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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 5 of 160

Marks: +1, -0

\Delta

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

K(a-b)(b-c)(c-a)

K

Solution: 👈: Video Solution

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

R_{2} \rightarrow R_{2} \;\; -R_{1}, R_{3} \rightarrow R_{3}-R_{1}

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

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

= (b-a)(c-a)[(c+a)-(b+a)]

R_{2} \rightarrow R_{2} \;\; -R_{1}, R_{3} \rightarrow R_{3}-R_{1}

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

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

= (b-a)(c-a)[(c+a)-(b+a)]

\text{ expanding along } C_{1}

\therefore \;\; K=1

= (b-a)(c-a)(c-b)

= (a-b)(b-c)(c-a)

K=1

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