Solve the following
Question:

If $a+x=b+y=c+z+1$, where $a, b, c, x, y, z$ are non-zero

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

  1. (1) $y(b-a)$

  2. (2) $y(a-b)$

  3. (3) 0

  4. (4) $y(a-c)$


Correct Option: 2,

Solution:

(2) Use properties of determinant

$\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|=\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|+y\left|\begin{array}{ccc}x & 1 & x+a \\ y & 1 & y+b \\ z & 1 & z+c\end{array}\right|$

$=0+y\left|\begin{array}{ccc}x & 1 & x+a \\ y-x & 0 & 0 \\ z-x & 0 & -1\end{array}\right| \quad\left[\begin{array}{l}R_{2} \rightarrow R_{2}-R_{1}, \\ R_{3} \rightarrow R_{3}-R_{1}\end{array}\right]$

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

Administrator

Leave a comment

Please enter comment.
Please enter your name.