Consider the determinant Δ=| array lll a_ 11 & a_ 12 & a_ 13 a_ 21 & a_ 22 & a_ 23 a_ 31 & a_ 32 & a_ 33 array | If a_ 13 =y z, a_ 23 =z x, a_ 33 =x y and the minors of a_ 13 , a_ 23 , a_ 33 are respectively

Correct answer: A. (z-y)(z-x)(y-x) Δ=| array lll a_ 11 & a_ 12 & a_ 13 a_ 21 & a_ 22 & a_ 23 a_ 31 & a_ 32 & a_ 33 array |a_ 13 =y z, a_ 23 =z x, a_ 33 =x y and M_ 13 =(z-y), M_ 23 =(z-x), M_ 33 =(y-x) Expanding the determinant along Column 3 , aligned & Δ=a_ 13 (a_ 21 a_ 32 -a_ 22 a_ 31 )-a_ 23 (a_ 11 a_ 32 -a_ 12 a_ 31 )+a_ 33 (a_ 11 a_2-a_ 12 a_1) & Δ=a_ 13 M_ 13 -a_ 23 M_ 23 +a_ 33 M_ 33 aligned where M_ i j denotes the minor of a_ i j element. aligned & Δ=y z(z-y)-z x(z-x)+x y(y-x) & Δ=y z(z-y)-z^2 x+z x^2+x y^2-x^2 y & Δ=y z(z-y)+(x y^2-z^2 x)+(z x^2-x^2 y) & Δ=y z(z-y)+x (y^2-z^2)+x^2(z-y) & Δ=(z-y) [y z-x(y+z)+x^2] & Δ=(z-y) [y z-x y-x z+x^2] & Δ=(z-y)[y(z-x)-x(z-x)] & Δ=(z-y)[(z-x)(y-x)] aligned

NDA 2022

Exam: NDA
Year: 2022
Subject: Mathematics
Chapter: Determinants

NDA 2022 · Mathematics · Determinants

Consider the determinant Δ=| array lll a_ 11 & a_ 12 & a_ 13 a_ 21...

Consider the determinant

\Delta=\left|

\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}

\right|

a_{13}=y z, a_{23}=z x, a_{33}=x y

and the minors of

a_{13}, a_{23}

a_{33}

are respectively

(z-y),(z-x),(y-x)

then what is the value of

\Delta

Options

A $(z-y)(z-x)(y-x)$
B $(x-y)(y-z)(x-z)$
C $(x-y)(z-x)(y-z)(x+y+z)$
D $(x y+y z+z x)(x+y+z)$

Correct answer

A. $(z-y)(z-x)(y-x)$

Step-by-step solution

\Delta=\left|

\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}

\right|a_{13}=y z, a_{23}=z x, a_{33}=x y

\text{\text{ and } }

M_{13}=(z-y), M_{23}=(z-x), M_{33}=(y-x)

Expanding the determinant along Column 3 ,

\begin{aligned} & \Rightarrow \Delta=a_{13}\left(a_{21} a_{32}-a_{22} a_{31}\right)-a_{23}\left(a_{11} a_{32}-a_{12} a_{31}\right)+a_{33}\left(a_{11} a_2-a_{12} a_1\right) \\ & \Rightarrow \Delta=a_{13} M_{13}-a_{23} M_{23}+a_{33} M_{33} \end{aligned}

where

M_{i j}

denotes the minor of

a_{i j}

element.

\begin{aligned} & \Rightarrow \Delta=y z(z-y)-z x(z-x)+x y(y-x) \\ & \Rightarrow \Delta=y z(z-y)-z^2 x+z x^2+x y^2-x^2 y \\ & \Rightarrow \Delta=y z(z-y)+\left(x y^2-z^2 x\right)+\left(z x^2-x^2 y\right) \\ & \Rightarrow \Delta=y z(z-y)+x \left(y^2-z^2\right)+x^2(z-y) \\ & \Rightarrow \Delta=(z-y) \left[y z-x(y+z)+x^2\right] \\ & \Rightarrow \Delta=(z-y) \left[y z-x y-x z+x^2\right] \\ & \Rightarrow \Delta=(z-y)[y(z-x)-x(z-x)] \\ & \Rightarrow \Delta=(z-y)[(z-x)(y-x)] \end{aligned}

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