If a+x = b+y = c+z+1, where a,b,c,x,y,z are non-zero distinct real numbers, then \(\begin{vmatrix} x &a+y & x+a\\ y & b+y & y+b\\ z & c+y & z+c \end{vmatrix}\) is equal to:

1) y(a-b)

2) 0

3) y(b-a)

4) y(a-c)

Solution:

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

Now, \(\begin{vmatrix} x &a+y & x+a\\ y & b+y & y+b\\ z & c+y & z+c \end{vmatrix}\)

= \(\begin{vmatrix} x &a+y & a\\ y & b+y & b\\ z & c+y & c \end{vmatrix}\) (C3 → C3-C1)

= \(\begin{vmatrix} x &y & a\\ y & y & b\\ z & y & c \end{vmatrix}\) (C2 → C2-C3) = \(y\begin{vmatrix} x &1 & a\\ y & 1 & b\\ z & 1 & c \end{vmatrix}\)

R2 → R2 – R1 and R3 → R3 – R1

= \(y\begin{vmatrix} x &1 & a\\ y-x & 0 & b-a\\ z-x & 0 & c-a \end{vmatrix}\)

= y[x×0-1{(y-x)(c-a)-(b-a)(z-x)}+a×0]

= y[bz-bx-az+ax-(cy-ay-cx+ax)]

= y[bz-bx-az-cy+ay+cx]

= y[b(z-x)+a(y-z)+c(x-y)]

= y[b{a-c-1}+a(c-b+1)+c(b-a)]

= y[ab-bc-b+ac-ab+a+bc-ac]

= y(a-b)

Answer: (1)

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