For any a,b,c R, the determinant {vmatrix} bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1 {vmatrix} is equal to

Correct Answer is : (a−b)(b−c)(c−a)(a-b)(b-c)(c-a)(a−b)(b−c)(c−a)

a(b²-c²)+b(c²-a²)+c(a²-b²)

Correct Answer is : (a−b)(b−c)(c−a)(a-b)(b-c)(c-a)(a−b)(b−c)(c−a)

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AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

For any

a,b,c \in R

\begin{vmatrix} bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1 \end{vmatrix}

is equal to

Solution:

a, b, c \in R

, the given determinant

\Delta=\begin{vmatrix} bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1 \end{vmatrix}

On applying

R_{2} \rightarrow R_{2}-R_{1}

and

R_{3} \rightarrow R_{3}-R_{1}

, we have

\Delta=\begin{vmatrix} bc & b+c & 1 \\ c(a-b) & a-b & 0 \\ b(a-c) & a-c & 0 \end{vmatrix}

=(a-b)(a-c)\begin{vmatrix} bc & b+c & 1 \\ c & 1 & 0 \\ b & 1 & 0 \end{vmatrix}

On applying

R_{3} \rightarrow R_{3}-R_{2}

, we have

\Delta \;\; =(a-b)(a-c)\begin{vmatrix} bc & b+c & 1 \\ c & 1 & 0 \\ b-c & 0 & 0 \end{vmatrix}

\;\; =(a-b)(a-c)[0-(b-c)]=(a-b)(b-c)(c-a)

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