Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)
Solution and Explanation
\(Δ=\)\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)
Applying \(C_1\rightarrow C_1+C_2+C_3\),we have
\(=\begin{vmatrix} a+b+c& -a+b & -a+c\\ a+b+c & 3b & -b+c \\a+b+c&-c+b&3c\end{vmatrix}\)
\(=(a+b+c)\)\(\begin{vmatrix} 1& -a+b & -a+c\\ 1 & 3b & -b+c \\1&-c+b&3c\end{vmatrix}\)
Applying \(R_2\rightarrow R_2-R_1\), and\( R_3\rightarrow R_3-R_1\)we have:
\( Δ=(a+b+c)\)\(\begin{vmatrix} 1& -a+b & -a+c\\ 0 & 2b+a & a-b \\0&a-c&2c+a\end{vmatrix}\)
Expanding along \(C_1,\)we have:
\(Δ=(a+b+c)[(2b+a)(2c+a)-(a-b)(a-c)]\)
\(=(a+b+c)[4bc+2ab+2ac+a^2-a^2+ac+ba-bc]\)
\(=(a+b+c)(3ab+3bc+3ac)\)
\(=3(a+b+c)(ab+bc+ca)\)
Hence,the given result is proved.
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