Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:

y(c-a) may be it is helpfully for you

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle\begin{vmatrix} x & a + y & a + x\\ y & b + y & b + y \\ z & c + y & c + z \end{vmatrix}$

$= \displaystyle\begin{vmatrix} x & a + y & a \\ y & b + y & b \\ z & c + y & c \end{vmatrix} \: \: (\because \: C_3'=C_3 - C_1 \: \: )$

$= \displaystyle\begin{vmatrix} x & y & a \\ y & y & b \\ z & y & c \end{vmatrix} \: \: (\because \: C_2'=C_2 - C_3 \: \: )$

$= y\displaystyle\begin{vmatrix} x & 1 & a \\ y & 1 & b \\ z & 1 & c \end{vmatrix} \: \: (taking \: \: y \:as \: common \: from \: 2nd \: column)$

$= y\displaystyle\begin{vmatrix} x + a & 1 & a \\ y + b & 1 & b \\ z + c & 1 & c \end{vmatrix} \: \: (\because \: C_1'=C_1 + C_3 \: \: )$

$= y\displaystyle\begin{vmatrix} x + a & 1 & a \\ x + a - 1 & 1 & b \\ x + a & 1 & c \end{vmatrix} \: \: (\because by \: the \: given \: condition)$

$= y\displaystyle\begin{vmatrix} x + a & 1 & a \\ - 1 & 0 & b - a \\ 0 & 0 & c - a \end{vmatrix} \: \: (\because \: R_2'=R_2 - R_1 \: \: and \: \:R_3'=R_3 - R_1 \: )$

$= y(c - a) \: \: \: ( on \: expansion \: )$