Using properties of determinants, prove that 

\begin{vmatrix} a^2+1 &ab &ac \\ab &b^2+1 &bc \\ac &bc & c^2+1 \end{vmatrix} = 1+a^2+b^2+c^2

 

 

 

 

 
 
 
 
 

Answers (1)

LHS = \begin{vmatrix} a^2+1 &ab &ac \\ab &b^2+1 &bc \\ac &bc & c^2+1 \end{vmatrix}

= \frac{1}{abc}\begin{vmatrix} a^3+a &a^2b &a^2c \\ab^2 &b^3+b &b^2c \\c^2a &c^2b & c^3+c \end{vmatrix}a\times R_{1},b\times R_{2}, c\times R_{3}

 

 

= \frac{abc}{abc}\begin{vmatrix} a^2+1 &a^2 &a^2 \\b^2 &b^2+1 &b^2 \\c^2 &c^2 & c^2+1 \end{vmatrix}

R_1=R_1+R_2+R_3

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

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

C_1=C_1-C_2, C_2= C_2-C_3

= \left ( 1+a^2+b^2+c^2 \right )\begin{vmatrix} 0 &0&1 \\-1 &1 &b^2 \\ 0&-1 & c^2+1 \end{vmatrix}

= \left ( 1+a^2+b^2+c^2 \right )\left [ 0-0+1(1-0) \right ]

1+a^2+b^2+c^2= RHS

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