Solveeit Logo
Login

Question

Mathematics Question on Determinants

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)

Answer

Δ=$$\begin{vmatrix} 3a& -a+b & -a+c\\\ -b+a & 3b & -b+c \\\\-c+a&-c+b&3c\end{vmatrix}
Applying C1→C1+C2+C3C_1\rightarrow C_1+C_2+C_3,we have
=∣a+b+c−a+b−a+c a+b+c3b−b+c\a+b+c−c+b3c∣=\begin{vmatrix} a+b+c& -a+b & -a+c\\\ a+b+c & 3b & -b+c \\\a+b+c&-c+b&3c\end{vmatrix}
=(a+b+c)$$\begin{vmatrix} 1& -a+b & -a+c\\\ 1 & 3b & -b+c \\\1&-c+b&3c\end{vmatrix}

Applying R2→R2−R1R_2\rightarrow R_2-R_1, andR3→R3−R1 R_3\rightarrow R_3-R_1we 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 C1,C_1,we have:
Δ=(a+b+c)[(2b+a)(2c+a)−(a−b)(a−c)]Δ=(a+b+c)[(2b+a)(2c+a)-(a-b)(a-c)]
=(a+b+c)[4bc+2ab+2ac+a2−a2+ac+ba−bc]=(a+b+c)[4bc+2ab+2ac+a^2-a^2+ac+ba-bc]
=(a+b+c)(3ab+3bc+3ac)=(a+b+c)(3ab+3bc+3ac)
=3(a+b+c)(ab+bc+ca)=3(a+b+c)(ab+bc+ca)

Hence,the given result is proved.