Let x, y, z 1 and A = [[1, log_x y, log_x z], [log_y x, 2, log_y z], [log_z x, log_z y, 3]]. Then

Q: Why can't we just calculate the adjoint directly? It seems like a lot of work.

Calculating the adjoint of a $3 \times 3$ matrix is tedious and prone to arithmetic errors. In JEE Advanced, we always look for properties. The property $|\text{adj}(\text{adj} M)| = |M|^{(n-1)^2}$ is a powerful shortcut that saves time and minimizes risk. Always prioritize properties over brute force!

Q: Wait, why do we reject the negative root here?

Actually, we don't reject a root here, but we must be careful with the determinant properties. Remember that $|M^k| = |M|^k$. Since $|A|=2$, $|A^2|=4$. Always verify your powers and ensure you are applying the exponent to the determinant value, not the matrix elements themselves.

Q: Is the formula $|\text{adj}(\text{adj} M)| = |M|^{(n-1)^2}$ always true?

Yes, for any square matrix $M$ of order $n$, this identity holds. It is derived from the fundamental property $|\text{adj} M| = |M|^{n-1}$. Applying it twice gives $(|M|^{n-1})^{n-1} = |M|^{(n-1)^2}$. It is a standard result you should memorize for the exam.

Question

Let $x, y, z > 1$ and $A = \begin{bmatrix} 1 & \log_x y & \log_x z \ \log_y x & 2 & \log_y z \ \log_z x & \log_z y & 3 \end{bmatrix}$. Then $|	ext{adj}(	ext{adj} A^2)|$ is equal to

Options:

(A) $6^4$

(B) $2^8$

(C) $4^8$

(D) $2^4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of logarithms and matrices. You see a matrix $A$ filled with terms like $\log_x y$ and $\log_z x$, and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for hidden simplicity. Our job is to peel back that mask.

### Phase 1: The Change of Base

Look at the matrix:

$$A = \begin{bmatrix} 1 & \log_x y & \log_x z \ \log_y x & 2 & \log_y z \ \log_z x & \log_z y & 3 \end{bmatrix}$$

The bases are all over the place: $x, y, z$. This is the primary source of our confusion. We cannot operate on a matrix where the language of the elements is inconsistent.

We need a common tongue. That language is the natural logarithm. Using the change of base property, $\log_a b = \frac{\ln b}{\ln a}$, we can rewrite every single element.

Suddenly, the matrix transforms into a beautiful, structured array of ratios:

$$|A| = \begin{vmatrix} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \ \frac{\ln x}{\ln y} & 2 & \frac{\ln z}{\ln y} \ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 3 \end{vmatrix}$$

### Phase 2: The Great Simplification

Now, look at the rows. The first row is dominated by $\frac{1}{\ln x}$, the second by $\frac{1}{\ln y}$, and the third by $\frac{1}{\ln z}$. This is not a coincidence; it is a design.

Let us factor these out. When we pull $\frac{1}{\ln x}$ from $R_1$, $\frac{1}{\ln y}$ from $R_2$, and $\frac{1}{\ln z}$ from $R_3$, we are left with a matrix where the numerators are $\ln x, \ln y, \ln z$.

But wait—the columns also have these terms! If we factor $\ln x$ from $C_1$, $\ln y$ from $C_2$, and $\ln z$ from $C_3$, the denominators and numerators cancel out perfectly. The logarithms vanish entirely.

We are left with the simple, elegant integer matrix:

$$|A| = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 3 \end{vmatrix}$$

### Phase 3: The Determinant Reveal

Evaluating this is a breath of fresh air. Expanding along the first row, we get:

$$|A| = 1(2 	imes 3 - 1 	imes 1) - 1(1 	imes 3 - 1 	imes 1) + 1(1 	imes 1 - 1 	imes 2)$$

This simplifies to:

$$|A| = 1(5) - 1(2) + 1(-1) = 5 - 2 - 1 = 2$$

The determinant of $A$ is $2$. This is the moment where the hard work pays off. We have tamed the beast.

### Phase 4: The Adjoint Power Play

We are asked for $|	ext{adj}(	ext{adj} A^2)|$. First, we handle the square. Using the property $|M^k| = |M|^k$, we find:

$$|A^2| = |A|^2 = 2^2 = 4$$

Now, we apply the property of the double adjoint: $|	ext{adj}(	ext{adj} M)| = |M|^{(n-1)^2}$. With $n=3$, the exponent becomes $(3-1)^2 = 4$.

Thus, we need to calculate $4^4$:

$$4^4 = (2^2)^4 = 2^8 = 256$$

We have arrived at the destination. The complexity was an illusion, and through systematic application of properties, we found the truth. The final answer is **256**.

Animated Solution for Mathematics - Matrices and Determinants: Let $x, y, z > 1$ and $A = 1 lo g_{y} x lo g_{z} x lo g_{x} y 2 lo g_{z} y lo g_{x} z lo g_{y} z 3$ . Then $∣ adj (adj A^{2}) ∣$ is equal to

Select Answer:

Visualized Solution

Introduction to Matrix $A$

Change of Base Property

Rewriting $∣ A ∣$

Factoring Denominators

Factoring Numerators

Evaluating the Determinant

Finding $∣ A^{2} ∣$

The Double Adjoint Property

Applying the Property

Final Computation

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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