Let P = [[-30, 20, 56], [90, 140, 112], [120, 60, 14]] and A = [[2, 7, ω²], [-1, -ω, 1], [0, -ω, -ω+

Q: Wait, why can't we just calculate $P^{-1}$ and then do the matrix multiplication? It seems straightforward.

While technically possible, it is a massive trap! In JEE Advanced, time is your most precious resource. Calculating the inverse of a $3 \times 3$ matrix with large elements is prone to arithmetic errors and will consume valuable minutes. Always look for algebraic properties first—they are designed to bypass the brute force.

Q: Why does $\det(P^{-1}MP) = \det(M)$? Is this always true?

Yes, it is a fundamental property of determinants! Since $\det(P^{-1}MP) = \det(P^{-1}) \cdot \det(M) \cdot \det(P)$ and $\det(P^{-1}) = \frac{1}{\det(P)}$, the determinants of $P$ and $P^{-1}$ cancel out perfectly. It is a beautiful piece of matrix algebra that simplifies complex transformations into something manageable.

Q: I keep getting stuck on the complex number part. How do I handle $\omega$ in the determinant?

Don't panic! Whenever you see $\omega$, immediately recall the two golden rules: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Use these to simplify your expressions at every step. If you find yourself with a high power of $\omega$, reduce it using $\omega^3 = 1$. If you have a sum, look for the $1 + \omega + \omega^2$ identity.

Question

Let $P = \begin{bmatrix} -30 & 20 & 56 \ 90 & 140 & 112 \ 120 & 60 & 14 \end{bmatrix}$ and $A = \begin{bmatrix} 2 & 7 & \omega^2 \ -1 & -\omega & 1 \ 0 & -\omega & -\omega+1 \end{bmatrix}$ where $\omega = \frac{-1+i\sqrt{3}}{2}$, and $I_3$ be the identity matrix of order 3. If the determinant of the matrix $(P^{-1}AP - I_3)^2$ is $\alpha\omega^2$, then the value of $\alpha$ is equal to ________

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: A Masterclass in Matrix Elegance

Imagine you are sitting in the exam hall. You see a matrix $P$ filled with large, intimidating numbers like $-30, 20, 56, 90, 140, 112, 120, 60, 14$.

Your first instinct might be to start calculating its inverse, $P^{-1}$. Stop! Take a deep breath.

In the world of JEE Advanced, whenever you see a matrix like $P$ inside a similarity transformation—specifically in the form $P^{-1}AP$—it is almost certainly a psychological trap. The examiners want to see if you will waste ten minutes on arithmetic or if you have the mathematical maturity to see the structure.

### The Algebraic Surgery

Let us perform some algebraic surgery on the expression $\det((P^{-1}AP - I_3)^2)$. We want to simplify the core of this expression.

We know that the identity matrix $I_3$ is special; it is the multiplicative identity. We can write $I_3$ as $P^{-1}I_3P$ without changing its value.

Now, look at our expression: $P^{-1}AP - P^{-1}I_3P$. By factoring out $P^{-1}$ on the left and $P$ on the right, we get $P^{-1}(A - I_3)P$. This is the moment of clarity. The structure has collapsed from a terrifying mess into a clean, elegant form.

### The Determinant Magic

Now, we need to square this. We are looking at $(P^{-1}(A - I_3)P)^2$, which is $(P^{-1}(A - I_3)P) \cdot (P^{-1}(A - I_3)P)$.

Notice the middle terms: $P \cdot P^{-1}$. That is just the identity matrix $I_3$! They vanish, leaving us with $P^{-1}(A - I_3)^2P$.

When we take the determinant, we use the multiplicative property: $\det(XYZ) = \det(X)\det(Y)\det(Z)$. The determinant of $P^{-1}$ and the determinant of $P$ are reciprocals, so they cancel out completely. We are left with $\det((A - I_3)^2)$, which is simply $(\det(A - I_3))^2$. The scary matrix $P$ is gone. It was never meant to be calculated.

### The Complex Dance

Now, we focus on $A - I_3$. We subtract $1$ from the diagonal elements of $A$. This gives us the matrix:

$$\begin{bmatrix} 1 & 7 & \omega^2 \ -1 & -\omega-1 & 1 \ 0 & -\omega & -\omega \end{bmatrix}$$

We expand this determinant along the first column to take advantage of the zero. The calculation leads us to $(\omega^2 + 2\omega) + (-7\omega + 1)$.

Combining these, we get $\omega^2 - 5\omega + 1$. Here is where the magic of complex numbers shines. We know $1 + \omega + \omega^2 = 0$, so $\omega^2 + 1 = -\omega$.

Substituting this in, our determinant becomes $-\omega - 5\omega = -6\omega$. Finally, we square this result: $(-6\omega)^2 = 36\omega^2$.

Comparing this to the given $\alpha\omega^2$, we find that **$\alpha = 36$**. You have just conquered a problem that was designed to break the unprepared. Keep this mindset—look for the structure, trust the properties, and never let the numbers intimidate you.

Enter Numerical Value:

Visualized Solution

Analyzing the Given Expression

Factoring the Identity Matrix

Squaring the Expression

Applying Determinant Properties

Simplifying the Goal

Constructing Matrix $A - I_{3}$

Setting up the Determinant Expansion

Computing the First Minor

Computing the Second Minor

Combining the Results

Using $ω$ Properties

Finding the Final Determinant

Comparing and Finding $α$

The Way Forward

The Sigma Insight: Properties of Determinants

The Illusion of Complexity

The Algebraic Surgery

The Determinant Magic

The Complex Dance

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