Matrices and Determinants 1 Question 20
Analytical and Descriptive Questions
20. If matrix $A=\begin{bmatrix}a & b & c \\ b & c & a \\ c & a & b\end{bmatrix}$, where $a, b, c$ are real positive numbers, $a b c=1$ and $A^{T} A=I$, then find the value of $a^{3}+b^{3}+c^{3}$.
$(2003,2 \mathrm{M})$
Integer Type Question
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Answer:
Correct Answer: 20. (4)
Solution:
- Given, $A=\begin{bmatrix}a & b & c \\ b & c & a \\ c & a & b\end{bmatrix}, a b c=1$ and $A^{T} A=I\quad $ …(i)
Now, $A^{T} A=I$
$\begin{bmatrix}a & b & c \\ b & c & a \\ c & a & b\end{bmatrix}$ $\begin{bmatrix}a & b & c \\ b & c & a \\ c & a & b\end{bmatrix}$ $ =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $
$ \begin{bmatrix} a^{2}+ & b^{2} + &c^{2} \\ a b+ & b c+ & c a \\ a b+ & b c+ & c a\end{bmatrix} $ $ \begin{bmatrix} a b+ & b c+ & c a \\ a^{2}+ & b^{2}+ & c^{2} \\ a b+ & b c+ & c a\end{bmatrix} $ $ \begin{bmatrix} a b+ & b c+ & c a \\ a b+ & b c+ & c a \\ a^{2}+ & b^{2}+ & c^{2}\end{bmatrix} $
$ =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $
$\Rightarrow a^{2}+b^{2}+c^{2}=1$ and $a b+b c+c a=0\quad $…(ii)
We know, $a^{3}+b^{3}+c^{3}-3 a b c$
$ =(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) $
$\Rightarrow a^{3}+b^{3}+c^{3}=(a+b+c)(1-0)+3$
[from Eqs. (i) and (ii)]
$\therefore a^{3}+b^{3}+c^{3}=(a+b+c)+3\quad $…(iii)
Now, $\quad(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$ =1 $
From Eq. (iii), $a^{3}+b^{3}+c^{3}=1+3 \Rightarrow a^{3}+b^{3}+c^{3}=4$
