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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Determinants

NDA 2021 · Mathematics · Determinants

If x=(a)/(b-c), y=(b)/(c-a), z=(c)/(a-b) then what is the value of...

If x=ab−c,y=bc−a,z=ca−bx=\frac{a}{b-c}, y=\frac{b}{c-a}, z=\frac{c}{a-b} then what is the value of the following? \left|1−xx11−y1z1\begin{array}{ccc} 1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1 \end{array}$$\right|

Options

  1. A0
  2. B1
  3. Cabca b c
  4. Dab+bc+caa b+b c+c a

Correct answer

A. 0

Step-by-step solution

\left|1−xx11−y1z1\begin{array}{ccc}1 & -x & x \\ 1 & 1 & -y \\ 1 & z & 1\end{array}\right|Applying \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1andand\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1=∣1−xx01+x−y−x0z+x1−x∣=1+xy+yz+zx\begin{aligned} & =\left|\begin{array}{ccc} 1 & -x & x \\ 0 & 1+x & -y-x \\ 0 & z+x & 1-x \end{array}\right| \\ & =1+x y+y z+z x \end{aligned}=1+ab(b−c)(c−a)+bc(c−a)(a−b)+ca(b−c)(a−b)=(a−b)(b−c)(c−a)+ab(a−b)+bc(b−c)+ca(c−a)(a−b)(b−c)(c−a)=0\begin{array}{r}=1+\frac{a b}{(b-c)(c-a)}+\frac{b c}{(c-a)(a-b)}+\frac{c a}{(b-c)(a-b)} \\ =\frac{(a-b)(b-c)(c-a)+a b(a-b)+b c(b-c)+c a(c-a)}{(a-b)(b-c)(c-a)}=0\end{array}
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