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TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 45 of 160

Marks: +1, -0

If

\Delta_1

\begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix}

\Delta_2

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

\frac{\Delta_1}{\Delta_2}

=

ab + bc + ca
abc
2 (ab + bc + ca)
$(a+b+c)^2$

Solution:

The determinant

\Delta_1

is expressed as,

\Delta_1

=

\begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix}

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

The determinant

\Delta_{2}

is expressed as,

\Delta_2

=

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

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

The ratio is calculated as,

\frac{\Delta_{1}}{\Delta_{2}}

= \frac{-(a-b)(b-c)(c-a)(a b+b c+c a)}{(a-b)(c-b)(c-a)}

=(a b+b c+c a)

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