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NDA Math Determinants

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Question : 11 of 68

Marks: +1, -0

If

x=\frac{a}{b-c}, \;\; y=\frac{b}{c-a}, \;\; z=\frac{c}{a-b}

then what is the value of the following?

\begin{vmatrix} 1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1 \end{vmatrix}

0
1
$a b c$
$ab+bc+ca$

Solution: 👈: Video Solution

Given,

x=\frac{a}{b-c}, y=\frac{b}{c-a}, z=\frac{c}{a-b}

\;\; \begin{vmatrix} 1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1 \end{vmatrix}

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

and

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

=\begin{vmatrix} 1 & -x & x \\ 0 & 1+x & -y-x \\ 0 & z+x & 1-x \end{vmatrix}

=(1+x)(1-x)-(-y-x)(z+x)

=1-x^{2}+x^{2}+(y+z)x+yz

=1+\left(\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{a}{b-c}\right)

\;\; +\left(\frac{b}{c-a} \times \frac{c}{a-b}\right)

=1+\left(\frac{ab-b^{2}+c^{2}-ac}{(a-b)(c-a)}\right)\left(\frac{a}{b-c}\right)

+\left(\frac{bc}{(c-a)(a-b)}\right)

=\frac{(a-b)(c-a)+a^{2}-ab-ac+bc}{(a-b)(c-a)}

=\frac{ac-a^{2}-bc+ab+a^{2}-ab-ac+bc}{(a-b)(c-a)}

=0

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