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If a^{2}+b^{2}+c^{2}=-2\; and

then f(x)  is a polynomial of degree

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Answers (2)

As we learnt in 

Property of determinant -

If each element in a row ( or column ) of a determinant is written as the sum of two or more terms then the determinant can be written as the sum of two or more determinants

- wherein

 

 f(x)= \begin{vmatrix} 1+a^{2} x&(1+b^{2})x &(1+c^{2})x\\ (1+a^{2})x&1+b^{2}x &(1+c^{2})x \\ (1+a^{2})x&(1+b^{2})x & 1+c^{2}x \end{vmatrix}   

    and a^{2} + b^{2} +c^{2}= -2

f(x)= \begin{vmatrix} 1+a^{2} x&x+b^{2}x &x+c^{2}x \\ x+a^{2}x &1+b^{2}x &x+c^{2}x \\ x+a^{2}x &x+b^{2}x & 1+c^{2}x \end{vmatrix}

c_1\rightarrow c_1+c_2+ c_3

\begin{vmatrix} 1+2x+x(a^{2}+b^{2} +c^{2}) &x+b^{2}x &x+c^{2}x \\ 1+2x+x(a^{2}+b^{2} +c^{2})&1+b^{2}x &x+c^{2}x \\ 1+2x+x(a^{2}+b^{2} +c^{2})&x+b^{2}x &1+c^{2}x \end{vmatrix}

\begin{vmatrix} 1+2x+x(-2) &x+b^{2}x &x+c^{2}x \\ 1+2x+x(-2)&1+b^{2}x &x+c^{2}x \\ 1+2x+x(-2) &x+b^{2}x &1+c^{2}x \end{vmatrix}

\begin{vmatrix} 1 &x+b^{2}x &x+c^{2}x \\ 1&1+b^{2}x &x+c^{2}x \\ 1&x+b^{2}x &1+c^{2}x \end{vmatrix}

R_1\Rightarrow R_1-R_2\ and\ R_2\Rightarrow R_2-R_3

\begin{vmatrix} 0 &x-1 &0 \\ 0 &1-x &x-1 \\ 1& x+b^{2}x &1+c^{2}x \end{vmatrix}

= -(x-1) \begin{vmatrix} 0 &x-1 \\ 1& 1+c^{2}x \end{vmatrix}

=(1-x) (1-x) = (1-x)^{2}= x^{2}-2x+1

so degree 2. 


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