Let A and B be two square matrices of order 3 such that |A| = 3 and |B| = 2. Then |AᵀA(adj(2A))⁻¹(ad

Q: Why can't I just calculate the matrices? I don't know what A and B are!

That is the beauty of this problem! You are not meant to find the matrices. The problem is designed to test your mastery of determinant properties. The specific values of the matrices are irrelevant; only their determinants matter.

Q: Why is $|kM| = k^3|M|$? Shouldn't it just be $k|M|$?

Great question! Remember that a determinant is a multilinear function of the rows. When you multiply a matrix by a scalar $k$, you are multiplying *every* row by $k$. Since there are 3 rows in a $3 \times 3$ matrix, you pull out a factor of $k$ from each row, resulting in $k \times k \times k = k^3$.

Q: Why did the $|A|$ and $|B|$ terms cancel out? Is that a coincidence?

It is not a coincidence at all! It is the hallmark of a well-constructed JEE problem. The expression is designed so that the variables cancel out, leaving you with a pure numerical result. If they didn't cancel, the answer would depend on the values of $|A|$ and $|B|$, which would make the problem ill-posed.

Question

Let $A$ and $B$ be two square matrices of order 3 such that $|A| = 3$ and $|B| = 2$. Then $|A^T A (	ext{adj}(2A))^{-1} (	ext{adj}(4B)) (	ext{adj}(AB))^{-1} A A^T|$ is equal to:

Options:

(A) 108

(B) 32

(C) 81

(D) 64

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating matrix expression. It looks like a tangled web of transposes, adjoints, and inverses. In the world of JEE Advanced, this is a classic psychological trap.

The problem asks for the determinant of $E = |A^T A (	ext{adj}(2A))^{-1} (	ext{adj}(4B)) (	ext{adj}(AB))^{-1} A A^T|$. Your first instinct might be to panic, but let us take a deep breath. We are not here to compute matrices; we are here to dismantle them using the elegant laws of linear algebra.

### The Toolkit

Before we touch the expression, let us sharpen our tools. We have five fundamental properties that will act as our scalpel:

1. The product property: $|XY| = |X| \cdot |Y|$.
2. The transpose property: $|A^T| = |A|$.
3. The inverse property: $|M^{-1}| = \frac{1}{|M|}$.
4. The adjoint property: $|	ext{adj}(M)| = |M|^{n-1}$.
5. The scalar property: $|kM| = k^n |M|$.

Since our matrices are of order $n=3$, the adjoint property simplifies to $|	ext{adj}(M)| = |M|^2$, and the scalar property becomes $|kM| = k^3 |M|$. These are the keys to the kingdom.

### The Simplification

Let us apply the product property to our expression. We can split the single large determinant into a product of several individual determinants:

$$E = |A^T| \cdot |A| \cdot |(	ext{adj}(2A))^{-1}| \cdot |	ext{adj}(4B)| \cdot |(	ext{adj}(AB))^{-1}| \cdot |A| \cdot |A^T|$$

Using the transpose property, we know $|A^T| = |A|$. So, we have $|A| \cdot |A| \cdot |A| \cdot |A|$, which is $|A|^4$.

Now, let us handle the inverses and adjoints. Using the inverse property, $|(	ext{adj}(2A))^{-1}| = \frac{1}{|	ext{adj}(2A)|}$ and $|(	ext{adj}(AB))^{-1}| = \frac{1}{|	ext{adj}(AB)|}$. Substituting these, our expression becomes:

$$E = |A|^4 \cdot \frac{1}{|	ext{adj}(2A)|} \cdot |	ext{adj}(4B)| \cdot \frac{1}{|	ext{adj}(AB)|}$$

Now, apply the adjoint property ($|	ext{adj}(M)| = |M|^2$):

$$E = |A|^4 \cdot \frac{1}{|2A|^2} \cdot |4B|^2 \cdot \frac{1}{|AB|^2}$$

### The Grand Cancellation

We are almost there. Now we use the scalar property. For $|2A|$, we pull out the scalar $2$, which becomes $2^3 = 8$. So, $|2A| = 8|A|$. Squaring this gives:

$$|2A|^2 = (8|A|)^2 = 64|A|^2 = 2^6 |A|^2$$

Similarly, for $|4B|$, we pull out $4$, which becomes $4^3 = 64 = 2^6$. So, $|4B| = 2^6 |B|$. Squaring this gives:

$$|4B|^2 = (2^6 |B|)^2 = 2^{12} |B|^2$$

Finally, $|AB|^2 = (|A| \cdot |B|)^2 = |A|^2 |B|^2$. Now, let us assemble the pieces:

$$E = |A|^4 \cdot \frac{1}{2^6 |A|^2} \cdot 2^{12} |B|^2 \cdot \frac{1}{|A|^2 |B|^2}$$

Look at the numerator and denominator. The $|A|^4$ in the numerator cancels perfectly with the $|A|^2 \cdot |A|^2$ in the denominator. The $|B|^2$ in the numerator cancels with the $|B|^2$ in the denominator.

We are left with:

$$E = \frac{2^{12}}{2^6} = 2^{12-6} = 2^6 = \mathbf{64}$$

The variables vanished, leaving behind a clean, elegant integer. This is the beauty of mathematics—the complexity dissolves when you apply the right principles.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ and $B$ be two square matrices of order 3 such that $∣ A ∣ = 3$ and $∣ B ∣ = 2$ . Then $∣ A^{T} A (adj (2 A))^{- 1} (adj (4 B)) (adj (A B))^{- 1} A A^{T} ∣$ is equal to:

Select Answer:

Visualized Solution

Analyze the Determinant Expression

Apply Product and Transpose Properties

Split the Determinant

Properties of Inverse and Adjoint

Substitute Inverse and Adjoint Properties

Scalar Multiple Property

Expand the Scalar Terms

Expand Remaining Terms

Assemble and Cancel (The JEE Trap)

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Toolkit

The Simplification

The Grand Cancellation

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