Animated Solution for Mathematics - Matrices and Determinants: Given that $x=-9$ is a root of $\left|\begin{array}{ccc}x& 3& 7\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0$ the other two roots are ......... and .........

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a tedious algebraic grind. We have a $3\times 3$ determinant, and we are asked to find its roots:

If you were to jump straight into expanding this determinant, you would be faced with a cubic equation that could easily lead to a calculation error. But in the world of JEE Advanced, we don't just solve; we observe. We look for the hidden structure.

The Power of Summation

Look closely at the matrix. If you sum the elements of each column, you get $x+2+7=x+9$, $3+x+6=x+9$, and $7+2+x=x+9$.

This is not a coincidence! This is the matrix whispering its secret to you. By applying the row operation $R_1\to R_1+R_2+R_3$, we transform the first row into $[x+9,x+9,x+9]$.

Suddenly, the problem changes from a complex expansion to a simple factorization.

Factoring and Zero-Creation

Once we factor out $(x+9)$, we are left with a much friendlier determinant:

Now, we have a row of ones. Whenever you see a row of ones, your instinct should be to create zeros. By applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, we clear out the first row:

This is the beauty of linear algebra—we have reduced a daunting cubic expression into a simple product of factors.

Expanding along the first row is now trivial. We are left with the following equation:

This simplifies to the equation $(x+9)(x-2)(x-7)=0$. The roots are staring us in the face: $x=-9$, $x=2$, and $x=7$.

We were given $-9$, so the other two roots are $2$ and $7$. Remember, in JEE, the path of least resistance is often the path of most insight. Keep looking for those symmetries, and you will find that even the most intimidating problems have a simple, elegant solution waiting to be discovered.