Animated Solution for Mathematics - Matrices and Determinants: 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 Journey Through Symmetry

Hello, future engineers! Today, we are going to tackle a problem that looks like a tangled mess of variables at first glance. We are given the condition $a+x=b+y=c+z+1$ and asked to evaluate a $3\times 3$ determinant.

If you feel intimidated, take a deep breath. In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions. We just need to find the right key to unlock them.

When you see a chain of equalities like $a+x=b+y=c+z+1$, your first instinct should be to introduce a constant. Let us define this constant as $k$.

By setting $a+x=b+y=c+z+1=k$, we have effectively decoupled the variables. Now, we can express $a,b,$ and $c$ in terms of $k$ and the other variables: $a=k-x$, $b=k-y$, and $c=k-z-1$.

This substitution is the turning point of our journey. It transforms a problem of six variables into a much more manageable structure.

Now, let us look at our determinant:

By substituting our expressions for $a,b,$ and $c$, the third column becomes $x+(k-x)=k$, $y+(k-y)=k$, and $z+(k-z-1)=k-1$. The second column becomes $a+y=(k-x)+y$, $b+y=(k-y)+y=k$, and $c+y=(k-z-1)+y$.

Suddenly, the determinant looks much more structured:

This is where the magic happens. We want to create zeros to make the expansion easier. Let us apply the column operation $C_2\to C_2-C_3$.

In the first row, the second column becomes $(k-x+y)-k=y-x$. In the second row, $k-k=0$. In the third row, $(k-z-1+y)-(k-1)=y-z$.

Our determinant is now:

We have successfully created a zero! Now, let us apply $R_1\to R_1-R_2$ to create another zero in the first row. The first row becomes $x-y$, $y-x$, and $0$.

Since $x-y=-(y-x)$, we can factor out $(y-x)$ from the first row. This leaves us with:

Expanding along the first row is now a breeze. We get:

Simplifying this expression, we find that the $ky$ and $kz$ terms cancel out beautifully, leaving us with $\Delta =(y-x)[y]$.

Finally, recalling our original condition $a+x=b+y$, we know that $y-x=a-b$. Substituting this back, we arrive at the final result:

And there it is! The complexity has vanished, leaving behind a simple, elegant solution. Remember, in JEE, always look for the symmetry, use the constant $k$ trick, and trust the process of row and column operations. You have got this!