If a+x=b+y=c+z+1, where a, b, c, x, y , z

Question:

If $a+x=b+y=c+z+1$, where $a, b, c, x$, $\mathrm{y}, \mathrm{z}$ are non-zero distinct real numbers, then

$\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to :

1. 0

2. $y(a-b)$

3. $\mathrm{y}(\mathrm{b}-\mathrm{a})$

4. $\mathrm{y}(\mathrm{a}-\mathrm{c})$

Correct Option: , 2

Solution:

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

$\left|\begin{array}{ccc}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right| \quad \quad C_{3} \rightarrow C_{3}-C_{1}$

$\left|\begin{array}{lll}x & a+y & a \\ y & b+y & b \\ z & c+y & c\end{array}\right| \quad \quad C_{2} \rightarrow C_{2}-C_{3}$

$\left|\begin{array}{lll}x & y & a \\ y & y & b \\ z & y & c\end{array}\right| \quad R_{3} \rightarrow R_{3}-R_{1}, R_{2} \rightarrow R_{2}-R_{1}$

$\left|\begin{array}{ccc}x & y & a \\ y-x & 0 & b-a \\ z-x & 0 & c-a\end{array}\right|$

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

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

$=(-y)[(a-b)(c-a)+(a-b)(a-c)+b-a)$

$=-y(b-a)=y(a-b)$