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# Engineering

Suppose that a, b, c are real numbers such that a + b + c = 1. If the matrix A=\begin{vmatrix} a &b &c \\ b & c& a\\ c& a& b \end{vmatrix}   be an orthogonal matrix, then

 

                    

  • Option 1)

    atleast one of a, b, c is negative

  • Option 2)

     |A| is negative

  • Option 3)

    a3 + b3 + c3 – 3abc = 1

  • Option 4)

    All of these

 

Answers (1)
G gaurav

As we have learned

Circulant determinant -

The elements of the rows (or columns) are in cyclic arrangement

- wherein

eg:- \begin{vmatrix} a & b & c\\ b&c &a \\ c & a & b \end{vmatrix} = -\left ( a^{3} +b^{3}+c^{3}-3abc\right )

 

 

AAT = ATA = I.  Also AT = A, so A2  = I  \Rightarrow  A  is involuntary matrix.

            \Rightarrow  |A2| = |A|2 = 1   or   |A| = ±1

            But |A| = \begin{vmatrix} a &b & c\\ b & c& a\\ c& a&b \end{vmatrix}=(a+b+c)\begin{vmatrix} 1 & b&c \\ 1 & c & a\\ 1&a & b \end{vmatrix}=(a+b+c)(ab+bc+ca-a^2-b^2-c^2)

            |A| = ab + bc + ca – a2 – b2 – c2  (\because a+b+c=1)

            \; \; \; \therefore \;  a2 + b2 + c2 – ab – bc – ca \; \geq  0

            So  |A| = -1. Hence a3 + b3 + c3 – 3abc = 1.

            Again a2 + b2 + c2 – ab – bc – ca = 1 \Rightarrow  1 – 3(ab + bc + c(A) = 1,  so ab + bc + ca = 0

            \Rightarrow atleast one of a, b and c is negative.

 


Option 1)

atleast one of a, b, c is negative

Option 2)

 |A| is negative

Option 3)

a3 + b3 + c3 – 3abc = 1

Option 4)

All of these

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