Question

If $a+x=b+y=c+z+1,$ where $a,b,c,x$ $y,z$ are non-zero distinct real numbers, then $\left|\begin{array}{ccc}x& a+y& x+a\\ y& b+y& y+b\\ z& c+y& z+c\end{array}\right|$ is equal to :

• A. 0
• B. $y(a-b)$
• C. $y(b-a)$
• D. $y(a-c)$

Answer 1

Answer: (B)

$a+x=b+y=c+z+1$
$\left|\begin{array}{ccc}x& a+y& x+a\\ y& b+y& y+b\\ z& c+y& z+c\end{array}\right|$
$C_3\to C_3-C_1$
$\left|\begin{array}{ccc}x& a+y& a\\ y& b+y& b\\ z& c+y& c\end{array}\right|$
$C_2\to C_2-C_3$
$\left|\begin{array}{ccc}x& y& a\\ y& y& b\\ z& y& c\end{array}\right|$
$R_3\to R_3-R_1,R_2\to R_2-R_1$
$\left|\begin{array}{ccc}x& y& a\\ y-x& 0& b-a\\ z-x& 0& c-a\end{array}\right|$
$=(-y)[(y-x)(c-a)-(b-a)(z-x)]$
$=(-y)[(a-b)(c-a)+(a-b)(a-c-1)]$
$=(-y)[(a-b)(c-a)+(a-b)(a-c)+b-a)$
$=-y(b-a)=y(a-b)$