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If the matrix \\\mathrm{\begin{bmatrix} a & b & c\\ b & c & a\\ c & a & b \end{bmatrix}} where a, b, c are positive real number such that abc = 1 and ATA = I. Then find the value of a3+b3+c3.

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

Both (A) and (C) are correct


Answers (1)

best_answer

 

 

symmetric and skew symmetric matrix -

symmetric matrix: A square matrix \\\mathrm{A=\left [ a_{ij} \right ]_{n\times n}} is said to be symmetric if A' = A,

    \\\mathrm{ i.e., a_{ij} = a_{ji}\; \forall\; i, j}

    \\\mathrm{A=\begin{bmatrix} a & h & g\\ h& b & f\\ g& f & c \end{bmatrix}\; then \; A'=\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c \end{bmatrix}}

 

Skew-symmetric matrix:  A square matrix  \\\mathrm{A=\left [ a_{ij} \right ]_{m\times n}} is said to be skew-symmetric if A’ = -A 

\\\mathrm{i.e. A' = -A , i.e., a_{ij} =- a_{ji}\; \forall \;i, j } \\\mathrm{Now \;if \;we\; put\; i\; =\; j,\; we\; have} \\\mathrm{ a_{ii} = -a_{ii},} \\\mathrm{ \therefore 2a_{ii}=0 \Rightarrow a_{ii}=0\; \forall \; i's}

That means all the diagonal element of the skew-symmetric matrix is 0 but not the other way around.

\\\mathrm{e.g.\;\; A=\begin{bmatrix} 0 & h & g\\ -h & 0 & f\\ -g & -f & 0 \end{bmatrix},\; then\; A' = \begin{bmatrix} 0 & -h & -g\\ h & 0 & -f\\ g & f & 0 \end{bmatrix} = -A}

-

 

 

Matrix is symmetric then A = AT

         ATA = A2

         |ATA| = |A||AT| = |A||A| = I

         |A|2 = 1

\\\mathrm{|A|=\pm1}

 

\\\mathrm{\left | A \right |=\begin{vmatrix} a &b & c\\ b& c&a \\ c &a &b \end{vmatrix}=-(a^3+b^3+c^3-3abc)} \\\mathrm{-(a^3+b^3+c^3-3abc) =\pm 1} \\\mathrm{given\; abc\; = \;1 } \\\mathrm{-(a^3+b^3+c^3-3abc) = \pm1} \\\mathrm{a^3+b^3+c^3=\pm1+3=2\:or\:4}

hence option (d) is correct

Posted by

Ritika Jonwal

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