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If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to
$0$
$y(a-b)$
$y(b-a)$
$y(a-c)$
Solution
$a+x=b+y=c+z+1$
$\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right| \quad \quad C_{3} \rightarrow C_{3}-C_{1}$
$\left|\begin{array}{lll} x & a + y & a \\ y & b + y & b \\ z & c + y & c \end{array}\right| \quad \quad C _{2} \rightarrow C _{2}- C _{3}$
$\left|\begin{array}{lll}x & y & a \\ y & y & b \\ z & y & c\end{array}\right| \quad R_{3} \rightarrow R_{3}-R_{1}, R_{2} \rightarrow R_{2}-R_{1}$
$\left|\begin{array}{ccc}x & y & a \\ y-x & 0 & b-a \\ z-x & 0 & c-a\end{array}\right|$
$=(-y)[(y-x)(c-a)-(b-a)(z-x)]$
$=(-y)[(a-b)(c-a)+(a-b)(a-c-1)]$
$=(-y)[(a-b)(c-a)+(a-b)(a-c)+b-a)$
$=-y(b-a)=y(a-b)$