Animated Solution for Mathematics - Matrices and Determinants: Let $x\in \mathbb{R}$ and let $P=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right],Q=\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]$ and $R=PQ{P}^{-1}$. Then which of the following options is/are correct ?

A Journey Through Matrices

Welcome, future engineers. Today, we are not just solving a matrix problem; we are exploring the deep, structural beauty of linear algebra.

When you see a relation like $R=PQ{P}^{-1}$, I want you to stop. Do not immediately reach for your pen to multiply matrices. Instead, take a breath and recognize the signature of similar matrices. This is the heartbeat of the problem.

The Invariance of the Determinant

In the world of linear transformations, similarity is a powerful concept. It tells us that $R$ and $Q$ are essentially the same linear operator, just viewed from a different coordinate system defined by $P$.

Because they represent the same underlying transformation, they share fundamental properties—most importantly, their determinant. We know that $\det (R)=\det (PQ{P}^{-1})$.

Using the multiplicative property of determinants, we can write this as $\det (P)\cdot \det (Q)\cdot \det (P^{-1})$. Since $\det (P^{-1})=\frac{1}{\det (P)}$, they cancel out beautifully, leaving us with the elegant result: $\det (R)=\det (Q)$.

This realization saves us from the nightmare of calculating $P^{-1}$ and performing triple matrix multiplication. We have bypassed the brute force and gone straight to the core.

The Art of Expansion

Now, let us calculate $\det (Q)$. We have:

A novice would expand along the first row, but you are JEE aspirants—you look for the path of least resistance. The second row is filled with zeros, making it the perfect candidate for expansion.

Expanding along the second row, we get:

This expression, $48-4x^2$, is the key to unlocking the options. When we check Option 3, we see a matrix that is almost identical to $Q$, but with a $5$ in the bottom-right corner.

Calculating its determinant and adding $8$ yields $4(10-x^2)+8=40-4x^2+8=48-4x^2$. It matches perfectly! Option 3 is correct, and we have arrived there not by guessing, but by understanding the structure of the matrix.

The Eigenvector Bridge

Finally, let us tackle the most intimidating part: the eigenvector relation. We are given $R\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=6\left[\begin{array}{c}1\\ a\\ b\end{array}\right]$ for $x=0$. This is the classic eigenvalue equation $Rv=\lambda v$, where $\lambda =6$.

As we discussed in the FAQs, we don't need to find $R$ explicitly. We use the similarity bridge: $v=Pu$, where $u$ is the eigenvector of $Q$. When $x=0$, $Q$ becomes a diagonal matrix:

For a diagonal matrix, the eigenvectors are simply the standard basis vectors. Since we are looking for the eigenvalue $\lambda =6$, which corresponds to the third diagonal entry, our eigenvector $u$ must be the third standard basis vector, $u=\left[\begin{array}{c}0\\ 0\\ k\end{array}\right]$.

Now, we transform back to our original vector $v$ using $v=Pu$:

Comparing this to the given vector $\left[\begin{array}{c}1\\ a\\ b\end{array}\right]$, we immediately see that $k=1$, which implies $a=2$ and $b=3$. Thus, $a+b=5$. The logic holds, the math is clean, and the result is undeniable.

This problem was not about grinding through calculations; it was about recognizing the symmetry and the relationships hidden within the matrices. When you approach JEE Advanced problems, remember: look for the structure first, simplify the algebra second, and execute with confidence. You have the tools; now go forth and master the matrix.