times c \right)+\left( c\times a \right) \\
   \left( c\times a \right)+\left( a\times b \right)+\left( b\times a \right) & \left( c\times c \right)+\left( a\times a \right)+\left( b\times b \right) & \left( c\times b \right)+\left( a\times c \right)+\left( b\times a \right) \\
   \left( b\times a \right)+\left( c\times b \right)+\left( a\times c \right) & \left( b\times c \right)+\left( c\times a \right)+\left( a\times b \right) & \left( b\times b \right)+\left( c\times c \right)+\left( a\times a \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   \left( {{a}^{2}} \right)+\left( {{b}^{2}} \right)+\left( {{c}^{2}} \right) & \left( ac \right)+\left( ba \right)+\left( cb \right) & \left( ab \right)+\left( bc \right)+\left( ca \right) \\
   \left( ca \right)+\left( ab \right)+\left( ba \right) & \left( {{c}^{2}} \right)+\left( {{a}^{2}} \right)+\left( {{b}^{2}} \right) & \left( cb \right)+\left( ac \right)+\left( ba \right) \\
   \left( ba \right)+\left( cb \right)+\left( ac \right) & \left( bc \right)+\left( ca \right)+\left( ab \right) & \left( {{b}^{2}} \right)+\left( {{c}^{2}} \right)+\left( {{a}^{2}} \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   {{a}^{2}}+{{b}^{2}}+{{c}^{2}} & ab+bc+ca & ab+bc+ca \\
   ab+bc+ca & {{a}^{2}}+{{b}^{2}}+{{c}^{2}} & ab+bc+ca \\
   ab+bc+ca & ab+bc+ca & {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$ ---(1).
We know that if two matrices are said to be equal, then the elements at the corresponding places must be equal i.e., if \[\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]=\left[ \begin{matrix}
   j & k & l \\
   m & n & o \\
   p & q & r \\
\end{matrix} \right]\], then $a=j$, $b=k$,……,$h=q$ and $i=r$. We apply this in equation (1).
So, we get ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ and $ab+bc+ca=0$. Let us find the value of $a+b+c$.
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$.
$\Rightarrow {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$.
Let us substitute ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ and $ab+bc+ca=0$.
$\Rightarrow {{\left( a+b+c \right)}^{2}}=1+2\left( 0 \right)$.
$\Rightarrow {{\left( a+b+c \right)}^{2}}=1+0$.
$\Rightarrow {{\left( a+b+c \right)}^{2}}=1$.
$\Rightarrow a+b+c=\pm 1$ ---(2).
We know that $\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$.
$\Rightarrow \left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-\left( ab+bc+ca \right) \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$.
$\Rightarrow \left( \pm 1 \right)\left( 1-0 \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$.
$\Rightarrow \left( \pm 1 \right)\left( 1 \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$.
According to the problem we have $abc=1$.
$\Rightarrow \left( \pm 1 \right)\left( 1 \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3\left( 1 \right)$.
$\Rightarrow \pm 1={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3$.
We have ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3=1$ or ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3=-1$.
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=1+3$ or ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=-1+3$.
$\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=4$ or ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=2$.
We have found the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$ as 2 or 4.

So, the correct answer is “Option D”.

Note: We can also solve the problem by taking determinant for the given condition ${{A}^{T}}A=I$. We use the fact that the determinant of the matrix is equal to the determinant of its transpose to get the value of k. If we are given a problem with having more than one correct answer in the options, we should check whether the second answer is present or not.