Let A be a 3 × 3 matrix such that |adj(adj(adj A))| = 12⁴. Then |A⁻¹ adj A| is equal to (1) 2√3 (2)

Q: Wait, why is the exponent $(n-1)^k$ and not just $(n-1) \times k$?

That is a fantastic question! Let us look at the logic. For one adjoint, we have $|\text{adj} A| = |A|^{n-1}$. When we take the adjoint again, say $B = \text{adj} A$, then $|\text{adj} B| = |B|^{n-1} = (|A|^{n-1})^{n-1} = |A|^{(n-1)^2}$. Each time we wrap the matrix in an adjoint, we raise the existing power to the $(n-1)$ power again, which leads to exponentiation, not multiplication.

Q: Why can we just split $|A^{-1} \text{adj} A|$ into $|A^{-1}| \cdot |\text{adj} A|$?

This is a direct application of the multiplicative property of determinants, which states that for any two square matrices $X$ and $Y$ of the same order, $|XY| = |X| \cdot |Y|$. Since $A^{-1}$ and $\text{adj} A$ are both $3 \times 3$ matrices, this rule holds perfectly.

Q: Is there any scenario where $|A|$ could be negative?

Yes, absolutely! The determinant can be negative. However, in this specific problem, we are given $|\text{adj}(\text{adj}(\text{adj} A))| = 12^4$. Since the exponent $8$ is even, $|A|^8 = 12^4$ implies $|A|^2 = 12$, so $|A| = \pm 2\sqrt{3}$. In most JEE problems of this type, we focus on the magnitude or the positive root unless specified otherwise, but keep in mind that the determinant can indeed be negative.

Question

Let $A$ be a $3 	imes 3$ matrix such that $|	ext{adj}(	ext{adj}(	ext{adj} A))| = 12^4$. Then $|A^{-1} 	ext{adj} A|$ is equal to

Options:

(A) $2\sqrt{3}$

(B) $\sqrt{6}$

(C) 12

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the journey of linear algebra. Today, we are going to demystify a problem that often intimidates students: the nested adjoint.

When you see $|	ext{adj}(	ext{adj}(	ext{adj} A))| = 12^4$, it is natural to feel a bit overwhelmed. But remember, in mathematics, complexity is often just a mask for a beautiful, underlying simplicity. Let us peel back that mask together.

### The Power of the Single Adjoint

Before we leap into the deep end, let us stand on solid ground. The adjoint of a matrix is a special beast, but it obeys a very elegant rule:

$$|	ext{adj} A| = |A|^{n-1}$$

Here, $n$ is the order of our matrix. For our $3 	imes 3$ matrix, $n=3$, so the determinant of the adjoint is simply $|A|^{3-1} = |A|^2$.

This is our key. It is the bridge between the world of adjoints and the world of simple determinants.

### The Generalization

Now, what happens when we nest these operations? Imagine we have a machine that takes a matrix and spits out its adjoint.

If we feed $A$ into it, we get $	ext{adj} A$. If we feed that result back into the machine, we get $	ext{adj}(	ext{adj} A)$. Mathematically, this is a recursive process.

As we derived in our FAQs, each layer of the adjoint operation raises the power of the determinant by $(n-1)$. Thus, for $k$ layers, we arrive at the general formula:

$$|	ext{adj}_k(A)| = |A|^{(n-1)^k}$$

In our problem, $n=3$ and $k=3$. So, the exponent becomes $(3-1)^3 = 2^3 = 8$.

Our equation transforms from a terrifying nested expression into the clean, manageable algebraic equation:

$$|A|^8 = 12^4$$

### The Final Simplification

Now, we solve for $|A|$. Taking the fourth root of both sides, we get $|A|^2 = 12$, which means $|A| = \sqrt{12} = 2\sqrt{3}$.

We have conquered the hardest part! Now, look at our target: $|A^{-1} 	ext{adj} A|$.

Using the multiplicative property of determinants, we split this into $|A^{-1}| \cdot |	ext{adj} A|$. We know that $|A^{-1}| = \frac{1}{|A|}$ and $|	ext{adj} A| = |A|^2$.

When we multiply these, we get:

$$\frac{1}{|A|} \cdot |A|^2 = |A|$$

The entire expression collapses beautifully into just the determinant of $A$. Since we already found $|A| = 2\sqrt{3}$, our **final answer is $2\sqrt{3}$**.

### Conclusion

You see? The problem was never about brute-forcing matrices. It was about recognizing the structure, applying the properties, and watching the complexity vanish.

Keep this logic in your toolkit, and no matrix problem will ever stand in your way again.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

Select Answer:

Visualized Solution

Introduction to the Problem

Property of a Single Adjoint

Generalizing for Nested Adjoints

Identifying Parameters $n$ and $k$

Setting up the Equation

Simplifying the Powers

Solving for $∣ A ∣^{2}$

Calculating the Determinant of $A$

Analyzing the Target Expression

Substituting Individual Properties

Final Simplification

Conclusion and Key Takeaway

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Power of the Single Adjoint

The Generalization

The Final Simplification

Conclusion

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 3×3 matrix such that ∣adj(adj(adjA))∣=124. Then ∣A−1adjA∣ is equal to