Evaluate the determinant: left| begin{array}{ | vedclass

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(A) Let $\Delta = \left| \begin{array}{lll} a-b & b-c & c-a \ b-c & c-a & a-b \ c-a & a-b & b-c \end{array} ight|$.Applying the column operation $

Vedclass · Accepted Answer

$0$

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(A) Let $\Delta = \left| \begin{array}{lll} a-b & b-c & c-a \ b-c & c-a & a-b \ c-a & a-b & b-c \end{array} ight|$.Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$:$\Delta = \left| \begin{array}{lll} (a-b)+(b-c)+(c-a) & b-c & c-a \ (b-c)+(c-a)+(a-b) & c-a & a-b \ (c-a)+(a-b)+(b-c) & a-b & b-c \end{array} ight|$.Simplifying the elements in the first column:$\Delta = \left| \begin{array}{lll} 0 & b-c & c-a \ 0 & c-a & a-b \ 0 & a-b & b-c \end{array} ight|$.Since all elements in the first column are $0$,the value of the determinant is $0$.

Evaluate the determinant: $\left| \begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array} \right|$

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