If A = 1/(5!6!7!) [[5!, 6!, 7!], [6!, 7!, 8!], [7!, 8!, 9!]], then |adj(adj(2A))| is equal to (1) 2²

Q: Wait, why is $|adj(adj(M))| = |M|^{(n-1)^2}$? Can't I just calculate the adjoint twice?

Calculating the adjoint of a matrix twice is a recipe for disaster in an exam! It is computationally expensive and prone to arithmetic errors. In JEE Advanced, we value properties over brute force. This identity is a powerful tool that reduces a complex matrix operation to a simple scalar power.

Q: Why is $|2A| = 8|A|$? Shouldn't it be $2|A|$?

This is a classic trap! Remember, when you multiply a matrix by a scalar $k$, you are multiplying *every* element by $k$. In a $3 \times 3$ determinant, you pull out $k$ from each of the three rows. Thus, $|kA| = k^3 |A|$. Since $k=2$ and $n=3$, we get $2^3 = 8$.

Q: How do I know when to use row operations versus just expanding the determinant?

Great question! If you see a column or row of ones, or if the numbers are large and share common factors, row operations are your best friend. They create zeros, which makes expansion along that row or column trivial. Never expand a $3 \times 3$ determinant directly if you can simplify it first!

Question

If $A = \frac{1}{5!6!7!} \begin{bmatrix} 5! & 6! & 7! \ 6! & 7! & 8! \ 7! & 8! & 9! \end{bmatrix}$, then $|adj(adj(2A))|$ is equal to

Options:

(A) $2^{20}$

(B) $2^8$

(C) $2^{12}$

(D) $2^{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we face a problem that looks like a nightmare: a matrix filled with factorials, nested adjoints, and scalar multipliers. It is designed to make you panic.

But here is the secret of JEE Advanced: the most intimidating problems often have the most elegant, simple solutions. We are going to strip away the layers of this problem one by one.

### The Power of Properties

Our target is $|adj(adj(2A))|$. If you try to calculate the adjoint of $2A$, then the adjoint of that result, and then the determinant, you will be trapped in a calculation loop for twenty minutes.

Instead, we use the property of the adjoint of an adjoint. For any $n 	imes n$ matrix $M$, the determinant of the adjoint of the adjoint is given by:

$$|adj(adj(M))| = |M|^{(n-1)^2}$$

Here, our matrix is $3 	imes 3$, so $n=3$. The power becomes $(3-1)^2 = 2^2 = 4$. Thus, our expression simplifies to $|2A|^4$. We have already reduced a massive matrix problem to a simple determinant calculation.

### The Scalar Trap

Now, we need $|2A|$. Many students make the mistake of thinking $|2A| = 2|A|$.

Remember, when you multiply a matrix by a scalar $k$, you multiply every element by $k$. In a $3 	imes 3$ determinant, you pull out $k$ from each of the three rows. So:

$$|2A| = 2^3 |A| = 8|A|$$

Our target is now $(8|A|)^4$. The only thing standing between us and the answer is the determinant of $A$.

### Taming the Factorials

The matrix $A$ is defined as:

$$A = \frac{1}{5!6!7!} \begin{bmatrix} 5! & 6! & 7! \ 6! & 7! & 8! \ 7! & 8! & 9! \end{bmatrix}$$

If you expand this, you will be dealing with numbers like $9!$, which is $362,880$. That is not the path to success.

Instead, look at the rows. We can factor out $5!$ from the first row, $6!$ from the second, and $7!$ from the third. These factors will cancel out the fraction outside perfectly! We are left with a simple numerical matrix:

$$\begin{vmatrix} 1 & 6 & 42 \ 1 & 7 & 56 \ 1 & 8 & 72 \end{vmatrix}$$

### The Final Victory

With a column of ones, we use row operations. Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$ gives us:

$$\begin{vmatrix} 1 & 6 & 42 \ 0 & 1 & 14 \ 0 & 2 & 30 \end{vmatrix}$$

Expanding along the first column, we get $1 \cdot (1 	imes 30 - 2 	imes 14) = 30 - 28 = 2$. So, $|A| = 2$.

Finally, substitute this back into our target:

$$(8 	imes 2)^4 = 16^4 = (2^4)^4 = 2^{16}$$

We have conquered the monster! The final answer is **$65536$** (or **$2^{16}$**).

Animated Solution for Mathematics - Matrices and Determinants: If $A = \frac{1}{5 ! 6 ! 7 !} 5! 6! 7! 6! 7! 8! 7! 8! 9!$ , then $∣ a d j (a d j (2 A)) ∣$ is equal to

Select Answer:

Visualized Solution

Problem Objective

Adjoint of Adjoint Property

Scalar Multiplication in Determinants

Determinant of Matrix $A$

Factoring Common Terms

Simplifying the Determinant

Row Transformations

Expanding the Determinant

Calculating the Final Value

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Power of Properties

The Scalar Trap

Taming the Factorials

The Final Victory

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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; } If A=5!6!7!1​​5!6!7!​6!7!8!​7!8!9!​​, then ∣adj(adj(2A))∣ is equal to