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

Answer 1

The correct option is A $y(a-b)$

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

Now,
$=\left|\begin{array}{ccc}x& a+y& x+a\\ y& b+y& y+b\\ z& c+y& z+c\end{array}\right|$

Applying$(C_3\to C_3-C_1)$
$=\left|\begin{array}{ccc}x& a+y& a\\ y& b+y& b\\ z& c+y& c\end{array}\right|$

Applying$(C_2\to C_2-C_3)$
$=\left|\begin{array}{ccc}x& y& a\\ y& y& b\\ z& y& c\end{array}\right|$
$=y\left|\begin{array}{ccc}x& 1& a\\ y& 1& b\\ z& 1& c\end{array}\right|$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=y\left|\begin{array}{ccc}x& 1& a\\ y-x& 0& b-a\\ z-x& 0& c-a\end{array}\right|$
$=y[x\times 0-1\left\{\right(y-x\left)\right(c-a)-(b-a\left)\right(z-x\left)\right\}+a\times 0]$
$=y[bz-bx-az+ax-(cy-ay-cx+ax\left)\right]$
$=y[bz-bx-az-cy+ay+cx]$
$=y\left[b\right(z-x)+a(y-z)+c(x-y\left)\right]$
$=y[b\{a-c-1\}+a(c-b+1)+c(b-a\left)\right]$
$=y[ab-bc-b+ac-ab+a+bc-ac]$
$=y(a-b)$