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If a+x=b+y=c+z+1 wherea,b,c,x,y,z are non zero distinct real nos then \begin{vmatrix} x &a+y &x+a \\ y &b+y &y+b \\z &c+y &z+c \end{vmatrix} is equal to:
Option: 1 y(a-c)
Option: 2 y(a-b)
Option: 3 y(b-a)
Option: 4 0

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best_answer

Given that a + x = b + y = c + z + 1 = 0

\\\begin{vmatrix} x &a+y &x+a \\ y & b+y &y+b \\ z &c+y &z+c \end{vmatrix}=\begin{vmatrix} x &a &x+a \\ y& b &y+b \\ z & c & z+c \end{vmatrix}+\begin{vmatrix} x &y &x+a \\ y & y &y+b \\ z & y &z+c \end{vmatrix}\\\text{Since,}\begin{vmatrix} x &a &x+a \\ y& b &y+b \\ z & c & z+c \end{vmatrix}=0\\\because \begin{vmatrix} x &a &x+a \\ y& b &y+b \\ z & c & z+c \end{vmatrix}=\begin{vmatrix} x &a &a \\ y& b &b \\ z & c & c \end{vmatrix}+\begin{vmatrix} x &a &x \\ y& b &y \\ z & c & z \end{vmatrix}=0+0

Now

\\\begin{vmatrix} x &y &x+a \\ y & y &y+b \\ z & y &z+c \end{vmatrix}=y\begin{vmatrix} x &1 &x+a \\ y & 1 &y+b \\ z & 1 &z+c \end{vmatrix}\\y\begin{vmatrix} x &1 &x+a \\ y & 1 &y+b \\ z & 1 &z+c \end{vmatrix}=y\begin{vmatrix} x &1 &x \\ y & 1 &y \\ z & 1 &z \end{vmatrix}+y\begin{vmatrix} x &1 &a \\ y & 1 &b \\ z & 1 &c \end{vmatrix}\\\Rightarrow y\begin{vmatrix} x &1 &a \\ y & 1 &b \\ z & 1 &c \end{vmatrix}\\C_1\rightarrow C_1+C_3

\\y\begin{vmatrix} x +a&1 &a \\ y+b & 1 &b \\ z+c & 1 &c \end{vmatrix}=y\begin{vmatrix} x +a&1 &a \\ x+a& 1 &b \\ x+a-1 & 1 &c \end{vmatrix}\\R_1\rightarrow R_1-R_2\\R_2\rightarrow R_2-R_3\\\Rightarrow y\begin{vmatrix} 0&0 &a-b \\ 1 & 0 &b-c \\ x+a-1 & 1 &c \end{vmatrix}=y((a-b))

 

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himanshu.meshram

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